*Welcome to www.educator.com.*0000

*In this lesson we are to take a look at basic types of numbers.*0002

*We will see that there are many different ways that we can take numbers and start to classify them.*0008

*I will go over all these types of numbers and more in detail as we see how a number gets into each of the groups.*0012

*You will also see how you can represent these numbers on a number line.*0020

*Be handy for say comparing numbers and figure out what it means to take the absolute value of a number.*0024

*We will also see some symbols on how you can compare numbers meaning to our inequalities.*0030

*When it comes to numbers you can really break them down into various different groups and classify them according to their properties.*0039

*The most common types of groups that we can use to classify numbers are the natural numbers, whole numbers, *0046

*integers, rational, irrational and imaginary numbers.*0052

*We will go over each of these groups in more detail.*0057

*In our first group will take a look at the natural numbers.*0064

*These are the numbers that do not contain any fractions or any decimals.*0068

*In fact they are sometimes called the counting numbers because they are some of the first number you learn when counting.*0073

*They contain the numbers 1, 2, 3, 4, 5 and it does go up from there so you know how we do not have a fractions and decimals and no negative numbers here.*0078

*In the next group we would start expanding on a lot of that last list just little bit and we also include the number 0.*0092

*Since we have all of the same numbers that we have before these natural numbers and we have that number 0,*0100

*you could say that all natural numbers are a type of whole number.*0107

*Watching a step in a few different times as we go through these groups of numbers, some numbers end up in more than 1 group.*0112

*An important part of this was that they contain a natural numbers and 0 to be a whole number.*0118

*Alright continuing around, we can also expand on those numbers by looking at the integers.*0126

*The integer is not only includes say the natural numbers but the negatives of all of our natural numbers and 0.*0133

*Again this makes all of our natural numbers and our whole numbers a type of integer.*0141

*You will see from the list that we got some nice numbers on here like -3, -2, - 1.*0148

*There is 0,1, 2, 3, 4 all the way up on that side.*0154

*The rational numbers are probably one of our most important groups.*0161

*These include all numbers that can be written as a fraction.*0166

*Now there is many different types of numbers that you can write as a fraction.*0171

*In fact all the numbers that we just covered previously can easily be turned into a fraction by putting them over 1.*0176

*A harder one says you determine whether you can write them as a fraction or not, or the one's that involve decimals.*0183

*Here is how you can tell if they are rational or not.*0189

*If that decimal terminates that means that stops, then you know you can write it as a fraction therefore it is rational.*0193

*If your decimal goes on and on forever and has a repeated block of numbers, then you may also write those as fractions, they are rational.*0200

*To help you figure out some of these, let us look at a few examples and see why they are all types of rational numbers.*0209

*The first one I'm looking at here is 3/17, we know how this one is already a fraction.*0217

*It is a pretty obvious choice that you can write as a fraction, it is rational.*0224

*This number 4 could have been one of our numbers on our natural number list and it is also one that we can write as a fraction fairly quickly by simply putting it over 1.*0231

*Since we can write as a fraction we know it is a type of rational number.*0246

*Some of the more difficult one, these are the ones that involve decimals.*0251

*In .161616 repeating of this one goes on and on forever and ever but it has about 16 to just keep repeating over and over again.*0255

*That is what I mean by repeated block of numbers.*0266

*Since it has a repeated block, it can be written as a fraction.*0269

*In fact, this one is written as 16/99, I know that it is a type of rational number.*0272

*The next one, 0.245 and then it stops, because it stops this is a type of terminating decimal.*0279

*It can be written as a fraction as well that we can count up the number of places in it and just put it over that number.*0289

*It is tens, hundreds, thousands, written as a fraction.*0296

*Look for these types of numbers when determining out your rational numbers. *0300

*If we know what numbers can be written as a fraction, then we must also talk about the numbers that cannot be written as a fraction.*0309

*These types of numbers are irrational numbers. *0315

*We saw many different types of numbers that could be written as a fraction.*0319

*They seem like they are might not be a whole lot that you can not write as a fraction*0322

*but it turns out there is many common numbers that simply cannot be written as a fraction.*0327

*More of the common ones are roots that cannot be reduced to any further.*0331

*If you have a decimal that goes on forever and does not have a repeated block of numbers in it, then that is a type of irrational number.*0336

*There are also many famous constants which happen to be irrational numbers.*0345

*They show up in many different areas.*0349

*Looking at my examples below to see why they are irrational numbers.*0351

*Here when looking at the square root of 57, I know that this does not reduce.*0357

*That gives you decide to punch this one into the calculator.*0365

*You will see that is has a decimal that just keep going on and on forever, if it does not have a repeated block of number.*0368

*That is how I know that that one is irrational.*0374

*That one is a little bit more clear to see because I can actually look at its decimal and see it has no repeated blocks and yet it goes on and on forever, it is irrational.*0378

*It is a very curious number and the variable it is one of those famous constants.*0388

*Pi is equal to 3.141592 and then it keeps going on and on forever.*0393

*And it does not have a repeated block of numbers in it, I know that it is irrational.*0401

*All the types of numbers we have cover those far are actually types of real numbers.*0409

*There is another group that is completely distinct from those real numbers.*0414

*Those are the imaginary numbers and you can usually recognize those ones because no contain an imaginary part with i in it.*0418

*The reasons why these will be so important is some equations might only have imaginary numbers as solutions.*0426

*We will learn more about these imaginary numbers in some future lessons.*0433

*As I said before, they are completely separate from our real numbers.*0439

*You would not have an imaginary number that also ends up on our list for real numbers, completely different things.*0443

*Here are some examples of some imaginary numbers.*0449

*I'm looking at 2i, I see that it has the (i) right next to it.*0452

*Definitely imaginary, 1/2 + 5/7i, I can see that (i) is in there.*0456

*This is one of our complex numbers but you know it is an imaginary number for sure.*0464

*At the end here I have the square root of -1, I do not see any (i) in there and why it could be an imaginary number.*0470

*We will learn that imaginary numbers come from taking the square root of negative quantities.*0478

*In fact, the square root of -1 is equal to (i), it is actually is an imaginary number. *0484

*To understand why some numbers get to be on multiple groups, you have to take a step back and look at the big picture for this classification.*0491

*I'm trying out a nice diagram so you can see what numbers end up in which groups.*0500

*The most important distinction that you could make between numbers is probably whether they are real or imaginary.*0506

*Since those groups are completely separate.*0512

*Those in the real category we can go further and start breaking that down into many other different types of numbers.*0515

*Again, we do that according to the properties.*0521

*The most important distinction we make is whether we can write it as a fraction, we call these rational.*0524

*Or whether we can not write those as fractions, we call these irrationals.*0529

*That is how I'm connecting things with arrows here.*0535

*I'm doing that to show how these categories break down.*0539

*Rational numbers are types of real numbers and irrational numbers are types of real numbers.*0543

*Continue on with those numbers that can be written as fractions, those are the rational.*0550

*We move on to integers.*0556

*You will notice at that stage at we can drop with all our fractions, we do not have decimals anymore.*0559

*Now we have numbers like -2 , -1, 0, 1 and we go on from there.*0563

*As we continue classifying them, we get to our whole numbers.*0571

*In these ones now, we do not have any more negatives.*0576

*We start at 0, we have 1, 2, 3 and we go up from there.*0579

*On to our primal simplest list, those are the natural numbers.*0585

*They start at 1,2, 3 and they go up from there.*0590

*Remember, these ones are known as our counting numbers.*0594

*One way that you can use this diagram to help you classify numbers*0599

*is to know that if a number ends up in one of these categories it is also in all of the categories above it.*0602

*We can see this happen for some of our numbers.*0609

*Let us take the number 2, I see that it is definitely on my natural number lists, but it is also a type of whole number.*0612

*In addition, is a type of integer and I can take 2 and write it as a fraction.*0621

*It is a type of rational number which is of course a type of real number.*0627

*2 gets to be in all of those categories above it.*0633

*I will also take one that is not in quite as many groups.*0638

*For example let us just take the square root of 3, it is an irrational number.*0640

*But it is also in a category above it, it is a square root of 3 and it is a type of real number.*0646

*Okay, not bad.*0653

*Now we know a little bit more about the different types of numbers.*0657

*We will show you how you can visualize a great way to compare them using what is known as a number line.*0660

*On a number line, we draw out a straight line and mark out some key values such as like -3, -2, all the way up from there.*0668

*We put the numbers that are smaller on the left and the larger numbers on the right.*0679

*In this way I can make good comparisons between numbers.*0688

*You can see that 0 is on the left side of 3, we could say that 0 is less than 3.*0692

*It is handy to be able to visualize numbers in this way when looking at their absolute value.*0701

*The absolute value of the number is its distance from 0 on a number line.*0707

*It is a quick example may be looking at the absolute value of 2.*0713

*Since 2 is exactly 2 away on a number line, I know that the absolute value of 2 is 2.*0718

*We will start another one, how about the absolute value of -3.*0727

*That one I can see is exactly 3 away on a number line, its absolute value is a +3.*0733

*We might develop some shortcuts and say wait a minute, the absolute value just takes the number and always makes a (+).*0741

*That is okay, that is the way it should work that is because our distances are always (+).*0746

*As long as we can go ahead and compare the numbers, we might as well pick up some new notation for doing this.*0753

*You can compare numbers using inequalities and use the following symbols.*0758

*You can use greater than, less then, greater than or equal to and less than or equal to.*0763

*The way these symbols work, is you want put the smaller number with the smaller end of the inequality sign.*0770

*And the larger end of the inequality sign with the larger number.*0780

*It could say something like -3 < 5, that would be a good comparison between the two.*0787

*We have seen a lot about classifying numbers and comparing them.*0798

*Let us go ahead and practice these ideas by classifying the following numbers.*0801

*Let us say from the list that all of the groups that the following numbers belong to.*0806

*Let me start with 2/3, first I think is 2/3 a real number or an imaginary number.*0810

*I do not see any (i) on it so I will call this a real number.*0817

*Now, I need to decide can I write it as a fraction or not.*0824

*This one is already a fraction I know that I can write as a fraction for sure.*0829

*I will call this a rational number.*0833

*Moving on from there, in my integers those containing numbers like -3, -2, -1, 0 end up from there.*0837

*That is how the integers, we do not have fractions, we do not have decimals.*0846

*This one does not get to be in the inter group or anything below that for that matter.*0849

*I could say 2/3 is a real number and I could say that 2/3 is a rational number.*0854

*Let us try another one of these, 2.666 repeating.*0861

*I do not see an imaginary part so I will say that this is definitely a real number.*0865

*We can not write it as a fraction, why do you see it has a repeated block of numbers that goes on and on forever, it is a type of rational number.*0871

*What else can I say? Is it an integer? No, it has the decimal part on it, it is not an integer.*0883

*I will leave that one as it is, moving on, the square root of 3.*0892

*This is a type of real number, it does not have any imaginary part on.*0898

*Can we write this one as a fraction or not? This one I can not.*0903

*In fact, when you look at the decimal, it goes on and on forever and it does not have that repeated block of numbers, irrational.*0907

*Since we do not have any more distinct groups of below irrational, we will go ahead and stop classifying that one.*0918

*Onto some other numbers, -5 that is a type of real number.*0924

*It looks good, can we write as a fraction?*0931

*You bet we will simply put it over 1, it is rational.*0934

*Is it an integer? it does not have any fractions, it does not have any decimals, I will say that it is an integer.*0941

*Is it a type of whole number? that is where I need to stop.*0952

*Whole numbers do not contain negative numbers.*0956

*-5 is real, it is rational and it is an integer.*0959

*On to the number 0, this one used to be in a lot of different groups.*0965

*0 is a type of real number.*0970

*You can write it as a fraction, we will say that it is rational.*0975

*It is a type of integer, since it is in between our negative numbers and our positive numbers.*0983

*It is definitely a whole number.*0993

*That is where this one stops getting classified because the natural numbers start at 1 and then go up from there.*1002

*One more, let us classify 9, this one is a type of real number.*1009

*We can write it as a fraction, I know that it is rational, it is definitely on our list of integers.*1016

*It is also a type of whole number and we can go just a little bit further with this one.*1028

*This is a type of natural number.*1035

*9 used to be in a lot of different groups.*1039

*It is a type of real number, a rational number, it is an integer, it is a whole number and is a type of natural number.*1041

*Let us try this in a slightly different way.*1050

*Here I have a giant group of numbers, we want to list out whether the numbers in some of our various different groups like imaginary, real, or irrational.*1052

*That way we can think of visualizing, classifying them in just a slightly different way.*1060

*Let us start out with the first one.*1067

*I want to figure out all the groups that -7 belongs to.*1068

*I know that it is a type of real number, let us go ahead and put it into that group.*1072

*Can we write this as a fraction or not, yes I can write it as a fraction.*1078

*Let us put it in our rational category.*1081

*Is it a type of an integer? Yes it is on my integer lists.*1086

*Is it a type of whole number? No, because our whole numbers do not contain negative.*1091

*We will stop classifying that number.*1096

*Let us try another one, negative the square root of 3, that is another type of real number.*1098

*However, that one I can not write as a fraction.*1107

*I better put it in the irrational category and then that one stop.*1111

*Moving on, -0.7 it is a type of real number.*1117

*This one can be written as a fraction, it is -7/10.*1125

*Let us go ahead and put it in our rational category.*1129

*Can we go any further from there?*1134

*Unfortunately not, because it contains those decimals and integers some contain decimals.*1136

*We can stop classifying that one.*1142

*Moving on to 0, 0 is a type of real number.*1146

*It is a type of rational, it is a type of integer and it is a type of whole number.*1151

*It gets to be in a lot of different groups.*1160

*Remember, it is not a natural number since that starts at 1 and goes up.*1162

*On the 2/3, that one is definitely a real number and since it is already a fraction, I know it is a rational number.*1167

*It is not an integer since it is a fraction, 2/3.*1178

*The square root of 11, it is a real number, it does not contain an imaginary parts.*1187

*This one cannot be written as a fraction and I will put it in the irrational category and then stop classifying that one.*1194

*On to our famous number here, pi.*1202

*Pi is a type of real number, even though it is a little unusual, it does go on and on forever.*1207

*It is a type of real number and it is irrational since I cannot write it as a fraction.*1212

*We will stop classifying them since there is no two groups below irrational.*1220

*On to the number 8, this one is going to be in a lot of different groups.*1226

*It is a type of real number, I can write as a fraction by putting it over 1.*1231

*It is on our integer lists, it is on our whole number list and it is a type of natural number, a lot of different things.*1237

*On to 15/2, I will say that that is a type of real number.*1247

*I can write it as a fraction, let us put it in our rational category.*1254

*Unfortunately it is not an integer so I will not put it in that one.*1259

*Then number12, 12 is a type of real number.*1266

*We can write it as a fraction by putting it over 1, let us put in rational.*1271

*It is a type of integer, it is a type of whole number and since the natural number starts at 1 and then goes 2, 3, 4.*1276

*All we have from there I know that it is a natural number.*1285

*Just one more to do, the number 3i.*1291

*I have to throw an imaginary number on my list so it will immediately drop that into the imaginary bin.*1295

*And that is all the more classifying we will do with that one.*1301

*Since again imaginary numbers and real numbers are completely distinct from one another.*1305

*What you will know is that most of these categories are all types of real numbers.*1310

*We have classified numbers, what gets better about comparing them on a number line or just being to plot them out.*1318

*The way we plot out a number on a number line is we find it.*1325

*Say using one of our markers below and put a big (dot) to where it is.*1330

*If I want to graph something out like 3 on a number line, I will find 3 and I will place a big old dot right at 3.*1333

*Once I applied it out, I can do some good comparisons.*1342

*We can see that since 3 is to the left of 4, that 3 is less than 4.*1346

*Since 3 is on the right side of -1, 9, 0, 3 is greater than -1.*1355

*Let us spot out a few more, -2 on our number line.*1361

*We would find -2 and go ahead and put up the big old dot there.*1366

*When it gets in to fractions and decimals it does get a little bit more difficult but you can still put these on the number line as well.*1371

*This one is 5/3 and I do not see any 5/3 in my markers here on the bottom.*1378

*What I can do is I can break down each little section into thirds and mark out the 5th one.*1384

*1/3 and more thirds and more thirds.*1390

*We are looking for 5/3, 1,2, 3, 4, 5, we put that big dot right here.*1395

*Now we can better compare where 5/3 is into other numbers.*1403

*5/3 < 2 but it is greater than 1.*1407

*Alright, -3.75 that would be the same as -3 and 75/100.*1414

*That can also be written as -3 and 3/4.*1425

*That tells me I need to break down my number line into quarters.*1430

*1/4, 1/4, 1/4 and 1/4.*1436

*I'm looking to mark out 3 whole sections and then 3 quarters.*1446

*And we are going the negative directions 3,1, 2, 3 and we will put up the old dot there.*1451

*We can see that -3.75 > -4 and it is also less than a -3.*1458

*Let us use our number lines so that we can actually line up various different numbers and see which ones are smaller than the other ones.*1470

*Be just a rough sketch of the number lines, I'm not going to be too accurate with my thick marks.*1481

*But I just used it so I know how they compare to one another.*1487

*Let us go ahead and start with our first number here and put -7 on a number line.*1493

*Since it is a negative number, I'm going to aim for somewhere on the left side here -7.*1498

*-3 is a little bit more than that, I will put it on the positive side over here.*1508

*Let me put a spot there for 3.*1515

*-0.7, that is not very big and is definitely larger than -7 and less than 3.*1519

*Let us go ahead and put it right here - 0.7.*1527

*0 is a good number and put it greater than -0.73.*1539

*2/3 is larger than 0, I will put on the right side.*1550

*Alright on to something little bit trickier, the square root of 15.*1561

*I know that that is less than 4, since the square root of that 16th is something a bit larger will be on the right side.*1566

*It is greater than 3, since the square root of 9 would be 3.*1573

*I'm going to put this one larger than 3, square root of 15.*1577

*It is a good one, definitely larger than square root of fifteenths.*1587

*-7/2, that one is about -7 1/2, I mean -3 1/2.*1597

*Let us put that one down here -7 1/2, -5 and one more number pi.*1606

*3.1415 a little bit larger than 3, put it a little bit larger than 3.*1625

*Now that we have used our number line, it gets some comparisons among all these.*1636

*We will simply list them from smallest all the way up to largest.*1641

*-7, -5, -7/2, -0.7, 0, 2/3, 3, pi, square root of 15 and 8, not bad.*1646

*For this last example, we will go ahead and use our inequality symbols like less than or greater than to go ahead and compare these 2 numbers.*1670

*If you want you can use a number line to plot them out before using these symbols.*1676

*Let us try the first one, comparing 6 and 2.*1684

*When I plot these out, 2 is on the left side of 6.*1690

*I know that 2<6, it is my smaller number, I will drop my inequality symbols so that I show that 2 is less than 6.*1695

*I can also say that 6>2.*1705

*Let us try another one, -7 and 5.*1711

*It is tempting to say that -7 is bigger but our negatives are on the left side and our positives are on the right side.*1722

*You can see that -7 is less than 5.*1730

*Let us write that out, -7<5.*1735

*-5 and -3, -5 is further down the -3, I know that it will be less than -3.*1745

*One more 2.3 and 5.7, 2.3, 5.7 will be much larger.*1765

*I know the 5.7 > 2.3 or in the order that they are in 2.3 <5.7.*1780

*These symbols are handy and in showing the comparison especially where they are on a number line.*1790

*One thing I did not use here is the or equals to symbol.*1796

*I could have put that in for all of these spots, 6 is greater than or equal to 2.*1802

*Or I could have said -7 is less than or equal to 5.*1807

*That is because it also takes into the possibility that the numbers could have been equal.*1812

*The reason why I did these is I can see that all of the numbers are not equal.*1819

*And it is a little bit more flexible when using this other one.*1826

*Watch for the or equal to symbol to show up when we are doing a lot of our inequalities, these ones are good.*1831

*Thank you for watching www.educator.com.*1838

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at operations on numbers.*0002

*When you hear me use that word operations, I'm talking about ways that we can combine together numbers or do something to a number.*0007

*It gets very familiar things such as adding, subtracting, multiplying and dividing.*0015

*As well as exponents and square roots.*0020

*To combine numbers together, we use a lot of familiar operations in order to do so.*0028

*Again, adding, subtracting, stuff like that.*0033

*Note that depending on the types of numbers being used, certain rules applies that will help us put them together.*0036

*The rules that I will focus on are the ones that involve positive and negative numbers.*0042

*How should we deal with those negative signs? Let us see what we can do with addition.*0048

*When you add numbers that have the same sign then you are looking at adding their absolute values together.*0055

*The results if they do have the same sign or have the same sign as the original numbers.*0062

*In other words, if you are adding the other two positive numbers its result will be positive.*0068

*For adding together two negative numbers, then your result will be negative.*0073

*Now if you happen to have two numbers that you are adding together and they are different in sign,*0079

*then we will actually handle these using subtraction.*0084

*Wait for the rules on what to do with positive and negative on my next slide.*0087

*When dealing with subtraction and what we want to do is subtract the number that is smaller in absolute value*0095

*from the number that is larger in absolute value and looking at each of their absolute values taking the smaller one from the larger one.*0102

*When looking at the final result, the answer will have the same sign as whichever number was larger in absolute value.*0111

*If your larger number was or the larger number in absolute value is negative, then the result will be negative.*0119

*Remember that we will also think of adding negative numbers using subtraction in this way.*0128

*In different way we are writing it but instead we are really using subtraction.*0134

*The other two familiar operations that we have are multiplication and division.*0142

*The way we handle a lot of signs with these ones are by thinking of this.*0147

*If we multiply a negative × a negative then the result will be positive.*0153

*A negative times a negative results positive.*0158

*If we take two things that are different in sign such as negative times a positive, then the result will be negative.*0164

*Now these are the same two rules that we end up using for division.*0175

*If we divide a negative by a negative, you will get a positive.*0180

*Negative divided by negative result positive.*0184

*If they are different in sign such as negative divided by positive, then the result is negative.*0189

*Keep these in mind when working with that your signs, multiplication and division.*0198

*Let us talk about exponents.*0207

*When we have an exponent, we can think of it as taking a number and multiplying it by itself in a certain number of times.*0210

*For example, maybe I have 3*^{7}, 7 would be our exponent.0218

*I can interpret that as multiplying that number by itself 7 times or some other key vocabulary you want to pick up on.*0224

*The exponent is that number raised next of the 3, exponent.*0236

*In the base of the number is the number that we are actually being repeatedly multiplied, base.*0244

*Once we see what number that is and how many times we need to multiply it, usually we can go ahead and simplify from there.*0252

*There are some common exponents that we usually give other names.*0258

*If I'm taking 5 and I'm raising it to the power of 2, 5 × 5, this is often said 5*^{2}.0262

*Another good common one would be something like 2*^{3}, 2 × 2 × 2 that would be 8.0278

*I could also say that this is 2*^{3}.0290

*Key on these special words for some of these other powers.*0294

*Another operation that we can do with numbers is taking the root of a number.*0303

*The principal square root of a number is the non negative number (n)*0309

*that you know if I were to say multiply it by itself, I would end up giving that (n).*0313

*This seems a little funny special worry but let us see if I can describe it using (n) as an example.*0319

*Let us say I wanted to find out the square root of 25.*0326

*What I'm looking for is what number would multiply it by itself in order to get a 25? That has to be 5.*0330

*It is what I'm talking about here, positive number such that when it is multiplied by itself you get that number underneath the root.*0340

*For this reason we have a little bit of a problem with our negatives underneath the root.*0353

*After all, what number would you multiply it by itself in order to get -16?*0360

*We learned from our rules of positive and negative numbers that -4 and +4 would work to get 16 *0364

*but unfortunately they are different in sign and we need them to be exactly the same, that is not going to work.*0373

*We are dealing with the principal square root because we are only interested in the positive numbers*0382

*that when multiplied by themselves would give us that number.*0387

*A perfect square is a number that is the square of a whole number and this one is usually reduced very nicely.*0393

*For example the square root of 9 would be the example of a perfect square, as it reduces down to a nice whole number, 3.*0401

*If you have a (a, b) being non negative real numbers, *0416

*then there are a few different ways that you can say combine or rip apart those roots.*0421

*3 multiplied underneath the same root and you can apply the root to each of those pieces.*0426

*If you are dividing and you have a root then you can put it over each of its pieces in the numerator and in the denominator.*0432

*You can use these rules in two different ways to simplify or combine words together.*0442

*We will see that a lot in some future lessons.*0446

*I'm pointing this out now so you do not make a common mistake.*0449

*Do not try and split up your roots over addition or subtraction.*0453

*We do not have a rule to do that yet or a good way to handle it.*0458

*In fact, in this example I have written below you can see that the two are not equal by simply evaluating each side.*0461

*9 + 16 would be a 25 and the square root of 25 is 5.*0469

*Looking at the right side square root of 9 is 3, square root of 16 is 4.*0475

*I'm putting those together I will get 7 and you can see that these things are not the same.*0482

*Be very careful when working with your roots.*0492

*Now that we know a few things about combining these, let us go through some examples and just practice with them.*0498

*The first one I want to add together a -3 and a -6.*0504

*I'm adding together two numbers that have exactly the same sign.*0510

*I will look at their absolute value and add those together.*0514

*The absolute value of 3 is 3, the absolute value of -6 is 6, if I combine those together I will get 9.*0519

*Since I'm adding together numbers that have exactly the same sign, the result will also have the same sign.*0531

*Adding two negative numbers my result is negative, -3+ -6 is a -9.*0537

*Moving on, 19 + a -12, I want to think of this as a subtraction problem since their signs are different.*0545

*How do I handle subtraction? Again I will look at their absolute value.*0558

*The absolute value of 19 and the absolute value of a-12, 19, 12.*0562

*I will subtract the smaller number from the larger number 19-12, what will that give me? I will get7.*0571

*I have to determine what sign this should be.*0584

*In subtraction we take the same sign as the larger number.*0588

*19 was larger in absolute value, it was positive, I know my result is 7.*0594

*If we do get a positive result as our answer, we do not write that positive sign up there.*0604

*We are just having 7, we will assume that that is a positive 7.*0609

*-8-11, an interesting way we can look at the problem, we could look at this as adding a -11.*0614

*The reason why I have looked at it in that way is that I could use my rules for addition.*0626

*If I'm adding things that have the same sign, I have looked at the absolute value of each of them, 8, 11.*0631

*Then I can simply add up those two values and get 19.*0640

*Since I'm adding two negative numbers, I know my result will also be negative, -19.*0645

*Let us try another one, 8 - -13.*0651

*When you subtract the negative this is another good situation that you could end up rewriting in a much simpler form.*0657

*When you subtract the negative, you can change it into addition.*0662

*This is 8 + 13 and now I'm adding together two positive numbers.*0667

*You made a positive 21 and one more, negative the absolute value of a -4 + 9.*0673

*Let us start inside those absolute values and see what we can do.*0683

*I'm adding together two things but they have different signs.*0687

*Let us look at the absolute value of a -4 and the absolute value of 9, 4, 9.*0692

*We want to subtract the smaller value from the larger value, 9-4 and that result would be 5.*0699

*I know what sign should that have, or the number that is larger in absolute value is 9.*0708

*And that was a positive value over here, I'm looking at a 5.*0716

*There are a lot of other things I have left out here so far, *0721

*those would be the absolute values and that leading negative sign.*0724

*Let us go ahead and put those in there now.*0727

*I want to do the absolute value of 5, all that is simply 5.*0731

*I still have that negative sign hanging out front and it is been there since the very beginning.*0740

*I can see that after done evaluating this one all the way, my answer is actually a 5.*0743

*Let us work on adding, multiplying and dividing the following numbers.*0752

*We only have one rule to take care of that is when we are multiplying together two negative numbers, we get a positive.*0757

*If we are multiplying together two numbers that are different in sign, the result should be negative.*0766

*4 × -7, I just want to think of 4 × 7 that would be 28.*0775

*Since they are different in sign, I know that this will be a -28.*0784

*Moving on, -6 × -5, 6 × 5 would give me 30 and now here I have a - × - I know the result will be positive.*0789

*But again we usually do not write that positive sign in there so just leave this as 30.*0801

*-12 ÷ 3, 12 ÷ 3 would be a 4, negative ÷ positive would be negative, our result is a -4.*0806

*One last one, -60 ÷ a -5, 60 ÷ 5 goes in there 12 times and negative ÷ negative is positive, my result is a 12.*0820

*You need to be very careful with these rules for multiplication, make sure you have these memorized.*0835

*Let us do a few involving our exponents.*0843

*Remember we can think of these as repeated multiplication.*0846

*2/3*^{4}, it can be the same as 2/3 × 2/3 × 2/3 × 2/3, we are doing it 4 times.0850

*I simply multiply it across the top and across the bottom 2 × 2 × 2 × 2 would be 16.*0862

*Then I have 3 × 3 × 3 × 3 = 81.*0871

*All the numbers here are positive so I know my result is positive.*0876

*Here is a tricky one, -5 ×-5 × -5, let us take this two other times so we can what is going on here.*0881

*Here I have a -5 × -5 the result of taking 5 × 5 would be 25.*0896

*Taking a negative × negative I would know that this would be 25.*0904

*That looks good, let us go ahead and work in this last value of -5.*0910

*We want to multiply 25 × -5, the result there would be 125.*0918

*Since I'm multiplying a positive × a negative result is -125, -125 would be my answer.*0929

*The next one looks very similar but it is actually very different.*0939

*That one is 2*^{3} and that negative sign is just out front of all that.0943

*We recognize that the 2 is the base and that the negative is not included in that base*0949

*since there is no parenthesis given to group it in that way.*0956

*We want to figure this one as 2 × 2 × 2, I have multiplied it out three times.*0960

*And as for what to do with that negative sign, if we have not put it up front it is a long pretty ride.*0966

*2 × 2 × 2 that would give me an 8, all of those numbers are positive, 8.*0973

*Of course let us put our negative sign out front since it was out front at the very beginning.*0979

*We can see that our result is -8.*0984

*Now on to some square roots, these ones we simply just want to break them down and simplify them as much as possible.*0990

*The first one I have the square root of 64.*0998

*Think for yourself what number would you have to multiply it by itself in order to get 64?*1001

*I have a couple of options that could be 8 and 8, that would have given me 64.*1008

*Or it could be -8 × -8, that would also give me 64.*1013

*We are only interested in the positive values that do so, let us not worry about those -8.*1018

*We will say that our answer for the square of 64 is 8.*1024

*Moving on, the square root of 169 ÷ 81.*1031

*This one we can use one of our rules and break up the root over the top and over the bottom.*1037

*Now we can end up taking the root of each of these individually.*1045

*What number multiplied by itself would give us a 169? That have to be a 13.*1048

*A number that would be multiply by itself to get an 81? 9.*1055

*The answer in this one is 13/9.*1060

*Continuing on, this one has a negative sign out front but that is not underneath the root.*1064

*I would not worry about it just yet instead let us just focus on the square root of 36.*1071

*That would be 6, of course we will go ahead and put our negative sign and see that our final result is a -6.*1079

*One more, this last one involves the square root.*1091

*What two numbers when multiplied together would give us a -49 remember they must be the same.*1095

*We got a few problems, do not we? If I try and use 7 and 7 that would give me 49, that does not work.*1102

*If I try and use a couple of-7, that does not work, that still gives me 49.*1109

*I can not use one positive and one negative even though those give me a -49.*1114

*Those are not the same sign, one is positive and one is negative.*1121

*What is going on here, if you remember about your types of numbers, these are imaginary numbers.*1125

*I will leave that one just as it is until we learn about simplifying imaginary numbers in some later lessons. *1137

*Some various different ways that you can go ahead and combine numbers using some very familiar operations.*1144

*Remember that most of the rules that I covered will give you some tips on what to do when they are different in signs.*1150

*Positive, negative, negative, positive and all those will be handy in figuring out the overall sign of your answer.*1155

*Thank you for watching www.educator.com.*1166

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at the order of operations.*0002

*As we will see the order of operations is a great way that we can start combining numbers and figure out what we should do first.*0009

*This one involve things like what should we do at parentheses and exponents*0015

*and when should we do our multiplication, division, addition and subtraction. *0020

*When trying to simplify much larger expression with many different types of operations present, we have to figure out what to do first.*0027

*Our order of operations gives us a nice run back on what we should be doing.*0037

*The very first thing that we should do is work inside our grouping symbols.*0042

*It means if you see parentheses or brackets work inside of those first.*0047

*Then move on to simplifying your exponents, things raised to a power.*0052

*Once you have both these in care, move on to your multiplication and division.*0058

*If you see lots of multiplication and division next to each other, remember to work these ones from left to right.*0063

*Now you have to do any remaining addition and subtraction.*0072

*And again when it comes to which of those is more important simply work those from left to right as well.*0074

*One handy way that you can remember of this entire list of that is the order of operations is to remember PEMDAS.*0081

*PEMDAS stands for parentheses, exponents, multiplication, division, addition and subtraction.*0090

*Let us try it out.*0096

*A great way that you can remember these is Please Excuse My Dear Aunt Sally.*0105

*I often heard a lot of my students use that one to make sure that they came out straight.*0110

*Start with your parentheses and then move on to exponents.*0116

*Be very careful if you are using this to memorize what to do first because sometimes when using it, it looks like multiplication is more important.*0120

*But work these ones from left to right.*0129

*The same thing applies to your addition and subtraction, work those from left to right.*0135

*Sometimes you will deal with a larger expression that has a fraction in it.*0150

*Even though you might not see some grouping symbols, think of the top and bottom as their own group.*0157

*That means work to simplify the numerator and get everything together up there.*0163

*And work to simplify your denominator, get everything together down there before we continue on with the simplification process.*0166

*As a quick example, let us look at this slide.*0173

*We have (-2 × 5) + (3 × -2) / (-5-3).*0176

*I'm going to work on the top part as its own group, and the bottom part as its own group.*0183

*Let us see what does this, -2 × 5 would give me -10, 3 × -2 =-6.*0191

*On the bottom in that group I have that -5 -3 =-8.*0202

*Okay -10 - 6=-16 and on the bottom I still have a -8.*0212

*We worked to look inside each of those groups and simplify them using our order of operations in there.*0220

*I simply have a -16 / -8 and that is a 2.*0226

*Watch for those large fractions to play a part.*0232

*Let us try some examples now that we know more about the order of operations and see how we can bring these into a much simpler expression.*0237

*This one is ((5 - 2)*^{2} + 1))/ -5, we also write down PEMDAS.0245

*This will act as our roadmap as we are going through the problem.*0257

*I want to look for grouping symbols or parentheses to see where I need to start.*0261

*That 5 - 2 looks like a good area, we will do that first, 5 - 2 is 3.*0266

*The only other grouping that I'm concerned with is the top and bottom of the fraction.*0278

*There is only one thing on the bottom so I'm just going to now focus on the numerator.*0283

*I can see that I have some exponents, I have a 3*^{2} in there.0290

*And now let us do that, 3*^{2} is 9, it is getting better.0294

*I want to move on to my multiplication and division.*0302

*Looking at the top and bottom of the fraction individually I do not see any multiplication or division, I can move on.*0306

*Addition and subtraction, why I do have some addition on the top, I put those together to get 10/-5.*0314

*We are looking at 10 ÷ -5 and now I can say that my result is a -2, this one is done.*0322

*You can see how we move through that order of operations as our road map.*0330

*In this next one we want to evaluate a (-12 × -4/3) - (5 × 6) ÷ 3, let us go over the map.*0338

*I do not see too much in terms of grouping but I do have this group of numbers over here.*0354

*Let us go ahead and take care of those.*0361

*Inside I have (5 × 6) ÷ 3, what should I do in there? I got multiplication and division.*0363

*Those ones remember we are working from left to right.*0370

*On the left side there I have multiplication then we actually do the division.*0374

*5 × 6 is a 30, now do the 30 ÷ 3 and get 10.*0383

*We have taken care of that grouping.*0394

*I'm just going to copy down some these other things and then we will continue on.*0396

*Our grouping is done, now on to exponents.*0406

*I do not see any exponents here so now on to multiplication and division.*0410

*We will do multiplication I got a -12 × -4/3.*0416

*A negative × a negative would give me a positive, multiplying on the top that would be 48/3.*0421

*Because of my fraction there, I do have some division I could take 48 and divided by 3 = 60.*0434

*On to addition and subtraction 16 – 10 = 6.*0444

*I have completely simplified this one and I can call it done.*0451

*This next one I have (12 ÷ 4) × (√5 - 1).*0458

*Starting with my grouping symbols and parentheses, I could consider everything underneath the square root as its own little group.*0469

*Let us work on simplifying that, I'm writing here 5-1 is a 4,12÷ 4 × √4, taking care of the square root entirely.*0475

*I'm looking at 12 ÷ 4 × 2, moving on do I see any exponents? No exponents.*0499

*On to multiplication and division, this is that tough one.*0509

*It is tempting to say that multiplication is more important but it is not.*0512

*Simply work these guys from left to right.*0516

*In this case, we are going to do the division first, 12 ÷ 4 is 3.*0519

*Then we are actually taking that and multiply it by 2 and get 6, this one is completely simplified.*0527

*Let us look at our example that involves lots and lots of different things.*0538

*I have (8 × 4) - (3*^{2} × 5) + (2 × the absolute value of -1) / (-3 × 2/3) +1 0542

*With so many different things in here we have to be careful in what to do first.*0562

*I'm dealing with a fraction here I want the top as its own group and the bottom as its own*0567

*and work inside each of those and try to simplify them.*0572

*Let us look at the top a little bit.*0575

*Inside of that I do not see any additional grouping symbols so I will try and do any exponents on the top.*0578

*I do have a 3*^{2}, let us change that into a 9.0587

*I have the absolute value of -1, might as well we go ahead and take care of that as well.*0594

*We are doing a little bit of simplifying on the top, let us see if there is any exponents in the bottom.*0602

*83 × 2*^{3} and change out into -3 × 8 and of course we still have the + 1.0607

*Continuing on, looking at the top I do not have any additional parentheses, I do not have any additional exponents, multiplication and division.*0621

*A lot of multiplication on the top, 4 × 8 would give me 32, 9 × 5 =45, 2 × 1=2.*0629

*On to the bottom, -3 × 8=-24 and then +1, multiplication and division done.*0644

*On to addition and subtraction and we are going to do this from left to right.*0655

*I will do 32 - 45, what do we got from there?*0660

*Let us imagine our technique for combining numbers that have different signs.*0665

*I'm just subtracting here, I get a result of 13.*0673

*The one that is larger in absolute value is the -45 so my result is a -13.*0677

*Looking at the bottom-23 almost done.*0685

*-11 at the top divided by -23, this one is completely simplified as 11/23.*0693

*When dealing with multiple operations it is important that we do have a roadmap in order to get through all of these.*0705

*Feel free to use PEMDAS also that you keep everything in order.*0711

*As you use PEMDAS, if you get down to your multiplication and division then use them from left to right.*0715

*If you get down to your addition and subtraction, again use those from left to right.*0720

*Thank you for watching www.educator.com*0725

*Welcome back to www.educator.com.*0000

*In this lesson we will take a look at the properties of real numbers.*0002

*Some of these properties involve the commutative, associative, distributive, identity, inverse and 0 properties.*0008

*It seems like that is quite a bit to keep track of but once we get into the nuts and bolts of it, you will see it is not so bad.*0015

*There are several properties that can help you put different numbers together.*0028

*Some of the very first properties are the commutative property and the associative property.*0034

*We will first deal with that commutative property first.*0041

*What the commutative property says is that the order of addition simply does not matter.*0044

*It also says that the order of multiplication does not matter.*0050

*It seems like you should, but a quick example shows that actually it does not make much of a difference.*0054

*For example, if I have 2 × 3 × 4, I'm just multiplying all those numbers together.*0060

*I actually do this from left to right, this would be 6 × 4, I will get 24.*0070

*Since the order does not matter, I can feel free to scramble up that order a little bit just like that.*0079

*3 × 4 × 2 and again do these from left to right, 3 × 4=12 × 2.*0087

*12 × 2 you will see that the answer is still the same.*0095

*When you have addition or multiplication, the order you have a little bit of freedom with, you can scramble them up just a little bit.*0102

*With the associative property, we are also changing something and we are saying that the grouping does not matter.*0110

*If you are dealing with the associative property of addition then changing the grouping of addition does not matter.*0117

*The associative property for multiplication then the grouping of multiplication does not matter.*0124

*Let me take a quick look at a nice quick example to see what I mean by the grouping.*0129

*Let us look at (2 + 3) + 4, here I'm adding a bunch of different numbers.*0135

*Up with parentheses to show that I'm grouping the first two together.*0143

*According to our order of operations, we would have 2 and 3 together first.*0147

*This would give us a result of 5 + 4, giving us a final result of 9.*0153

*If I take those same numbers and the same exact order and this time I decide to group together the 3 and the 4 together,*0160

*you will see that you will get exactly the same answer.*0170

*We will do what is inside parentheses first, 3 + 4 is 7 and then we will go ahead and we will put that 2 with the 7.*0173

*I'm sure enough you will notice that our answers are exactly the same.*0184

*The associative property says that the grouping for addition and multiplication does not matter.*0188

*The distributive property is a neat one, it deals with multiplication and addition.*0203

*It says that we can distribute our multiplication over addition.*0207

*We looked at several different examples of this but one of the classic things is something like this.*0212

*I have 2 × (3 + 4), according to the distributive property I will take this multiplication with the 2 and distribute it over addition.*0221

*I actually multiplied it by the 3 and then multiply it by 4.*0233

*This will be 2 × 3, this will be 2 × 4 and I can take care of each of those individually.*0239

*Get a 6 + 8, combine those together and get a 14.*0248

*Just to highlight that this property is valid, I will also do the same problem using my order of operations to take care of the part inside parentheses first.*0256

*3 + 4 is 7 and 2 × 7 is 14, I'm sure enough just like we should we get the same answer.*0268

*This is a unique one involves multiplication and addition.*0281

*Some other properties that you want to keep in mind have to deal with 0 and 1.*0288

*With the division of property of 0, we say that 0 divided by any number we get the results is 0.*0295

*If we are trying to divide by 0, that is something that we can not do.*0305

*For that one we say that it is undefined.*0311

*Be very careful for this one, many people will try and divide by 0 later on, simply can not do it.*0315

*When it comes to 1, 1 is a little bit easier to deal with.*0323

*When you divide a number by 1, the number remains unchanged, nothing happens.*0326

*Think of 5 ÷ 1, you will still get 5, or 7 ÷ 1 =7.*0332

*A number divided by itself is 1, that will be like 5 ÷ 5, number divided by itself results to 1.*0338

*All of these properties seem like they might be arbitrary, but all of them play a very important role.*0348

*On to the multiplication property of 0 and let us see what we can do with that.*0360

*With the multiplication property of 0, when you multiply it by 0 the result is simply 0.*0366

*Think of stuff like 3 × 0 result 0.*0372

*Multiplication property of 1 says that when you multiply a number by 1, the number remains unchanged.*0377

*3 × 1 now 3 stays exactly the same, it is still 3.*0383

*With the addition property of 0, when you add 0 to a number, the number remains unchanged, 3 + 0 is still 3.*0389

*These last few properties that deal with 0 and 1, they can be a little confusing to keep straight.*0400

*But again the most important is you can not divide it by 0, watch out for that one.*0407

*You may be thinking that all of these properties and say okay what is the big deal?*0416

*Why do I necessarily need to know these? *0419

*Will they become extremely important when you are manipulating large expressions, especially when you get into a lot of the solving stuff later on.*0421

*What they do is they allow us to take an expression change in many different ways without changing what it is.*0429

*For example, if I want to look at an example of the students work,*0436

*what allows them to do many of these steps is some of those of properties that we covered earlier.*0441

*In fact, let us take a close look at this and see what properties we will use.*0446

*In the very first part of this I see (7 + 2) × (x + 9) and in the next part I know that the 2 is moved inside parentheses and now I have 18.*0451

*What happened there is they took the 2 and they distributed it over the x and the 9.*0462

*What allows a person to do that part of this work is our distributive property.*0469

*In the next little bit of work, I see that they have actually switch the order of the 2x and the 18.*0478

*They have moved around where things are.*0483

*Are they allowed to do this? The answer is yes.*0487

*We are only dealing with addition right there so they can change the order just fine.*0489

*This is our commutative property, let us see what is going on in the next step.*0496

*In the next step, everything is in the same exact order but now the parentheses have moved into a different spot.*0504

*Now they are actually around the 7 and 8.*0510

*Since only addition is being shown here, then I know that this is a valid step, this is our associative property.*0514

*It looks like they did some simplifying to combine the 7 and 18 together and they switched the location of the 25 and 2x.*0527

*It looks like in that one they have used the commutative property again.*0536

*The big thing to take away from this is that we will be manipulating expressions quite a bit.*0541

*These properties that work in the background that allows to do many of these manipulations.*0546

*Switching the order of things or switching our grouping as we go along.*0551

*It is time to get into some examples and see what we can do about identifying many of these great properties.*0557

*Here I have lots of different examples and we just simply want to identify the property that is being used.*0564

*The first one I have 6 + - 4 = -4 + 6, carefully look at that and see what is changed.*0570

*One thing that picked up in my mind is I'm looking at the order of things and the order has been switched.*0579

*I will call this my commutative property.*0586

*Since this is dealing with addition, I could take this a little bit further and say this is the commutative property of addition.*0598

*The next one let us see what we got, 7 × 11 × 18 = (11 × 8) × 7, it is tricky here.*0617

*Let us see what is changed, the grouping is actually exactly the same.*0626

*The 11 and 8 are being grouped together and over here the 11 and the 8 are still being grouped together.*0632

*It has nothing to do with grouping.*0637

*This is another one where the order has changed.*0639

*Since the order has changed, it is still our commutative property.*0642

*This one has nothing to do with addition.*0651

*It is actually multiplication so I will say commutative property of multiplication.*0652

*Nice, I like it, next problem (11 × 7) + 4 = (11 × 7) + (11 × 4) more interesting.*0663

*This one has multiplication and addition, in fact it looks like they took the 11 and multiplied it by the 7 and by the 4.*0672

*We can recognize that as our distributive property.*0681

*That is the unique one, it has both multiplication and addition.*0694

*Continuing on, pi × √2 =√2 × pi, someone has been messing around with the order of things, another one of my commutative properties.*0699

*Since I have multiplication here, commutative property of multiplication.*0720

*One more for this example 6 + 2 + 7 = 6 + 2 + 7.*0728

*We noticed with that one all the numbers are in exactly the same order.*0736

*The order is untouched but what is different here is that the parentheses are showing up in an entirely different spot.*0741

*They are changing the grouping, this is our associative property.*0749

*I have addition, associative property of addition, not bad.*0765

*If you change the order of the numbers present, then you are dealing with your commutative property.*0773

*If you change the grouping, you have your associative property.*0780

*With these ones we have some sort of property present and we want to fill in the blank so that the property ring is true.*0786

*The first one it looks like they are trying to display the commutative property.*0796

*I have 3 + 4 + 5= 3 + 5+ _, the commutative property changes the order of things.*0799

*It looks like we are changing the order of the 4 and the 5 here.*0808

*I see I got the 5, we will put the 4 in there and that would be a good example of the commutative property.*0812

*I technically, the commutative property of addition.*0818

*The next one is the associative property that one changes grouping so the order should stay exactly the same.*0821

*Let us write the same exact numbers in the same exact order,4, 5, and 3, it looks good.*0829

*Now I can see that since the parentheses are in a different spot, the grouping has changed.*0837

*Alright distributive property, that one involves multiplication over division.*0844

*I mean multiplication over addition.*0850

*I can see that the two has been taken to the 7 and the 3, I'm just missing the 7.*0856

*We have a good example of all 3 properties.*0866

*Example 3 deals with our division by 0, remember that we can not divide 0.*0873

*It is important to recognize what numbers can we sometimes not use, those are known as restricted values.*0878

*With these ones we are going to try and come up with a number that our letter x in this first one can not be.*0886

*Let us say we do not want the bottom to be 0, if x was 4 then what we will have on the bottom is 4 - 4.*0896

*That would definitely give us a 0.*0911

*In terms of our restricted value, we would say that x can not equal 4, since 4 is restricted.*0915

*It can not be 4 when we get that 0, let us try the next one.*0926

*Looking at the bottom of that fraction, I have 5 + a, what would (a) have to be to give me a 0 on the bottom?*0931

*I'm adding to 5 and thinking about 0 but it is possible if we start thinking negative numbers.*0942

*In fact what happens if a =-5, based on what you have seen there on the bottom would be 5 - 5 and that would give you 0.*0948

*My restricted value, we would say that (a) can not be -5, I want that 0 on the bottom.*0959

*Let us try this one, it looks a little bit more difficult.*0968

*What can (y) not being what is restricted?*0973

*I will be subtracting 1 from it but I'm also multiplying by 4.*0980

*The one thing that is going to do it, I have to borrow my fractions but 1/4 is actually my restricted value.*0988

*To see this, I'm thinking of taking 1/4 and multiply it by 4, that gives you 1.*0995

*When we subtract 1 you get 0, you know that (y) can not equal a 1/4.*0999

*Keep in mind that you can not divide by 0 and sometimes there are simply values that are restricted, you can not use them.*1009

*One last example here, we will look at the distributive property and how we can use it to simplify some expressions.*1020

*You may have see me use these lines on top of the number here, actually show the multiplication of what I need to do.*1028

*Definitely a good practice you should adopt.*1035

*In the first one, I looked at 2 being multiplied by (a) and 2 being multiplied by (k).*1039

*This will give me (2 × a) + (2 × k), I will leave that one as it is.*1046

*Moving on to the next one, here is -5 × (4 - 2x), this one used multiplied by both parts inside those parentheses.*1056

*-5 × 4 =-20, we will do that first, -5 × -2x.*1069

*Negative x negative is a positive, we will say10x.*1077

*One more, you may be looking at that one the same way then why are we not using the distributive property.*1085

*All I see is a negative sign but imagine that as -1 and you will see that the distributive property will come in handy just fine.*1090

*We will take -1 and distribute it among both parts on the insides of those parentheses.*1098

*-1 × -2 = 2 and -1 × 10q= -10q.*1104

*In all of these examples, we want to take the multiplication and distribute it out over addition.*1117

*It even works for some of these subtraction problems here because subtraction is just like adding a negative number.*1122

*Thanks for watching www.educator.com.*1130

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at how you can multiply and divide rational expressions.*0003

*There is a lot to know when it comes to rational expressions.*0011

*We will first start off by looking at what values would be undefined or restricted in some of these rational expressions.*0013

*We will also take quite a bit of time to figure out to how simplify them and put them into lowest terms.*0021

*There will be a few situations where you have to recognize that there are equivalent forms of these different rational expressions.*0028

*They may look different but they are the same.*0035

*Finally we will get into that multiplication and division process so you can see it all put together.*0038

*A lot of things that we will do these rational expressions ties into rational numbers in general.*0046

*In terms of the techniques and how we simplify them.*0052

*Remember that a rational number is any number that can be written as a fraction.*0056

*The number 2/3 is a good example of one of these rational numbers. *0061

*We are working with rational expressions you can consider those of the form p/q when p and q are both going to be polynomials.*0066

*Let me give you some examples of what these rational expressions look like.*0076

*2x + 3 ÷ 5x*^{2} - 7.0080

*Since both of those are polynomials and they are simply being divided that will be a good rational expression. *0086

*How about x*^{3} - 27 ÷ 4x + 3? That will also be a good example of another rational expression.0093

*What this rational expression since we have some variables in them, we are always concerned about dividing by 0. *0107

*We do not want use any values for x that would make the bottom 0.*0113

*Think of this expression up here x + 5 ÷ 2x - 4.*0117

*I can plug in many, many different values for my x as long as I do not get a 0 on the bottom.*0123

*In fact if I try and plug in a 2 into this that would give me 0 on the bottom.*0130

*2 × 2 - 4 and that is bit of a problem. *0136

*Since it gives me that 0 the bottom this expression is undefined at 2 or I might simply make a note somewhere and say x can not equal 2.*0141

*For many of the expressions that you will see on the future slides here, *0151

*we will assume that it can take on many different values, but not those restricted values.*0156

*When we are into the multiplying and dividing we will always be concerned with taking our answer down to its simplest form.*0165

*How is it that we go ahead and reduce a fraction to it simplest terms?*0173

*Remember how you do this process with some similar type of rational expressions.*0178

*We will try and cancel out a lot of the extra p’s that are present.*0183

*If I’m looking at 6x*^{2} ÷ 2x^{2} I could look of this as 2 × 3 × x × x × x and in the bottom would be 2 × x × x.0188

*As long as I have multiplication there I can simplify it by canceling out a lot of these extra factors and get something like 3x.*0204

*The good news is that type of simplifying is the same process we want to do for rational expressions.*0215

*We want to look for their common factors. *0222

*These are polynomials we will probably have to factor first, so that we have the actual factors in their p’s that are multiplied.*0226

*Watch how that works for this guy 4y + 2 ÷ 6y + 3.*0233

*Let us factor the top and bottom.*0238

*On the top there is a 2 in common and I can take it out from both parts.*0241

*That would be 2y + 1 and on the bottom it looks like there is a 3 in common, we will take that out, 2y + 1.*0246

*Notice we have this common piece 2y + 1 that is multiplied on the top and bottom.*0259

*We can cancel that out and we would just be left with 2/3.*0266

*Remember that we can only cancel out factors, pieces that are multiplied.*0272

*If you attempted to do some canceling at the very beginning, you can not do that, not yet because you are dealing with addition.*0277

*Be very careful as we try and simplify your expressions.*0287

*Some answers may look a little different from other answers, but actually they may just be equivalent.*0294

*One thing that can make equivalent expression is where you put the negative sign. *0301

*You could put it in the top, the bottom or out front of one of these rational expressions.*0306

*In all cases, they represent the same exact quantity.*0313

*You will say that they are all equivalent.*0317

*This also applies to your rational expressions, but sometimes it is not quite as obvious. *0320

*For example, maybe I'm looking at x - 7 ÷ 2.*0328

*Let us go ahead and put a negative sign on the top.*0334

*If I do that, that would be -x and may be distribute it through with that negative + 7 ÷ 2.*0339

*I could have also given that negative sign to the bottom x – 7 ÷ 2 and give that to the bottom and that would give me x - 7 ÷ -2.*0350

*These two expressions are equivalent.*0367

*They do equal the same thing, even though they look a little different.*0369

*Be careful if you are working on your homework, working with other people and you do not get exactly the same answer, *0375

*you might actually have equivalent expressions just inside a different form.*0380

*Now that we know about simplifying and some things to watch out for how do you get into the multiplication and division process.*0389

*I want to think back on how you do this with fractions.*0396

*Before multiplying fractions together and it is a nice process of multiplying across the top and multiplying across the bottom. *0400

*The same thing applies to your rational expressions.*0409

*We will be dealing with polynomials for sure, but will just multiply across the top and across the bottom.*0412

*In order for this to work, we must know how to factor our polynomials. *0418

*That way we can end up just multiplying their factors together.*0423

*When we are all done, we want to make sure that we have written it in lowest terms.*0427

*Try and cancel out any extra factors after you are done multiplying.*0431

*If you know how the multiplication process works, then you will also know a lot about the division process.*0438

*Think about how this works when you divide fractions.*0445

*From looking at something like 2/3 ÷ 5/7 and we have been taught to flip the second fraction and then multiply.*0448

*Which is of course, multiply across the top and multiply across the bottom.*0456

*There is some good news and this also applies to our rational expressions. *0460

*If you want to multiply them together, flip your second rational expression and then multiply the two.*0465

*Factoring will definitely help in this process that way we have to keep track of the individual factors, and where they go.*0472

*Always write your answer in lowest terms when you are all done.*0479

*That is quite a bit of information just on simplifying and multiplying and dividing.*0485

*We will look at some quick examples and see how this works out.*0491

*We want to take all of these rational expressions and put them into lowest terms.*0497

*We will be going through a simplification process.*0501

*Notice how in a lot of these we are dealing with addition and subtraction, do not cancel out yet until you get it completely factored.*0505

*Let us start with the first one.*0512

*I have (x*^{2} - y^{2}) ÷ (x^{2} + 2xy + y^{2}).0514

*These look like some very special formulas that we had earlier.*0520

*I have the difference of squares on top. *0523

*I have a perfect square trinomial on the bottom so I can definitely factor these.*0525

*I have x + y x – y on the top and on the bottom x + y, x + y.*0533

*Notice how we have a common factor the x + y.*0549

*We can go ahead and cancel that out. *0554

*This will leave us with an x - y ÷ x + y.*0558

*I can be assured that this is in the lowest terms because there are no other common factors to get rid of.*0565

*The next ones are very tricky.*0571

*Notice how the top and bottom almost look like the same thing.*0574

*It is tempting to try and cancel out.*0578

*Be careful, we cannot cancel them out unless they are exactly the same thing.*0581

*One thing to notice here is a -5 and here is 5. *0586

*Those are not the same thing, they are different in sign.*0590

*And same thing over here, this is w*^{2} and this one is –w^{2}.0593

*Those are not the same in sign.*0597

*If you end up with a situation like this where they are almost the same, you are dealing with subtraction and the order is just reversed.*0600

*You can factor out a -1 from either the top or bottom.*0607

*If I factor out a -1 from the top then what is left over?*0614

*-1 × what will give me a w*^{2}, is a -w^{2} and let us see if I take out 5, that should do it.0620

*-1 × -w = w, -1 × 5 = -5.*0633

*All of that is on top and I still have my 5 - w*^{2} on the bottom.0641

*We are getting a little bit closer and things are starting a matchup in sign a little bit better.*0646

*I’m just simply going to reverse the order of these and you will see that they are common factors.*0652

*-1 is still out front, (5 - w*^{2}) (5 + w^{2}) and now I can go ahead and cancel these out.0657

*The only thing left here is a -1.*0677

*That looks much nicer than what we started with.*0680

* In the next one I have 25q*^{2} - 16/12 – 15q.0684

*This one is going to take a little bit more work but I see I have one of those special cases on the top.*0691

*That is another difference of squares.*0697

*(5q + 4) (5q – 4) *0700

*Let us see if we can do anything with the bottom.*0712

*Does anything go into 12 and 15?*0714

*These both have a 3 in common, let us take that out.*0718

*We are looking pretty good and we can see that this is getting pretty somewhere to the previous example.*0725

*These look almost the same we are dealing with subtraction, but the order is just reversed. *0732

*We are going to take out a negative from the bottom so that they will be exactly the same.*0736

*(5q + 4) (5q – 4)*0741

*There you will take out the 3, let us take out -1 as well.*0750

*That will give us -4 + 5q.*0753

*It is better starting to match the top we just have to reverse the order.*0758

*Reading on the bottom 5q - 4 and now we can see we have a common factor to go ahead and get rid of.*0773

*The answer to this one would be 5q + 4 ÷ -3..*0782

*The last one involves 9 – t ÷ 9 + t.*0791

*In this one, another one that looks very close.*0796

*Unfortunately there is not a whole lot we can do to simplify it.*0800

*It is already simplified.*0802

*You might be wondering why cannot we just cancel out some 9 and call it good from there.*0805

*We can only cancel out common factors, things that are multiplied.*0810

*We cannot cancel out the 9 nor we can cancel out the t’s.*0814

*Unfortunately there is nothing to factor from the top or factor from the bottom. *0818

*This one is simplified just as it is.*0822

*Be on the watch out for cases like this and know when you can cancel out those extra terms.*0834

*In this next few we are going to go through the multiplication process and then try and bring it down to lowest terms.*0844

*We just have to multiply across the top and then multiply across the bottom. *0850

*That will make our lives a little bit easier.*0854

*Let us start off with this first one.*0856

*Multiplying across the top I will have 8x*^{2} × 9 / 3xy^{2}.0859

*From here I can cancel a lot of my extra stuff.*0873

*I will cancel out extra 3 that it is the 9.*0876

*I can cancel out one of these x’s here and I can cancel out one of these y’s.*0880

*Let us see what is left over.*0886

*I still have 8x, 3, a single y on the bottom.*0888

*This is 24x ÷ y and that only multiply the two together, but I brought it down to lowest terms.*0895

*Onto the next one, multiplying the top together will give me 3t – (u × u) / (t × 2) × (t – u).*0905

*One obvious common term is that t – u, let us go ahead and get rid of that.*0926

*We would have left over 3u t × 2 or just make my brain feel better, 2 × t.*0933

*We have multiplied those together and reduce it to its lowest terms.*0944

*I want to point out something, back here it is tempting to go through the distribution process and put the 3 and t and the 3 and u, *0949

*but actually you do not want to do that just yet. *0958

*Go ahead and leave them into your factors because it will make it much easier to cancel them out.*0961

*If you do end up distributing them, you have to pull them back a part into their factors later on.*0966

*You are not saving yourself any work.*0971

*Leave the factors in there or if is not factored already go ahead and factor it so you can easier multiply and reduce.*0974

*Let us get into a much bigger one.*0984

*In this one we want to multiply together and then put into its lowest terms.*0987

*(x*^{2} + 7x + 10 / 3x + 6) × (6x – 6 / x^{2} + 2x – 15)0992

*This is a rather large one, but I'm not going to multiply together the tops first or the bottoms just yet.*1005

*I’m going to work on factoring just for little bit. *1011

*How would my first polynomial here factor?*1017

*My first terms would need to be an x that will give me x*^{2} and my other terms would have to be 2 and 5.1023

*That will be the only way that I can get that 10 and they would add to be the 7 in the middle.*1031

*That would factor that and we will also factor the bottom.*1037

*It looks like there is a common 3, let us pull that out.*1041

*We have just factored that first rational expression.*1047

*Onto the next one I see it has a common 6, x – 1 and on the bottom I see a trinomial.*1051

*Let us go ahead and break that down into two binomials.*1061

*X*^{2} + 2x – 15.1065

*What would that factor into?*1068

*How about 5 and -3?*1069

*Now that I have all the individual factors, I can multiply them together much easier.*1075

*I will simply write all of the factors on the top and multiply that out.*1079

*All the factors in the bottom, 3x + 2 x + 5 and x – 3.*1092

*In that step they are all multiplied together.*1102

*Let us cancel out a lot of those extra terms.*1105

*x + 5 were gone, x + 2 those were gone.*1109

*Let us cancel out one of the 3 and 6.*1116

*The only thing we have left over is 2 and x – 1 / x -3.*1120

*Since there are no more common factors I know that this one is finally in lowest terms.*1129

*This takes care of a lot of multiplication let us get into dividing these and also putting them into lowest terms.*1138

*With this one needs one need one extra step, we have to flip the second rational expression and then multiply and reduce.*1144

*Let us start off with this first one.*1154

*I have 9x*^{2} ÷ 3x + 4.1156

*I will flip the second one, 3x + 4 ÷ 6x*^{3}.1162

*If I’m going to multiply the two together, I need to write all of their factors, the top ones together.*1173

*We will write all the factors of the bottom one together.*1184

*Now they are multiplied.*1189

*We can go ahead and cancel a lot of those extra terms.*1192

*We have a common 3x + 4 in the top and bottom, let us get rid of that, we do not need that.*1195

*We cancel out an extra 3 in the top and bottom.*1201

*On top we have x*^{2} and in the bottom we have x^{3} so an x^{2} will cancel out leaving me x in the bottom.1207

*That is quite a bit of canceling.*1216

*Let us see if we can figure out what is left over.*1217

*I still have a 3 on top, we still have a 2 and x on the bottom.*1219

*3 / 2x is the only thing left when we reduce it to lowest terms.*1225

*The next one looks like a do- see, let us go ahead and take this one nice and easy.*1231

*We will first write the first rational expression, just as it is.*1237

*We will go ahead and flip the second rational expression.*1247

*4r -1*^{2} and -r^{2} r + 31250

*If I'm going to multiply this together we will go ahead and write all of the factors on the top.*1263

*4r*^{2} – 1 on the bottom 2r + 1 -r^{2} and r + 31272

*If we multiply things together let us see what we can cancel out.*1290

*One initial thing, I see an r + 3, let us get rid of that.*1294

*I got an r on top and r*^{2} so let us get rid of some r.1299

*I think there is even a little bit more we can cancel out as long as we recognize that we have a very special polynomial on top.*1305

*Notice how we have the difference of squares on top the 4r*^{2} -1.1318

*We can write that as 2r + 1 and 2r - 1.*1325

*We can see that we do have an extra factor hiding in there that we can get rid of. *1333

*We can get rid of that 2r + 1.*1338

*What is left over, I have 4 × 2r -1 on the top / -r in the bottom.*1346

*This one is completely factored.*1354

*I got one more example and this is another large one.*1365

*We will have to just walk through it very carefully.*1369

*I have xw - x*^{2} / x^{2} – 1 then we are dividing that by x - w / x^{2} + 2x +1.1371

*Let us see what we can do.*1381

*Let us go ahead and rewrite it and have the second one flipped over.*1383

*Multiply it by x*^{2} + 2x + 1x – w.1395

*We will go ahead and start putting things together. *1407

*Let us multiply (xw - x*^{2}) (x^{2} + 2x + 1) ÷ (x^{2} – 1) (x – w)1410

*This one looks like it is loaded with lots of things that we can go ahead and factor.*1428

*Let us go ahead and do that and we are going to do that bit by bit.*1432

*I’m looking at this very first factor here and notice how they both have terms in there that have x.*1436

*We can factor out an x from each of those.*1443

*That will leave us with w – x.*1449

*In this one over here we can factor it into an x + 1 and x + 1.*1454

*It is one of our perfect squares.*1460

*(x + 1)(x +1)*1465

*On to the bottom, this one right here is the difference of squares x + 1 x -1.*1473

*Unfortunately this guy we do not have anything special about it, I will just write it.*1486

*That will allow me to at least cancel out a few things.*1493

*I have an extra x + 1 that I will go ahead and get rid of.*1496

*It looks like I can almost get rid of something else.*1500

*Notice how we have this w – x and x – w.*1503

*It is almost the same thing, and it is involving subtraction. *1507

*I need to factor out a -1 from one of these.*1511

*Let us do it to the top -w + x.*1515

*I still have an x + 1 on top / x – 1 and x – w.*1522

*-x will rearrange the order of these guys and switch them around.*1537

*That will give us (x – w) (x + 1) / (x – 1) (x – w).*1544

*We can see sure enough, there is another piece that we can go ahead and get rid of.*1556

*We only have one thing left, -x × x +1 / x – 1.*1563

*Since we have no other common factors, this is finally in its lowest terms. *1575

*When it comes to working with these rational expressions, remember how all of these processes work, which is normal fractions.*1580

*Whether that means multiplying fractions by going across the top and bottom*1587

*or dividing fractions by flipping the second one.*1591

*Then cancel out your extra factors so you can be assured that is brought down to lowest terms.*1593

*Thank you for watching www.educator.com.*1599

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at adding and subtracting rational expressions. *0003

*In order for this process to workout correctly I have to spend a little bit of time working on finding a least common denominator. *0009

*This will help us so we can rewrite our rational expressions and actually get them together. *0017

*Then we will get into the addition process.*0022

*We will look at the some examples that have exactly the same denominator.*0025

*We will look at others which have different denominators.*0029

*Once we know more about the addition process, the subtraction process will not be too bad.*0032

*You will see that it will involve many of the similar things.*0036

*So far we have covered a lot about multiplying and dividing rational expressions, *0044

*but we also need to pick up how we can add and subtract them.*0048

*One key that will help us with this is you want to think of how you add and subtract simple rational number.*0053

*Think of those fractions, how do you add and subtract fractions?*0059

*One key component is that we need a common denominator before we can ever put those fractions together.*0063

*Think of how that will play a part with your rational expressions. *0070

*Let us first look at some numbers okay.*0076

*Let us suppose I had 2/1173 and 5/782 and I was trying to add these or subtract them.*0078

*No matter what I'm trying to do, I have to get a common denominator. *0088

*One thing that will make this process very difficult is that I chose some very large numbers in order to get that common denominator.*0093

*One thing that could help out when searching for this common denominator is not to take the numbers directly since they are so large.*0101

*You have to break them down into their individual factors.*0108

*Down here I have the same exact numbers, but I have broken them down into 3 × 17 × 23.*0112

*I have broken down the 782 into 2 × 17 × 23.*0119

*What this highlights is that the numbers may not be that different after all. *0123

*After all, they both have 17 and 23 in common and the only thing that is different is this one has a 3 and this one has a 2.*0128

*When trying to find that common denominator, I want to make sure it has all the pieces necessary *0139

*or I could have built either one of these denominators.*0145

*It must have the 2, 3 and also those common pieces of the 17 and the 23.*0147

*This is the number that you would have been interested in for getting these guys together.*0153

*When working with the rational expressions, we will be doing the same process.*0161

*We want them to have exactly the same denominator, but I would not be entirely obvious to look at the denominator and know what that is.*0166

*We will have to first factor those denominators, we can see that the pieces present in which ones are common and which ones are not.*0176

*We will go ahead and list out the different factors in the denominator first.*0184

*We will also list out the variables, when it appears the greatest number of times.*0192

*We will list the factors multiplied together to form what is known as the least common denominator.*0200

*Think of the least common denominator as having all the factors we need and we could have built either one of those original denominators.*0206

*Let us take a quick look at how this works with some actual rational expressions.*0216

*I want to first look at factoring the bottoms of each of these.*0221

*8 is the same as 2 × 2 × 2 and y*^{4} we will have bunch of y all multiplied by each other.0225

*For 12 that would be 3 × 2 × 2 and then a bunch of y, 6 of them.*0236

*When building the least common denominator, I will first gather up the pieces that are not the same.*0245

*Here I have the 3 and this one has an extra 2.*0254

*I need both of these in my least common denominator 2 and 3.*0259

*This one has a couple of extra y.*0267

*Let us put those in there as well. *0270

*Once we have spotted all the differences between the two then we can go ahead and highlight everything that is the same. *0274

*We have a couple more twos and 1, 2, 3, 4 y.*0282

*Let us package this altogether. *0290

*2 × 2 × 2 = 8 × 3 = 24 and then I have 6y, y*^{6}.0292

*Let us do some shortcuts here. *0303

*One, you could have just figured out the least common denominator of the 8 and the 12 that will help you get the 24.*0305

*You can take the greatest value of y*^{6} and gotten the y^{6}.0312

*We use techniques like that to help us out when looking for that least common denominator.*0321

*Some of our expressions may get a little bit more complicated than single monomials on the bottom. *0330

*Let us see how this one would work.*0335

*This is 6 /x*^{2} - 4x and 3x – 1/ x^{2} – 6.0336

* In order to figure out what our common denominator needs to be, we are going to have to factor first.*0343

*Let us start with that one on the left and see if we can factor the bottom.*0350

*It looks like it has a common x in there.*0353

*We will take out and x from both of the parts.*0356

*For the other rational expression that looks like the difference of squares.*0363

*x + 4 and x – 4*0370

*We just have to look with these individual factors. *0376

*I can see that what is different is this x + 4 piece and the x piece.*0381

*Let us put both of those into our common denominator first.*0388

*I have x and x + 4 then we can go ahead and include the pieces that are common.*0392

*Any common pieces will only include once.*0402

*This down here represents what our least common denominator would be. *0406

*We need an x, x + 4, and x -4. *0411

*Finding the least common denominator is only half the battle. *0419

*Once you find the least common denominator, you have to change both of your rational expressions *0422

*so that they contain this least common denominator. *0428

*Once you have identified it go one step farther and rewrite the expressions so that they have this least common denominator. *0433

*Let us watch how this works with our numbers.*0441

*That way we could get a better understanding before we get into the rational stuff.*0443

*Here are these fractions that I had earlier and you will notice that the bottom is already factored.*0447

*Our LCD in this case was 2 × 3 × 17 × 23.*0454

*Now suppose I want them to both have this as their new denominator.*0463

*2 × 3 × 17 × 23, 2 × 3 × 17 × 23.*0472

*When looking at the fraction on the left here the only difference between this and the new LCD that I wanted to have is it is missing a 2.*0483

*I could give it a 2 on the bottom but just to balance things out I will also have to give it a 2 on the top. *0495

*On the top of this one will be 2 × 2.*0504

*I better highlight that this 2 was the one we put in there.*0510

*For the other one, it needs to have that 3, I will give it a 3 on the top and there is where the 3 came from in the bottom.*0515

*You can see that we give the missing pieces to each of the other fractions. *0526

*If I was looking to add or subtract these I will be in pretty good shape since they have exactly the same denominator.*0532

*We want to do the same process with our rational expressions, give to the other rational expressions its missing pieces *0538

*so it can have that least common denominator.*0545

*We can find a least common denominator now, which means we can get to the process of adding and subtracting our rational expressions. *0551

*Think of how this works with our fractions.*0560

*If I have two fractions and I have exactly the same denominator then I will leave that denominator exactly the same *0562

*and I will only add the tops together.*0570

*This works as long as my bottom is not 0.*0574

*This will be the exact same thing that we will do for our rational expressions, the only difference is that this P, Q, and R *0578

*that you see is my nice little example, all of those represent polynomials instead of individual numbers.*0585

*As soon as we get our common denominator we will just add the tops together. *0591

*If they do not already have a common denominator, we have to do a little bit of work.*0599

*It means we want to find a common denominator and often times we will have to factor first before we can identify what that is. *0603

*Then we will have to rewrite the expression so that both of them have this least common denominator. *0613

*Once we have that then we will go ahead and add the numerators together and leave that common denominator in the bottom. *0619

*Even after that we are not necessarily done.*0627

*Always factor at the very end to make sure that you are in the lowest terms.*0629

*Sometimes when we put these together we can do some additional canceling and make it even simpler. *0633

*The subtraction process is similar to the addition process.*0644

*You will go through the process of finding the common denominator. *0650

*Make sure they both have it, and then you will end up just subtracting the tops.*0652

*Remember though you want to subtract away the entire top of the second fraction.*0660

*Often to do this, it is a good idea to use parentheses and distribute through by your negative sign on the top part.*0665

*That way we would not forget any of your signs. *0671

*It is usually a very common mistake when subtracting these rational expressions.*0674

*Let us get to business.*0681

*Let us look at this example and add the rational expressions.*0682

*I have (3x / x*^{2} – 1) + (3 / x^{2} – 1). 0686

*The good news is our denominators are already exactly the same.*0691

*I will simply keep that as my common denominator in the bottom and we will just add the tops.*0697

*Even though we have added this and put into a single rational expression, we are not done. *0707

*We want to make sure that it is in lowest terms.*0711

*Let us go ahead and factor the top and bottom, see if there are any extra factors hiding in there.*0716

*As I factor the top, this will factor into 3 × x +1 and then we can factor the bottom, this will be x + 1 and x – 1.*0722

*I can definitely say yes there is a common piece in there, it is an x +1 and we can go ahead and cancel that out.*0735

*I'm left with a 3 / x - 1 and now I have not only added the rational expressions, this is definitely in lowest terms.*0743

*Let us try this on another one.*0758

*We want to add together the two rational expressions I have (-2/w + 1) + (4w/w*^{2} -1).0760

*This one is a little bit different.*0769

*The denominators are not the same. *0771

*Let us see if we can figure out what the denominator should have in the bottom by factoring them out and seeing what pieces they have.*0775

*w*^{2} - 1 is the difference of squares which would break down into w + 1 and w -1.0785

*It looks like that first fraction is missing a w -1.*0796

*We have to give it that missing piece.*0802

*We will write it in blue.*0808

*I will give it an extra w -1 on the bottom and on the top just to make sure it stays the same.*0809

*Our second fraction already has our least common denominator, so no need to change that one.*0819

*Now that it has a common denominator we will keep it on the bottom and we will simply add the numerators together.*0826

*I got -2w -1 + 4w.*0835

*We cannot necessarily leave it like that, I’m going to go ahead and continue combining the top, *0841

*maybe factor and see if there is anything else I can get rid of.*0846

*Let us distribute through by this -2.*0851

*-2w + 2 + 4w / (w +1) (w -1)*0853

*(-2w + 4w) (2w + 2)*0868

*I think I already see something that will be able to cancel out.*0876

*Let us factor out a 2 in the top.*0880

*Sure enough, we have a w + 1 in the top and bottom that we can get rid of.*0887

*That is gone. *0895

*Our final expression here is 2 / w -1 and now we have added the two together and brought it down to lowest terms.*0898

*Let us do a little bit of subtraction.*0910

*This one involves (5u /u – 1 – 5) + (u /u -1).*0912

*This is one of our nice examples and that we are starting off and has exactly the same denominator. *0919

*That is good so we can go ahead and just subtract the tops.*0925

*I will have 5u - the other top 5 + u.*0931

*Note what I did there, I still have the entire second top and I put it inside parentheses and I'm subtracting right here. *0940

*One common mistake is not to put those parentheses in there and you will only end up subtracting the 5.*0950

*You do not want to do that.*0956

*You want to subtract away the entire second top.*0956

*To continue on, I want to see if there is anything that might cancel.*0961

*I’m going to try and crunch together the top a little bit and see if I can factor. *0964

*Let us distribute through by that negative sign.*0969

*5u - 5 - u / u -1*0974

*That will give me 5u – 1 = 4u – 5/ u – 1.*0981

*It looks like the top does not factor anymore.*0995

*This guy is in lowest terms.*0997

*Let us tackle one more last one and these ones involve denominators that are not the same.*1008

*We are going to have to do a lot of work on factoring and seeing what pieces they have before we even get into the subtraction process. *1019

*Let us go ahead and factor the rational expression on the left.*1026

*On the bottom I can see that there is an (a) in common.*1032

*Over on the other side I can factor that into a - 5 and another a – 5.*1042

*They almost have the same denominator.*1056

*They both have that common a - 5 piece and the one on the left has an extra a.*1059

*And one on the right has an additional a – 5.*1063

*Let us give to the other one the missing pieces. *1066

*Here is our rational expression on the left, we will give it an additional a – 5.*1078

*With this one it already has a - 5 twice, we will give it an additional a on the bottom and on the top.*1091

*Now that they have exactly the same denominator we can focus on the tops 3a a– 5 - 4a /(a – 5) (a- 5).*1104

*It looks like we can do just a little bit of combining on the top. *1123

*I have (3a*^{2} - 15a – 4a) / (a)(a – 5)(a-5)1127

*When combined together the -15a and the -4a, 3a*^{2} – 19a.1142

*We have completely subtracted these we just need to worry about factoring and canceling out any extra terms.*1156

*On the top I can see that they both have an extra a in common, 3a-19.*1168

*Let us take that out of the top.*1174

*We will cancel out that a and now we brought it down into lowest terms. *1183

*I have 3a - 19 / (a – 5) (a -5)*1189

*Whether you are adding or subtracting rational expressions make sure that you have your common denominator first.*1202

*Once you do, you just have to focus on the tops of those rational expressions by putting them together.*1209

*Once you do get them together remember you are not done yet, feel free to factor one more time and reduce it to lowest terms. *1215

*Thank you for watching www.educator.com.*1222

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at some complex fraction.*0002

*We will first have to do a little bit about explaining what a complex fraction is *0012

*and then I’m going to show you two techniques and how you can take care of them.*0015

*In the first one we will write out these complex fractions as a division problem *0018

*and then we will go ahead and use the method of this common denominator in order to simplify them.*0023

*Each of these methods has their own advantages but they both should work when dealing with simplifying complex fraction. *0029

*What a complex fraction is, it is either the numerator or the denominator is also a fraction.*0040

*Here is a good example of a complex fraction using numbers. *0047

*You will notice that the main division bars is actually sitting right here.*0050

*But in the numerator I have 2/5 and in the denominator I have 1/7. *0055

*If I have fractions made up of other fractions. *0061

*These are exactly the types of things that we are looking to simplify.*0064

*Since we are on a lot of rational expressions, then we will not only look at just numbers but we will look at more complex fractions like this. *0072

*One of the first techniques that we can use to clean this up is to use division.*0082

*We will use that main division bar and write this as a division problem. *0089

*Here I have (x + ½) ÷ (6x + 3)/4x.*0095

*It is the main division bar right there that will turn into division.*0101

*That means we will have to use all of our tools for simplifying the left and right, and eventually be able to get them together.*0106

*In a previous lesson we learn that we need to flip the second rational expression then multiply across the top and bottom.*0115

*That is exactly what you will see with these.*0120

*Let us grab on these rational expressions and give it a try.*0125

*You have (t*^{2} u^{3}) / r ÷ t^{4}u/r^{2}.0127

*We want to identify what is on the top and what is on the bottom?*0135

*We want to write those again as a division problem.*0140

*(t*^{2} u^{3}) / r ÷ t^{4}u/r^{2}0144

*It looks pretty good.*0157

*We want to turn this into a multiplication process by flipping that second rational expression t*^{2} u^{3}/ r will now be multiplied by r^{2} ÷ t^{4}u.0158

*That looks much better.*0177

*We can go ahead and multiply across the top.*0178

*I will just put all of these on the top and multiply across the bottom.*0182

*I will put of all of these on the bottom.*0187

*Now that we have this we would simply go through and cancel out our common factors.*0190

*We will get rid of t*^{2} on the top and t^{2} in the bottom.0198

*Making a t*^{2} in the bottom.0204

*We can cancel out u, bring this down to u*^{2} and we can cancel out on r.0207

*We have r in the top.*0217

*Let us write down everything that is left over.*0220

*u*^{2} r ÷ t^{2} and then we could consider this one simplified.0222

*It changes it into a problem that we have seen before.*0231

*You just have to do a lot of work with simplifying.*0234

*When it gets to simplifying a complex fraction, even that process is not necessarily the easiest to go through.*0240

*In fact, you will find that this next problem is quite lengthy. *0248

*We have (1/x + 1 + 2/y – 2) ÷ (2 /y – 2 – 1/x + 3).*0251

*Let us identify everything on the top and everything in the bottom.*0262

*That way we can simply rewrite this 1/x + 1 + 2/y – 2 all of this is being divided by everything on the bottom 2/y – 2 – 1/x + 3.*0269

*If I have any hope I'm doing this as a division problem that I need to normally flip that second fraction.*0298

*Notice how in this one I do not have a single fraction.*0305

*It is tempting to say, hey why we just flip both of them but that is not how division works.*0309

*We need to combine it into a single fraction before we can flip it and then do the multiplication.*0313

*Let us see if we can get these guys together with some common denominators. *0322

*The common denominators on this side are the x + 1 and y – 2.*0327

*In order to get those together, I would have to give this fraction, y - 2 on the top and bottom.*0343

*To make it workout over here we will give the top and bottom of that one x +1 and the 2 is still up there.*0353

*It looks like that first piece will turn into y – 2 + let us go ahead and distribute this guy in there (2x + 2) ÷ (x + 1) (y – 2).*0365

*That just takes that and crunches it down a little bit.*0384

*Let us focus on this other one.*0388

*We need a common denominator and I see there is a y - 2 and x + 3.*0390

*I will need to give the fraction on the left an additional x + 3.*0408

*We still have the 2 in there.*0412

*2 × x + 3*0414

*Over on the other side let us give the top and bottom of that one y -2.*0417

*When those are put together, we will do a little bit of distributing here.*0424

*We will have (2x + 6) – (y + 2) ÷ (y – 2) (x + 3).*0431

*Let us go ahead and write this again and see if we can do the actual division.*0447

*It looks like I can cancel out a few things in here.*0455

*Let us save ourselves a little bit of work.*0458

*y + 2x ÷ (x + 1) (y -2), we are dividing it by the second fraction.*0461

*Here is when I’m going to flip and multiply it, multiplied by 2x - y + 8.*0476

*I have combined the 6 and the 2 together, y - 2 x + 3.*0487

*I can just combine the tops and bottoms. *0498

*Quite a lot of pieces in here that is okay, at least I see one piece will cancel out and that is the y – 2.*0516

*We are left with y + 2x and (x + 3) ÷ (x +1) and 2x - y + 8. *0528

*Even though it can be a lengthy process, by rewriting it as a division process and using our tools from before*0543

*you will see is that it is possible to reduce and simplify this complex fractions. *0549

*The other method which can often be a lot cleaner is using the least common denominator *0556

*in order to clear out all the fractions in the top and bottom.*0561

*In order for this method to work, you must find the least common denominator of all the little fractions present in your complex fraction.*0566

*Go ahead and look at your numerator and denominator and think of all the least common denominator for all those fractions. *0573

*Once you find that LCD, then you are going to multiply on the top and the bottom of the main fraction *0580

*or on the top and bottom of the main division bar.*0586

*This will clear up things immensely but you have to be careful on canceling out.*0591

*You do not want to accidentally cancel out something that you should not.*0596

*You will see that you will clear out a bunch of stuff and then you end up simplifying just as you would normally.*0598

*No matter what method you use, you should get the same answer. *0608

*Use the method that you are more comfortable with. *0614

*I like to recommend a method two because it is usually much cleaner than using the first method.*0620

*However, anything that is cleaner use less opportunities for mistakes. *0627

*One downside to the second method is usually happen so quickly it is hard to keep track of everything that was in there.*0633

*Let us try this second method with the following complex fraction. *0643

*I have (2/s*^{2} t + 3 /st^{2}) ÷ (4/s^{2} t^{2} – 1/st).0647

*Let us first see if we can identify the least common denominator.*0656

*Try and pick out all these little denominators here, see what the LCD would be.*0661

*They all have some s and the largest one I see there is an s*^{2}. 0668

*They all have t’s but the largest one in there is a t*^{2}.0675

*I’m going to take this and I’m going to multiply it on the top and the bottom of this expression so I can rewrite it.*0680

*There is our main division bar.*0698

*Let us get to multiplying.*0708

*On the top of this entire main division bar I will multiply it by this LCD.*0709

*I will do the same thing on the bottom to keep things nice and balanced. *0715

*The top and bottom both have two terms, so we will definitely make sure that we distribute each of these parts.*0719

*We are going to write the result of this multiplication and then watch how many things will cancel out in the next step.*0730

*(2s*^{2} t^{2} / s^{2} t) + (3s^{2} t^{2} / st^{2}) ÷ (4s^{2} t / s^{2} t^{2} - s^{2} t/st).0737

*It looks messy and it looks I have actually made things even more complicated.*0768

*Watch what is going to go take a vacation here soon.*0772

*I have these s*^{2} they will go away. 0776

*Then I have a single t and t up top, they will go away.*0779

*On the next one, there is my s one of those will be gone and both my t*^{2} are gone.0783

*Onto the bottom s*^{2} is gone, s^{2} is gone.0790

*I think I am missing some of my squares.*0797

*Our t*^{2} are gone and we can get rid of one of these t’s.0805

*The denominators of all those little fractions, these guys that we are so worried about at the very beginning, *0817

*all of them have been cancel out in some way or another. *0823

*This means as soon as we write down our leftover pieces, this one is simplified.*0826

*2t + 3s, those are the only things that survived up here ÷ 4 - st and those are the only things that survived on the bottom. *0831

*This one is in its most simplified form.*0847

*It is nice, quick and easy method you just have to properly identify the LCD first.*0852

*You will know you are using the method right if all of these denominators end up going away.*0858

*If any of them are still in there double check to see what LCD you used.*0862

*Let us try this one more time but something a little bit more complicated. *0869

*This one is (2y + 3 / y – 4) ÷ (4y*^{2} – 9) + (y^{2} – 16).0873

*Let us examine these denominators so we can find our LCD.*0882

*We are looking at this one this is the same as y - 4 and y + 4.*0888

*Over here it already has the y -4 in it.*0897

*The only piece that I am missing is the y + 4.*0900

*My LCD will contain both of these parts.*0905

*I have the y – 4 and y + 4, both of those in there.*0908

*I’m going to take that and we are going to multiply it on the top and bottom of the original.*0917

*Let me just quickly rewrite this and I’m going to rewrite it with the factored form on the bottom.*0923

*We will take our LCD and we will multiply it on the top and on the bottom of our main fraction here.*0939

*y – 4 y + 4.*0952

*Let us go ahead and put everything together and let us see how this looks.*0959

*I have (2y + 3 ) (y – 4) (y + 4) ÷ y – 4.*0965

*Then comes our main division bar right there.*0980

*On the bottom is (4y*^{2} – 9) × (y – 4) × (y + 4) ÷ (y – 4) (y + 4).0985

*Watch how many things will cancel in this next step.*1005

*y – 4 and y – 4 those are gone.*1009

*y – 4, y – 4, y + 4, y + 4, 4 those are gone.*1013

*All of these problems that we had at the very beginning, they are no longer problem.*1017

*They are gone.*1022

*We will simply write down all of the left over pieces.*1023

*2y + 3 y + 4 4y*^{2} – 9.1028

*Be careful, there is still some additional reducing that you can do even after using your LCD like this.*1044

*One thing that I can see is that I can actually continue factoring the bottom.*1050

*Let us write that out.*1056

*2y + 3/ y + 4 and this will be over I have different squares on the bottom so, 2y + 3, 2y – 3.*1058

*I’m sure enough now we can more easily that I have an extra 2y – 3 in the bottom and that is gone as well.*1074

*This one finally reduces down to y + 4 / 2y -3 and now we are finally done.*1081

*The second method is definitely handy and clears up a lot of fractions very quickly.*1092

*Use whatever method you are more comfortable with.*1097

*Thank you so much for watching www.educator.com. *1101

*Welcome back to www.educator.com.*0000

*In this lesson we are going to work on solving rational equations.*0003

*Two things that we will look at is I can solve a equation that contains a rational expression.*0009

*I will throw in an example or we deal with a formula that also involves a rational expression.*0016

*Recall that when you are working with equations that have a lot of fractions in them we will often use the least common denominator *0025

*to go ahead and clear out all of those fractions. *0032

*These particular types of equations you definitely want to make sure that you check your solutions. *0036

*This will be especially important for these types of equations this is because we are dealing with expressions that involve fractions,*0041

*we will have some restricted values, values that we cannot have.*0051

*Some of these values may try and show up as our solution.*0056

*We simply have to get rid of them.*0059

*They are not valid and they will make our denominator 0.*0061

*When working with these expressions that have a lot of fractions we like to keep the least common denominator throughout the entire process*0068

*or that way we can go back in and make sure that makes sense in the original.*0076

*Let us go ahead. *0081

*When you see my example for formulas here again and know that the same process plays out. *0086

*We like to use that least common denominator to go ahead and try and clear things out*0092

*Of course, our goal is to isolate which ever variable we are trying to solve for.*0096

*When we are all done with that formula we still have an equation of these words *0103

*and for more variables still floating around in there.*0107

*That is okay as long as we get that variable we are looking for completely isolated. *0109

*That is how we will know it is solved.*0114

*Let us take a look at one of these.*0119

*I have 1-2 / x + 1 = 2x / x +1 and we can see that this is a rational equation because looking at all these rational expressions I have, *0121

*think of those fractions. *0131

*To help out I’m going to work on getting a common denominator.*0134

*Already these 2 parts of it have an x + 1 in the bottom.*0139

*We are just going to focus on trying to give an x + 1 to the one over there on the left.*0144

*I can do that by taking 1 / 1 and then multiplying the top and bottom of that by x + 1.*0151

*(x + 1)(x +1) - 2 / (x + 1) = 2x /(x +1)*0162

*On the left side that allows us to finally combine things now that we have a common denominator. *0175

*In doing so, we just put the top together.*0182

*x + 1 - 2 / x +1*0184

*Doing a little bit of simplifying let us go ahead and subtract 2 from 1, x - 1 now x + 1 = 2x / x +1.*0196

*At this stage notice how we have two fractions that are set equal to one another and the bottoms are exactly the same.*0211

*Since the bottoms are the same then I know that the tops of these must also be the same.*0219

*Let us just focus on the top part for a little bit and see if we can find a solution out of that.*0225

*x - 1 = 2x this one is not so bad to solve.*0230

*Let us just go ahead and subtract an x from both sides and I think we will be able to get a possible solution.*0236

*It looks like a possible solution is that -1 = x.*0245

*Be careful this can sometimes happen with these rational equations. *0249

*It looks like we have done all of our steps correctly and it looks like we have found a solution, but this guy does not work.*0253

*To see why it does not work, take this possible solution all the way back to one of our original denominators *0260

*and you can see that if you try and plug it in and it makes the bottom 0.*0267

*We have to get rid of that possible solution.*0272

*Since we do not have any other possibilities left, we can say that this particular equation has no solution. *0275

*With these types of equations, even when you find something that looks like it is a solution you have to go back *0284

*and check to make sure that it make sense in the original.*0291

*Oftentimes you may end up getting rid of things that will simply make that denominator 0.*0294

*Let us try a different one.*0301

*This one is 2 /(p*^{2} - 2p) = 3 /(p^{2} – p)0302

*Before working with this, I need to see what is in the denominator.*0311

*To get a better idea we are going to go ahead and factor.*0316

*In this side I can see that they both have a p in common, we will go ahead and take that out.*0321

*I get p × p -2.*0326

*On the right side here, it looks like it has a p in common as well, p × p -1.*0330

*They almost have the same denominator, but I need to essentially give each one its missing factor.*0339

*Let us give the left side p -1 and we will give the right side p -2.*0346

*Just copy this down and see where the extra pieces are coming from.*0353

*On the left I will put in p -1 on the bottom and on the top.*0365

*The one on the right we will give p - 2 in the bottom and p -2 on top.*0374

*Distributing this will give us (2p -2) / and now we have this nice common denominator equals (3p – 6) / (p) (p -1) (p -2)*0385

*This is right back to that situation we had before.*0406

*The bottoms of each of these fractions are exactly the same I know that the tops of the fractions will also end up being the same. *0409

*Let us extract out just the tops and focus on those.*0418

*If we have to solve this what can we do?*0425

*I will go ahead and add 2 to both sides 2p = 3p - 4 and let us go ahead and subtract 3p from both sides, -p = -4.*0428

*One last step, let us go ahead and multiply both sides by -1.*0448

*My possible solution is that p = 4.*0455

*We cannot necessarily just assume that that is the solution, not until we go back to the original *0459

*and make sure that it is not going to give us 0 in the bottom.*0464

*Let us check to see what factors are in the bottom.*0468

*Some of our restricted values that we simply cannot have is 1p = 0, we cannot have that one. *0473

*We also cannot have p = 2 and we cannot have p = 1.*0480

*Those three values are restricted.*0487

*Fortunately when we look down at what we got, it is not any of these restricted values.*0490

*We will go ahead and keep it as our solution.*0497

*The solution to this one is p = 4. *0499

*Let us look at a little bit large one, one that has a few more things in the denominator.*0505

*This one is 1 / x -2 + 1/5 =2/5 × (x*^{2} – 4)0511

*Quite a bit going on the bottom and it looks like we definitely need to factor one of our pieces *0518

*just so we can see what total factors we have in here.*0523

*1/x – 2 + 1/5 = 2/ I have 5 and then the x*^{2} – 4 that happens to be the difference of squares.0528

*(x + 2)(x – 2)*0545

*It looks like this will be our LCD and we are going to give it to these other fractions here with missing pieces.*0550

*The 1/x – 2 and see what we can put into that one. *0560

*It needs to have a 5, we will put that down below and up top and it also needs to have an x + 2 and below and up top.*0569

*Onto the next fraction here, this is 1/5 and it looks like it is missing an x - 2 piece so we will throw that in there.*0585

*We are missing x + 2 missing both of those pieces.*0597

*On the other side of our equal sign it already has the LCD, so we just leave it just as it is.*0607

*We spread out quite a bit but now we are going to focus on the tops of everything since all the bottoms are now exactly the same.*0616

*We will also use our distribution property a little bit to help us out by multiplying 5 × 2 *0626

*and we will take the two binomials over here and we will foil them so we can put those together.*0633

*5x + 10 would be that first piece on the top of the first fraction.*0641

*We have our first terms x*^{2}, outside terms 2x, inside terms – 2x and last terms -4.0649

*We want that all equal to 2.*0658

*You can notice this is just the tops of all of our fractions.*0661

*Let us go ahead and combine some things.*0666

*My + 2x and my - 2x will go ahead and cancel each other out and we will have x*^{2}, I will take care of that one.0669

*5x took care of that one and combining the 10 and -4, +6 = 2.*0680

*We are almost there. Let us just go ahead and subtract 2 from both sides and see that we need to solve a quadratic.*0689

*We are going to solve x*^{2} + 5x + 4 is all equal to 0.0698

*This quadratic is not too bad I can end up factoring it using reverse foil without too much problem.*0703

*Our first terms x and x and two terms that need to multiply to give us 4 but add to give us 5.*0710

*I am thinking of 1 and 4, those will do it, both positive.*0719

*This will give us two possible solutions.*0725

*When I take x + 1 = 0 and x + 4 = 0, x could equal -1 or x could equal -4.*0727

*There are two possible things that I'm worried about.*0739

*Before we put our seal of approval on them, *0744

*let us go back to some of those original fractions and see what some of our restricted values are.*0746

*Looking at this first one, the bottom would be 0 when x is equal to 2 so I know one of my restricted values x can not equal 2.*0752

*The bottom of this one is a number 5, so that one is not is not going to be 0.*0762

*Over here I can see that when x =-2 I will run into a 0 on the bottom, x cannot equal -2.*0768

*This factor here is exactly the same as the first one.*0776

*We already have that list as one of our restricted values. *0780

*As long as I know my possible solutions are 2 and -2, it looks like we are okay.*0783

*Since I'm looking at -1 and -4, both of these are going to be valid solutions, none of them are restricted.*0789

*The process of solving these rational equations you just have to work on combining them and focus on the tops for a bit.*0796

*Let us look at that formula as I promised.*0807

*To see how you could solve the formula involving some these rational expressions.*0809

*To make things interesting, let us go ahead and solve for y.*0814

*In order to get rid of a lot of these things in the bottom we will identify our lowest common denominator.*0822

*One has an x another one has y and the other one has z.*0830

*It looks like we will need all three of those parts, x, y and z in our new denominator.*0833

*We will go through and give each of them their missing pieces.*0840

*Here we have our originals, let us throw in the missing.*0846

*The first one it already has an x in the bottom, we will give it y and z on the top and the bottom.*0852

*The next one already has a y, so give it the x and z.*0860

*For the last one we have z so xy.*0868

*This will ensure that all of the bottoms are exactly the same. *0878

*We will go ahead and turn your attention to just the tops of all of these and we will continue moving forward.*0882

*2yz = 1 × x × z that will be just xz + xy.*0889

*We are trying to get these y isolated all by himself and looks like we have two of them, one on the left side and one on the right side.*0899

*Let us get them on the same side by subtracting an xy from both sides.*0907

*2yz – xy*0914

*We can factor out that common y that way we will have 1y to deal with.*0922

*Let us go ahead and factor out the common y out front.*0928

*y and then left over we will have 2z – x.*0931

*We are almost free with this one.*0941

*To get the y completely isolated, let us finally divide both sides by that 2z – x.*0946

*y = xz / 2z - x and this one is solved because we have completely isolated the variable we are looking for.*0952

*In a sense we have created a new type of formula.*0970

*Be careful in solving many of these different types of rational equations.*0973

*Make sure you always check your solutions and use that least common denominator to help yourself out. *0976

*Thank you for watching www.educator.com.*0982

*Welcome back to www.educator.com.*0000

*In this lesson we are going to go ahead and take care of rational inequalities.*0002

*There are a few different techniques that you could use for solving rational inequalities,*0007

*but I’m just going to focus on looking at a table and keeping track of the sign for these.*0011

*Recall that when we are trying to solve an inequality that involves a polynomial, we want it in relation to 0. *0020

*The reason why we are doing that is we just have to focus on whether it is positive or negative, a lot less to handle.*0029

*This will make looking at the intervals that represent our solution a little bit easier to find.*0039

*You will see that the process for solving these rational inequalities looks a lot like *0047

*the process for solving our polynomials that we covered earlier.*0051

*We will still look at factoring it down, figuring out where each of those factors are equal to 0 *0055

*and looking at tables so we can test the values around each of those 0.*0060

*The actual process for solving a rational inequality looks like this. *0067

*The very first thing that we are going to is we are going to set the inequality in relation to 0 on one side.*0072

*It means get everything shifted over, so you have that 0 sitting over there. *0078

*Then we will go ahead and solve the related rational equation.*0084

*Solve the same thing and put an equal sign in there and see when it is equal to 0.*0088

*We will also figure out where the denominator can be equal to 0.*0093

*Those are some of our restricted values that we cannot use.*0097

*The reason why it is important to solve it and find out where the bottom is 0 is at those points it could change sign. *0100

*We will call those particular points our critical values.*0108

*It is around those points that it could change in signs, we are interested in what happens.*0113

*To determine what it does around those we will use a few test points *0119

*that will help us to determine which individual interval satisfy the over all inequality. *0124

*Until we get to finding our intervals, we are not quite done yet. *0132

*We also have to pay close attention to the end points to see which ones should be included and which ones should not be included.*0135

*I will give you a few tips on that, watch for that.*0143

*Here are my tips.*0149

*Remember that when you are working with these rational inequalities and get everything over onto one side, *0150

*we might combine it into one large rational expression.*0155

*If we got 3 or 4 of them, put them all down into one.*0159

*Make sure you factor both the top and the bottom.*0164

*You need to see where each of the factors is equal to 0.*0166

*A lot of factoring.*0169

*You will never include any values that make the denominator 0.*0172

*We cannot divide by 0 even if it has an or equal in there, never include anything that makes the bottom 0.*0176

*What that said, if we are dealing with a strict inequality like greater than or less than *0184

*then we will not include any points on the end points.*0189

*If we have greater than or equal to, less than or equal to, we will include where the numerator or the top will equal 0. *0196

*Those are the only end points we have to worry about including.*0209

*That is a lot of information, let us jump in and look at some of these examples.*0215

*I want to solve when -1/x + 1 is greater than or equal to -2/x - 1. *0219

*Rather than worrying about clearing up fractions or anything like that, *0226

*let us get everything over onto one side and get it in relation to 0.*0229

*I'm going to add 2/x - 1 to both sides.*0236

*My goal here is to combine these fractions into one large fraction.*0251

*Do not attempt to clear out these fractions like you would with a rational equation. *0256

*If you clear out the fractions you will lose information about the denominators *0261

*and we want to check around points where the bottoms could be 0.*0265

*Do not clear those out.*0269

*My LCD that I will need to use will be (x + 1)( x – 1)*0271

*Let us give that to each of our fractions.*0280

*-1 will give this one x -1.*0288

*We will do that on the bottom and on the top then we will give the other one x +1.*0292

*Let us combine this into a single fraction here.*0311

*We would have to do a little bit of distributing so - x + 1 + 2x + 2 / x + 1 x - 1 greater than or equal to 0.*0317

*Just a little bit more combining on the top that is a - x + 2 would be a single x and then 1 + 2 would be 3.*0337

*It is quite a bit of work but we have condensed it down into a single rational expression on the left there and we also have it in relation to 0.*0355

*That is a good thing.*0363

*I want to figure out where would this thing be equal 0? *0365

*It will equal 0 anywhere in the top would equal to 0. *0369

*The top is equal to 0 at x = -3.*0375

*We also want to check where would the denominator be 0?*0383

*The bottom equal 0 at two spots when x =- 1, 1x =1.*0389

*All three of these values are what I call our critical values, *0399

*and it is around these values that we need to check the sign of our rational expression here.*0403

*That way we can see whether it is positive or negative. *0408

*Just like before when we are dealing with those polynomial inequalities, this is where our table come into play. *0412

*First start out by drawing a number line and putting these values on a number line.*0419

*I want to start with the smallest ones so -3 and then we will work our way up from there - 1 and 1.*0425

*Along one side of this table we will go ahead and we will write down the factors that our rational inequality here.*0435

*I have x + 3 x +1 and x – 1.*0442

*Now comes the part where we simply grab a test values and see what it is doing around these.*0450

*Our first test value we need to pick something that is less than -3.*0457

*Let us choose -4, that is on the correct side and we will put it into all of our factors to see what sign they have.*0461

*-4 + 3 = -4 + 1 still - and -4 -1=-5, negative.*0469

*We will select a different test value between -1 and -3.*0484

*-2 will work out just great and we will put it into all of our factors.*0490

*Let us see what that does.*0494

*-2 + 3 that would give us something positive, -2 + 1 that will be negative and -2 – 1, negative.*0497

*We are doing pretty good and now we need a test value between -1 and 1.*0512

*I think 0 is a good candidate for this.*0518

*That will be a nice one to test out.*0519

*0 + 3 = 3, 0 + 1 =1, 0-1=-1.*0522

*One last test value we need something larger than 1.*0532

*I will put 2 into all of these.*0536

*2 + 3= 5, 2 + 1 =3, 2 -1 =1*0539

*My chart here is keeping track of the individual sign of all the factors.*0547

*We will look at this column by column, so we can see how they all package up for our original rational expression here.*0553

*In this first column I have a negative and it is being divided by a couple of other negative parts.*0561

*What I’m thinking of visually in my mind here is that this will look a lot like the following.*0572

*I have a negative value on top and a couple of negative values in the bottom.*0578

*When those negative values in the bottom combine they will give us another positive value.*0583

*We have a negative divided by a positive.*0591

*That means the overall result of that first interval is going to be negative.*0595

*In the first interval it is a negative.*0603

*For the next one I will have a positive ÷ negative × negative and that will end up positive.*0607

*Then I will have a positive ÷ positive × negative = negative.*0616

*My last I will have positive ÷ couple of other positive values, everything in there is positive.*0623

*I know what my rational expression is doing on each of the intervals.*0630

*I know when it is positive, I know when it is negative and things are looking pretty good.*0634

*Looking at our rational inequality over here I can see that I'm interested in the values that are greater than or equal to 0. *0639

*That means I want to figure out what are the positive intervals.*0647

*In this chart that we have been keeping track of all that so I can actually see where it is positive. *0652

*I have between -3 and - 1. *0659

*I have from 1, all the way up to infinity. *0662

*Both of those would be some positive values.*0665

*Let us write down those intervals.*0669

*I'm looking at between -3 and -1 and from 1 all the way up to infinity.*0670

*There is one last thing that we need to be careful of what endpoints should I include, which endpoints should I not?*0679

*Notice how we are dealing with or equals to. *0688

*I want to include places where the top of my rational expression could have been 0.*0695

*It is a good thing I highlighted it early on, it fact they are right here. *0700

*I know that the top will be equals 0 at -3.*0703

*I'm going to include that in my intervals.*0708

*We never include spots where the bottom is 0. *0712

*We have marked those out and I will use parentheses to show that those should not be included.*0716

*Let us go ahead and put our little union symbol so we can connect those two intervals.*0723

*This interval from -3 all the way up to -1 and -1 to infinity *0728

*that would be our interval that represents the solution for the rational inequality.*0734

*Let us try one that looks fairly small, but actually has a lot involved in it.*0743

*This one is x -5 / x -10 is less than or equal to 3.*0748

*Like before let us get everything over on to one side first. *0754

*That way it is in relation to 0.*0757

*x -5 / x -10 -3 is less than or equal to 0.*0760

*We need to worry about getting that common denominator.*0773

*I see I have x -10 in the denominator, let us give that x -10 to the top and bottom of our 3.*0776

*Looking much better from now, we will go ahead and combine these rational expressions right here.*0799

*I think we need to do a little bit of work, let us go ahead and distribute this -3, as long as we have it there.*0807

*x -5 - 3x + 30 / x – 10 you want to know where is that less than or equal to 0.*0813

*Just a few things that we can combine we will go ahead and do so.*0827

*-2x from combining our x and let us see a 25 will be from combining 5 and 30.*0832

*We want to know where is that less than or equal to 0.*0850

*We are in good shape so far.*0855

*Now that we have crunched it down into one rational expression, we are interested in where is it equal to 0. *0857

*This will be equal to 0 wherever the top is equal to 0.*0865

*We are going to do a little bit of work with this one, but we can figure out where the top is equal to 0.*0869

*We will have to move the 25 to the other side and divide both sides by -2.*0874

*I have 25/2.*0881

*Let us make a little note is where the top is equal to 0. *0885

*Okay, that looks pretty good.*0890

*Let us figure out where the bottom equal 0, that is not so bad that will be x = 10.*0892

*It is around those two values that we will go ahead and check to see what the sign is, around 10 and 25/2.*0903

*Let us go to our table here.*0915

*The smaller of these values would actually be our 10 and 25/2.*0922

*We will put that over there. *0929

*Let us check the sign and see how these turnouts.*0933

*- 2x + 25 and the other one would be x – 10.*0936

*Let us grab some sort of test value that is less than 10. *0947

*One thing I can see here is I will plug in a 0 that is less than 10.*0952

*If I plug it in the first one I will get 25.*0958

*If I plug in 0 for the other one I will get -10.*0961

*That is my sign for that one.*0965

*On to the middle interval I need something between 10 and 25/2*0969

*I think 11 will work out just fine.*0974

*If I put 11 into that top one I will get -22 + 25 that will be positive.*0979

*If I put 11 into the bottom one, 11 - 10 would be 1. *0987

*Okay, interesting.*0994

*I need to pick something larger than 25 / 2.*0997

*25/2 is about 12 ½ let us go ahead and test out something like 20.*1002

*If I put 20 into the top one I will have -2 × 20 that will be -40 + 25 that will give me a negative value. *1010

*If I put 20 into the bottom one it will be 20 – 10 and that will be a 10.*1019

*These two things are being divided, to look at our over all sign we will take each of these signs and go ahead and divide them.*1026

*Positive ÷ negative = negative, positive ÷ positive = positive and negative ÷ positive = negative.*1035

*What am I interested in for this particular one I want to know where it will be less than or equal to 0. *1046

*Let us look for those negative values. *1053

*I can see one interval here everything less than 10 and I have another interval here where it is greater than 25/2.*1057

*Let us write down those intervals.*1066

*From negative infinity up to 10, from 25/2 all the way up to the other infinity.*1067

*One last thing, we need to figure out what end points we should include and which ones we should not include.*1076

*This one does have or equals 2 so I will include the places where the top is equal to 0.*1082

*We already marked those out, the top is equal to 0 at 25/2 so we will include it. *1089

*We never include spots where the bottom is equal to 0, so I will not include the 10.*1097

*We will finish off by putting a little union symbol in between. *1104

*The intervals that represent the solution are from negative infinity up to 10 and from 25/2 all the way up to infinity. *1108

*And then we can consider this inequality solved.*1118

*It is a lengthy process, but we will use that table to help you organize your information *1121

*and watch at what point you can and cannot include in the solution intervals. *1126

*Thank you for watching www.educator.com.*1132

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at applications of rational equations.*0002

*In some of the examples I have cooked up we will look at how some examples involves a just numbers.*0009

*There are ones that involve motion and some of my favorite ones that involve work.*0015

*This one is going to be a little bit trickier to think about, but once you get the process done *0020

*you will see that the work problems are not so bad.*0024

*Depending on the unknowns in the problem and depending on how we should go ahead and package everything up *0031

*there could be a few situations that actually lead to rational expressions. *0037

*Think about those fractions or where an unknown ends up on the denominator. *0041

*What we want to do is be able to solve these using many of the techniques that we have picked up for rational equations *0050

*and see how we can recognize these in various different situations, such as numbers, motion, and especially work.*0056

*The work problems are kind of unique and that you want to know how to represent the situation. *0067

*If you know how long it takes to complete an entire job then you know the rate of work is given by the following formula 1/T.*0075

*The way you can read this is by using that T for the amount it takes to do that one job.*0084

*Here is a quick example. *0090

*Let us say it takes Betty 7 hours to paint her entire room. *0092

*Well, that means that every single hour 1/7 of the room is going to be painted.*0098

*We are going to make a little chart for just keeping track of everything.*0103

*Maybe x will be my time and let us say how much of the room has been painted so far.*0109

*One hour, two hours, three hours and make sure you jump all the way to 7.*0120

*After she is in the room for 7 hours, she will have the entire room painted, the whole thing.*0127

*If you scale it back, what if she is only working for one hour then only 1/7 of the room is painted.*0135

*If she is in there for two hours, 2/7 of the room is painted and if she is in there for three 3/7.*0144

*You can see that we are just incrementing this thing by exactly 1/7 every time.*0151

*We can make some adjustments to our formula here and say that T would be the amount it takes to do that one for the person *0156

*and may be multiply it by x, that would represent how long they have been doing that particular job.*0165

*Watch for that to play a key component with our work problem in just a bit.*0173

*Let us first see our example of numbers and just see something where our variable ends up in the denominator.*0180

*In this one we have a certain number and we are going to add it to the numerator and subtract it from the denominator of 7/3.*0186

*The result equals the reciprocal of 7/3 and we are interested in finding that number.*0194

*Let us first write down our unknown. *0202

*x is the unknown number.*0205

*We construct a model situation here.*0217

*Take that unknown number and add it to the numerator, but subtract it from the denominator of 7/3.*0219

*Here is 7/3, so we are adding it to the numerator and subtract it from the denominator.*0227

*The result equals the reciprocal of 7/3, that is like 7/3 but we flipped it over.*0238

*This will be our rational inequality here.*0245

*What we have to do is work on solving it.*0249

*To solve many of our rational equalities we work on finding a least common denominator *0253

*which I can see for this one will be 3 - x and 7.*0261

*Let us give that missing piece to each of the fractions.*0268

*Here is my original (x + x) (3 – x) = 3/7 let us use some extra space in here.*0272

*The one on the left, it could use 7, let us put that in there.*0286

*The one on the right it is missing the 3 – x.*0294

*At that point, the denominators will be exactly the same.*0302

*We will just go ahead and focus on the tops of each of these.*0307

*7 × 7 + x = 3, 3 – x.*0312

*Continuing on and solving for our x we will go ahead and distribute our 7 and 3.*0321

*That will give us 49 7x = 9 - 3x.*0328

*Moving along pretty good. *0336

*Let us go ahead and add 3x to both sides giving us a 49 + 10x = 9.*0338

*We will go ahead and subtract 49 from both sides.*0351

*I have x = 10x is equal to -40.*0359

*Dividing both sides by 10, I have that x = -4.*0368

*Just like when we are working with equations, it has to make sense in our original.*0374

*Looking at the original rational expression I have a restricted value of 3 and I know that it is not 3 that will make the bottom 0.*0379

*I do not get any restricted values from the other faction because it is simply always 7 on the bottom.*0388

*Since the -4 is not 3, I'm going to keep it as a valid solution.*0394

*That one looks good.*0402

*Let us look at one that involves motion.*0407

*We will set these up using a table and also use that same idea to help us organize this information.*0409

*A boat can go 10 miles against a current in the same time it can go 30 miles with the current.*0416

*The current flows at 4 mph, find the speed of the boat with no current.*0422

*We have an interesting situation.*0428

*We have a boat that looks something like this and we have the flow of the river. *0430

*Now in one situation, it is fighting against the current, and the way you want to think of that in relation to its speed *0439

*is that the speed of the river is taking away some of the speed of the boat.*0447

*You will see a subtraction process.*0452

*If the boat is going in the same direction of the river, they are both helping each other out *0456

*and you will see an addition problem with both of their speed.*0461

*You will know they are both helping each other out.*0464

*Let us see if we organize this information so I can see how to connect it.*0468

*We need to think of two different situations.*0492

*We are either going against, or we are going with the river.*0495

*We will look at the rate, the time, and the distance.*0502

*This will help us keep track of everything. *0511

*And of course we are leaving unknown in here.*0513

*Since we are finding the speed of the boat with no current, let us set that as our unknown.*0516

*x is the speed of the boat with no current.*0522

*I think we have a good set up and we can start organizing our information.*0540

*In the first bit of this problem we know that it can go 10 miles where it is going against the current.*0545

*That is its distance.*0553

*It went 10 miles when it is going against the current.*0554

*It can do that in the same time it can go 30 miles with the current.*0558

*A little bit of different information, this one will be 30 when it is going that way.*0564

*The current of the river flows at 4 mph. *0570

*If we are looking at the speed of the boat and it is going against that river, probably this will be the boat - the current.*0575

*If we are looking at it going with the river, that will be the speed of the boat + the speed of the current.*0585

*They are helping each other out.*0590

*The only thing we do not know in here is the time, but I do know that the time was exactly the same for both of these situations.*0593

*Let us see what do we got here.*0603

*I will go ahead and create an equation for each of these.*0604

*x -4 × time = 10 and x + 4 × time = 30.*0607

*I know the times are exactly the same for each of these, let us solve them both for time.*0620

*This one I will go ahead and divide both sides by x - 4.*0628

*In this one I will divide by x + 4.*0631

*And I'm ready to develop that rational equation. *0638

*I will set each of these equal to each other since the times are equal to each other.*0641

*I have a rational equation then I can go ahead and try and solve.*0651

*10 ÷ x – 4 = 30 ÷ x + 4*0655

*To get through solving process, we find our common denominator.*0661

*I'm going to give x + 4 on the left side here and I will give x - 4 to the other side.*0668

*We will note that we made the denominators exactly the same.*0687

*We just need to focus on the tops of each of these.*0691

*Continuing on, you will distribute 10 and we will distribute 30.*0702

*10x + 40 = 30x -120*0710

*Subtracting a 10x from both sides will give us 40 = 20x -120 and let us go ahead and add 120 to both sides.*0720

*160 = 20*0739

*We can divide both sides by 20 and I have that x is equal to 8.*0745

*Let us make sure that it makes sense.*0754

*Some restricted values I have for my equations here I know that x cannot equal 4 and -4 *0757

*and fortunately both that we have found for a possible solution is neither of those.*0766

*We can say the speed of the boat in still water would be 8 mph and then this guy is done.*0772

*Do not be afraid to use those tables from earlier to organize your information.*0790

*Joe and Steve operated a small roofing company and Mario can roof an average house alone in 9 hours.*0800

*Al can roof a house alone in 8 hours.*0809

*We want to know how long will it take them to do their job if they work together.*0812

*First we need to figure out the rates of each of them individually.*0818

*Let us go ahead and focus on Joe.*0821

*Joe can roof an average house alone in 9 hours.*0830

*Looking at just Joe we know that every hour he will get 1/9 of that house done.*0834

*Steve over here can roof the house in 8 hours.*0844

*He is working every single hour 1/8 of that house will be done. *0853

*We can put in that time on it.*0858

*Let x be the number of hours.*0861

*You have 1/9 × however many hours they work and 1/8 × qualified by however many hours Steve works.*0871

*We want to know how long it will take them to do the job if they work together.*0881

*We will take each of their work that they are doing and we will add them since they are working together, *0889

*we want to know when they will complete one job.*0895

*We have all of our information here and we can go ahead and try and solve this.*0899

*Our LCD would be 72.*0906

*We will multiply that through on all parts.*0911

*Doing some reducing 72 and 9 = 8, 72 and 8 = 9 and now we have an equation 8x + 9x = 72.*0923

*Combining together what we have on the left side this would be 17x is equal to 72 *0942

*and we can divide both sides of that by 17 to get x = 72 ÷ 17.*0951

*It is looking pretty good.*0959

*That represents how long it will take them to work together.*0961

*If you want to represent that as a decimal, you could go ahead and take 72 ÷ 17.*0964

*When I did that I got about 4.24 hours, I did round it.*0971

*That gives me a better idea of how long it took them when working together.*0980

*Anytime when you are working with these types of problems, it should be less than any one of them working by themselves,*0985

*since they are helping each other out.*0992

*This one looks good.*0995

*In this last example we are going to look at a water tank that has two hoses connected to it.*0998

*Even though this is not a work problem you can see that we can set it up in much the same way.*1003

*Let us go ahead and give it a read.*1008

*The information that we have is that the first hose can fill the entire tank in 5 hours. *1011

*The second hose connected to this tank it can empty it in 3 hours. *1018

*If we start with a completely full tank and then we turned both of them open, *1023

*the question is how long will it take it to empty the entire tank?*1029

*Let us have an accrued picture of what we are dealing with here.*1034

*This would be our water tank and we have one hose that is going in and one hose that is going out. *1039

*The hose that is going in, it can fill the tank in 5 hours.*1054

*That means if we just leave it on every hour 1/5 of that tank will fill up.*1058

*I know its rate is 1/5.*1063

*If I look at emptying the tank it is a much smaller time to empty it, 1/3.*1068

*If I did have them both open I would know that the second one *1076

*would be able to empty the tank since it empties faster than the first one can fill it.*1080

*Let us see what that we can do to set this one out.*1086

*The first thing I want to consider was how much the tank is going to be empty every single hour.*1091

*Starting with the 1/3 I know that every hour that passes by the 1/3 of it will be emptied out.*1097

*x is the number of hours.*1106

*The 1/5 coming in is not emptying the tank but it is actually filling it back up.*1118

*We will say it is the opposite of emptying the tank by 1/5 and it will do that for every hour.*1125

*We want to know is when will one tank be completely empty. *1132

*We have a lot of similar components for this one and it often look like a lot of our work problems.*1137

*All we have to do is go ahead and solve it.*1145

*Our LCD here that would be 15.*1148

*Let us multiply all parts by 15 and see what that does.*1153

*15, 15 and 15.*1159

*Canceling out some extra stuff here I get 5x - 3x = 15.*1163

*Doing a little bit of combining on the left side I have 2x =15 or x = 15 ÷ 2 or 7 ½ hours.*1173

*Notice how in this case it is taking longer since both of them are open. *1190

*The reason for this is that they are not working with each other.*1194

*They are working against each other.*1198

*One was trying to fill the tank, and one is trying to empty the tank.*1199

*Use those bits of information you can find along the word problem to give you a little bit of intuition about your final solution. *1205

*In that way you can be assured that it does make sense in the context of the problem.*1212

*Thank you for watching www.educator.com.*1217

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at a very interesting application of our rational.*0002

*We will look at variation and proportion.*0006

*We will have to do a little bit of work just to explain what I mean by the word variation.*0012

*We will break this down into few other things.*0017

*We will look at direct variation, inverse variation, and combined variation. *0019

*This particular section is filled with lots of different word problems all of them involving variation.*0025

*Let us see what we can do.*0032

*When you hear that word variation we are talking about a connection between two variables, *0036

*such that one is a constant multiple of the other.*0041

*You will see some nice handy formulas that will highlight how one is a constant multiple of the other.*0044

*If you are looking for little bit more intuition on the situation then you can grasp on to that. *0050

*Intuitively a variation can be thought of as a special connection and the two basic types are direct and inverse.*0055

*In this connection when one quantity goes up then the other one would go down.*0063

*In that one I'm talking about an inverse variation.*0069

*If I’m thinking about those two quantities when one goes up and the other one goes up as well, that will be an example of a direct variation. *0075

*It is how one affects the other one and they could be moving in the same direction our going in the opposite directions.*0084

*To model this type of situation that has variation in it, you can end up setting up a proportion.*0094

*Be very careful on where you put each of the individual components and make sure you line them up correctly.*0100

*To model a direct variation, you could use the following proportion x1 / y1 = x2 / y2.*0107

*The way want to interpret those subscripts that you are seeing on everything is that all of the values with a subscript of 1 involved one situation.*0116

*Everything with the 2 involves a second situation.*0126

*Since the x are both on the top that will be from one particular type of thing and y will be from another type of things.*0133

*We will line those up to make sure that they at least agree.*0142

*It seems a little vague but let us get into an example *0148

*and you will see how I do set this up using a proportion and how everything does line up.*0151

*This one says that an objects weight on the moon varies directly as its weight on earth.*0156

*Neil Armstrong, the first person to step on the mood weighed 360 pounds on Earth, when he is on the moon he was only 60 pounds. *0163

*The question is if we take just an average person who weighs 186 pounds, what will their weight be if they go to the moon?*0174

*We have that this is a direct variation because it pointed out and says it is a direct variation.*0186

*Let us just think a little bit and see if that makes sense.*0192

*Intuitively in a direct variation when one quantity goes up then should be the other one.*0196

*When one quantity goes down, then so should the other one.*0201

*That is what is going on with our moon weight.*0204

*If I decide to go on and get heavier and heavier on Earth, that same thing is going to happen on the moon.*0207

*I will get heavier and heavier on the moon. *0214

*This is a type of direct proportion.*0217

*Let us see what we can set up.*0223

*x1 / y1 = x2 / y2*0224

*To help out I’m going to highlight two things, we will go ahead and keep our Earth weights on top *0231

*and we will go ahead and put our moon weights on the bottom. *0238

*Let us look at our first situation with our Neil Armstrong.*0245

*He weighed 360 pounds on Earth, and only 60 pounds on the moon.*0250

*360 on Earth and 60 on the moon.*0255

*I will compare that with our average person there.*0267

*They weigh 186 pounds on the Earth, but we have no idea how much they weigh on the moon.*0270

*I'm going to put that as my unknown.*0277

*Everything on the tops of these fractions represents an earth weight and everything on the bottom represents a moon weight.*0280

*You can see that we have developed one of these rational equations and now we simply have to solve it.*0287

*This one is not too bad if you go ahead and reduce that fraction on the left, it goes in there 6 times.*0292

*We can multiply both sides by x.*0301

*I have 6x = 186. *0308

*Finally dividing both sides by 6 I will get that x =31.*0316

*This reveals how much that average person, the 186 pounds would weigh when they are on the moon.*0327

*There are lots of other types of situations that could involve a direct proportion and they might not come out and say its direct proportion.*0338

*Look for clues along the way to help out.*0344

*This one says that a maintenance bill for a shopping center containing 270,000 ft*^{2} is $45,000.0348

*What is the bill of the first store in the center that is only 4800 ft*^{2}?0356

*Let us think of what type of a variation this should be.*0361

*If I have a large store as a part of the shopping mall and I will have to end up paying a much larger maintenance or bill because of my larger store. *0366

*If I have a small store then my bill should be very small.*0378

*Those quantities are moving in the same direction. *0382

*I’m going to say this is an example of direct variation.*0385

*We will set it up using our proportion.*0392

*We want to keep things straight. *0403

*What do each of these quantities represent?*0405

*Let us say the top will be our square footage and we will make the bottom the bill.*0409

*How much it cost?*0421

*Let us go ahead and substitute some of this information we have. *0424

*If I’m looking at the entire shopping center I know how big it is and I know the bill for the entire place.*0427

*270,000, 45,000 and now I can look at my much smaller store, it only has 4800 ft*^{2} 0435

*and this one we have no idea what the bill is so I will leave that one as my unknown.*0453

*I think we can do a little bit of simplifying with this one as well. *0459

*I will divide these two and we get 6 and I need to just solve this rational equation which is not too bad as long as we multiply both sides by x.*0464

*I have 6x = 4800. *0481

*Divide both sides by 6 and then I think we will have our solution x =800.*0488

*Even if they do not come out and tell you what type of variation is going on here, *0497

*look at the values present to see if things are moving in the same direction, or in opposite directions.*0502

*For inverse variation we want to make sure that if one quantity goes up, the other one goes down.*0510

*One way that we can model it now with our proportions is to set up like this.*0516

*We will go ahead and use us a subscript for one situation and we will put them on opposite sides of our equal sign.*0521

*We will get that inverse relationship. *0528

*When one goes up, the other one will go down.*0531

*For our second situation we will put these on two opposite sides of the equal sign.*0535

*One thing that will not change is that both of these x will represent the same type of thing *0541

*and both of these y will still represent the same type of thing.*0546

*We can definitely note that the things are on opposite sides of the equal sign for these given situations.*0550

*For my inverse variation problem I will look at the current in a circuit and we are told that it varies inversely with the resistance. *0562

*If I have a current that is 30 amps when the resistance is 5 ohms, find the current for resistance of 25 ohms.*0572

*I have a lot of things in here and we just want to keep track of each of them.*0581

*Let us see if we can highlight each of these situations. *0586

*The current is 30 amps when the resistance is 5.*0589

*We will be looking for the current when the resistance is 25.*0596

*Let us get our proportion out here.*0605

*I’m using these little subscripts and put them on opposite sides.*0611

*Onto the first situation, we will keep everything on the top. *0615

*Let us say that is our resistance, we will put everything on the bottom as the current.*0624

*Our first situation, we know that the current is going to be 30 when the resistance is 5.*0635

*In that situation I will put my 5 and 30.*0641

*For the other situation I want to figure out what the current is.*0647

*I have no idea what that is but I know that the resistance is 25 ohms.*0651

*This is my initial proportion that I have set up.*0657

*All we have to do is go through the process of solving it.*0661

*This one is not too bad. *0664

*Let us multiply both sides by 30x and see what is left over.*0666

*30 × 5, since the x will cancel out equals 25 × x since the 30 cancel out.*0673

*Okay looking pretty good. *0688

*Also 5 × 30 =150 equals 25x and we can simply divide both sides by 25 to get our answer. *0690

*The current is 6 amps.*0708

*When dealing with these variation problems always look at your solution and see if it makes sense in the context of the problem. *0722

*One thing that I'm noticing in here is that I start off with a resistance of 5 and a current of 30.*0729

*One thing I'm doing with my current is that I'm lowering it, and I'm taking it from 30 down to 25.*0736

*Since I have an inverse relationship that is going to have an effect on our amps is the current.*0744

*That should bring it up.*0751

*Originally I started with 5 and I can see my answer is 6.*0753

*It did go up and things are working in the right direction.*0757

*Since earlier, we said that it variation is relationship when one is a constant multiple of the other.*0764

*We can model this using some nice formulas.*0769

*For direct variation you can use the formula y = k × x.*0773

*For your inverse variation, you could use y = k ÷ x.*0779

*We have two variables in each of these equations that we are looking at connecting are the x and y.*0785

*The k in each of these is known as our constant of proportionality.*0791

*It connects how they are related. *0795

*A lot of problems that you will end up doing with these variations, if you want to use these formulas *0799

*then you will end up going through three steps to figure out what is going on. *0806

*You might first go ahead and identify what type of variation you are using so that you can take one of these formulas.*0811

*Then you will end up using a little bit of known information to go ahead and solve for that constant of proportionality, k.*0818

*Once you know what k is, then you can usually figure out some new information by substituting it into the formula.*0825

*I got a couple of examples where we are going through these 3 steps.*0832

*Let us first just see an example of using an inverse of variation using one of these formulas.*0840

*This one says that the speed of water through a hose is inversely proportional to a cross section of the area of the hose. *0846

*If a person places their thumb on the end of the hose and decreases the area by 75%, what does this do the speed of the water?*0855

*Let us think about what we got here.*0867

*We are looking for a connection among the cross area section of the hose and the speed of the water.*0868

*We are told that they are set up inversely so when one goes down, the other one should go up.*0876

*For decreasing the area then we should expect it to increase the speed of the water.*0879

*In fact we can be a little bit more precise, but despite playing around with those formulas for a little bit. *0887

*Our relationship is that the speed of the water is inversely proportional to a cross section of the area.*0893

*Maybe x = k/a.*0900

*I'm going to decrease the area by 75%, one way that I could model that is by simply multiplying our area by what is left over, 25%.*0905

*Let us relate this to the original before I manipulated it.*0923

*25 is the same as ¼ I'm dividing by ¼.*0928

*When you divide by a fraction that is the same as multiplying.*0936

*This is multiply by 4.*0943

*Now we can see the difference between the original situation.*0947

*The area decreased by 75%.*0956

*They are almost exactly the same but in the second situation, it is 4 times as much.*0966

*It will actually increase the speed by 4 times.*0971

*There are lots of other types of variations that you can go ahead and work into a problem.*0978

*Some of these will look similar to previous types that we covered.*0982

*Let us go through how each of these are connected. *0984

*If I say that y varies directly to some power of x then there are some k that connects those two things. *0990

*You will notice that it looks like our direct proportion only since we are dealing with an nth power, *0997

*like a 3rd power 4th power that we put that exponent right on the other variable.*1004

*Our y varies inversely as the nth power looks much like our inverse relationship as well.*1013

*But since we are doing it to the nth power, notice how we put that exponent right on the other variable.*1020

*In this one is a new one that involves a few more variables, this is our joint variation.*1029

*y varies jointly as x and z so more than one variable now.*1034

*If there some sort of positive constant such that y = k × x × z.*1039

*In this one the k is our constant and the variables are being connected, the y, x, and z.*1046

*You can start mixing and matching these various different types of variation.*1057

*The reason why that we do not want to start mixing and matching them is we can get some more complicated models.*1064

*With these more complicated models we are hoping to be able to more accurately model what is happening in a real life situation.*1071

*Usually it does a better job than any one variation could do by itself. *1079

*The key when putting all of those different types of variations together is to look for clues in the actual problem itself.*1085

*Here is a formula that I have it is T = k × x × z*^{3} /√y .1092

*Here is how I could describe that problem using the language of these variations. *1100

*The k right here is my constant of proportionality and we are looking to get it connect together the t, x, z, and y.*1107

*Here is what I can say, the t varies jointly with x and z*^{3}.1116

*It is telling me that I have my constant of proportionality but the x and z should end up in the top and the z should be cubed.*1123

*t also varies inversely with the √y.*1133

*Since it is an inverse relationship I will put that one in the bottom and make sure to put in that square root.*1138

*Watch for those keywords so we can figure out where we should put things.*1145

*Let us try some of these variation problems and watch as I walk through the three steps of *1151

*starting with some template solving for our constant of proportionality, k and the last part figure out some new information. *1155

*What do I know so far?*1167

*I know that y varies jointly as x and z.*1168

*Just on that little part, let us go ahead and make a nice formula. *1173

*y varies jointly as x and z.*1177

*y = 12 when x = 9 and z = 3.*1184

*I can use that information right there to go ahead and solve for my constant of proportionality. *1191

*y =12, x = 9, z= 3*1198

*The only unknown I have in there is that k value.*1206

*We will combine things I will get 27 divide both sides by 27 and then maybe reduce that by dividing the top and bottom by 3.*1210

*Now that we know much more I can end up revising our formula. *1228

*y = xz and now I can say that the k value is this 4/9.*1236

*Now that we know about our constant of proportionality, let us find z when y = 6 and x = 15.*1244

*We will substitute those directly in there and you will see the only unknown I have left will be that z.*1253

*Y= 6, 4/9 × 15 and we are not sure what z is now we have to solve for it.*1259

*A little bit of work.*1270

*I think I can cancel out an extra 3 here.*1272

*I have 6 = 20/3 × z we could multiply both sides by 3.*1276

*I will get rid of those 3, 18 = 20z.*1292

*Lastly let us go and divide both sides by 20. *1298

*18 ÷ 20 =z*1301

*We will just go ahead and put that into lowest terms.*1306

*Divide the top and bottom by 2 and I know that 9/10 is equal to z.*1310

*If you follow along we start with some sort of template for our variation, we solved for our unknown constant of proportionality k.*1316

*We have used it to figure out some new information about the other variables.*1325

*For this last type of example let us look at one that is a little bit more of a word problem.*1331

*This one says that we have the maximum load of the beam and how much you can support varies jointly *1336

*with the width of the beam and the square of its depth, but it varies inversely as the length of the beam.*1343

*Some of that initial information there is just giving us a formula for how everything is connected.*1352

*We will definitely set up the equation that will model that situation.*1358

*Let us continue on.*1364

*Assume that a beam that is 3 feet wide, 2 feet deep, 30 feet long and it can support 84 pounds.*1365

*What is the maximum load a similar beam can support if it is 2 feet wide, 5 feet deep and 100 feet long?*1371

*A lot of information is floating around there.*1378

*Let us start off by seeing if we can set up this formula.*1381

*The maximum load that a beam can support varies jointly as the width of the beam and the square of its depth.*1385

*Since this is a jointly, we are talking about k × width × the square of its depth or d*^{2}.1395

*It varies inversely with the length, I will put the length on the bottom.*1408

*This part here is our formula that we will end up using.*1413

*Assuming that a beam is 3 feet wide, 2 feet deep and 30 feet long, and it can support 84 pounds, *1421

*let us use that to figure out our unknown constant sitting right here that k.*1426

*I know it can support 84 pounds, we do not know k.*1432

*3 feet wide, I put that in for width, 2 feet deep and it is 30 long.*1440

*The only unknown is that k.*1448

*Let us go ahead and simplify it just little bit in here.*1451

*I have 3 × 2*^{2} and 2^{2} is 4 = 121455

*I can get that k a little bit more by itself if I multiply both sides by 30, 84 × 30.*1463

*Let us go ahead and put those together and see that would be 2520.*1473

*Divide both sides by 12.*1483

*That is where we have been able to figure out what that unknown constant is.*1488

*Now that I have that I can go back to the original formula.*1494

*M = 210w × d*^{2} / l1498

*We will use this new formula to figure out a little bit of known information.*1506

*We will look at it for the other beam where we are looking for its maximum load if we know its width, depth, and its length.*1512

*Let us give it a try.*1520

*That formula we developed was its maximum load = 210 × width × d*^{2} / length.1525

*Let us put in our information.*1536

*For this new beam, its width is 2 feet and its depth is 5. *1541

*It also has a length of 100. *1549

*As soon as we simplify all that we will be able to figure out what its maximum load should be.*1551

*210 × 2 = 420 × 25 ÷ 100*1558

*Let us do some simplifying here 420 × 25 = 10500 ÷ 100 and looks like this reduces to 105.*1573

*This new beam can support a maximum of 105 pounds.*1593

*Being able to recognize various different types of variation *1599

*and combine them all together is essential for some more complicated problems.*1602

*Remember that if you have those nice formulas then you can go through 3 steps.*1607

*One, build a formula or equation that models the situation, solve for your unknown k *1611

*and finally use some new information to figure out some more information about the problem. *1617

*Thank you for watching www.educator.com.*1622

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at rational exponents. *0002

*The neat part about rational exponents is you will see that we will develop a way and connect them to our radicals.*0011

*You will see that there are a few rules for working with these rational exponents and a lot of them come for just our rules for exponents.*0019

*We will look at a few ways that you can combine terms that have some these rational exponents on them.*0027

*We have seen many different types of radicals. *0034

*We have seen exponents but there is actually a great connection between the two. *0039

*If you have a radical of some index like a square root or third root and it is raised to a power, *0044

*you can write this in one of the following two ways.*0051

*You can write it as the root of that expression ratio power or you can write it as the expression raised to a fractional power. *0054

*One thing to notice were the location of everything has gone to.*0066

*The power in each of these problems I have marked that off as a that shows up on the top of the fraction. *0070

*The root is going to be the bottom of the fraction.*0083

*You can take any type of radical and end up rewriting it as a fractional exponent or as a rational exponent.*0089

*To get some quick practice with this let us try some examples.*0098

*I have 36^½, -27*^{1}/3, 1^{3}/2 and -9^{3}/2.0102

*We are going to evaluate these up by turning them into a radicals. *0110

*Okay, so this first one, the way we interpret that is that I'm going to put it under a root with an index of two.*0116

*This is like looking at the √ 36, which is of course just 6.*0123

*For the other one, if I see -27*^{1}/3 that will be like taking -27 to the 3rd root so this one is a -3.0132

*In the next one notice how we have a power and a root to deal with. *0146

*81*^{3}/20149

*There are two ways you could look at this.*0153

*You could say this is 81*^{3} and we are going to take the square root or you can take the square root of 81, 0155

*then raise the result to the third power.*0163

*Both of them would be correct, but I suggest going with the one that is a lot easier to evaluate.*0166

*I’m thinking of this one. *0171

*If you take 81 and raise it to the third power we are going to get something very large *0173

*and try to figure out what is the square of that is going to be a little difficult.*0177

*But look at the one on the right, I can figure out what the √ 81 is, I get 9.*0180

*We can go ahead and take 9*^{3} it would be 729.0189

*One last one, I have 9*^{3}/2 and a negative sign out front.0201

*Let us first write that as the √9*^{3} as for this negative sign it is still going to be out front.0206

*I have not touched it.*0215

*I'm not including that in everything because there is no parenthesis around the -9 in the original problem.*0217

*I’m starting to simplify this.*0224

*The √9 would be 3 then I will take 3*^{3} and get -27.0226

*In all of these situations I'm looking at the top of that fraction make it a power, we get the bottom to see what the root needs to be. *0236

*The good part about taking all of our radicals and writing them using these exponents, *0247

*it means that we can use a lot of our tools that we have already developed for exponents and we have done quite a bit of them.*0252

*In fact a quick review, we have a product rule for exponents, a power rule *0258

*and we have gotten different ways that we could go ahead and combine them. *0264

*We also have rules on how to deal with fractions like adding subtractions, subtracting fractions. *0269

*We have our zero exponent rule, our quotient rule and negative exponent rules. *0273

*All of these rules will help us when working with our radicals.*0278

*Just watch on what our base is and what we need to do from there. *0284

*Now using some of these rules if you do have radicals you might be able to combine them under a single root.*0293

*That involves using a common denominator most of the time.*0305

*You can use this tool for working with rational expressions.*0311

*Watch how I find the common denominator for some of my problems and actually get everything under one root.*0315

*Let us do these guys a try.*0323

*64/27*^{-4}/3 0325

*Before I get to that 4/3 part I’m going to apply some of my other rules for exponents and specifically that negative exponent. *0329

*One way I can treat a negative exponent is it will change the location of the things that it is attached to.*0339

*I’m going to write this as (27/64)*^{4}/3.0344

*I’m going to give that 4/3 to the top and to the bottom that is using my quotient rule.*0352

*I will end up rewriting the top and bottom using my radicals. *0362

*I’m looking at the 3rd root of 27 and we will take the 4th power of that *0369

*and then we will take the third root of 64 and take the 4th power of that one.*0376

*Let us see what this gives us.*0390

*On top the 3rd root of 27 that will be 3 and 3rd of 64 =4 and both of these are still raised to the 4th power.*0393

*We will go ahead and take care of that multiplication.*0403

*81 / 256*0407

*Being able to take it and write it as radicals, it meshes well with all the rest of our rules.*0413

*Let us use the quotient rule on this next one.*0421

*4*^{7}/4 ÷ 4^{5}/40423

*I need to subtract my exponents.*0429

*Good thing both of these have the same denominator this will simply be 4*^{2}/4.0434

*That continues to simplify this would be 4^½ which written as a radical is the √4 which all simplifies down to 2.*0442

*Be very comfortable with switching back and forth between those rational exponents and your radicals.*0451

*On to ones that are a little bit more difficult. *0461

*These will involve trying to simplify much larger expression and some of the terms will have those rational exponents. *0464

*Okay, so here I have (r^¼ y*^{5}/7)^{28} ÷ r^{5}.0474

*I can apply my 28 to both of the parts on the insides since they are being multiplied.*0486

*Let us see what this looks like.*0493

*28 × 1/4 = 28/4 and I have y*^{5} × 28 ÷ 7 and all of that is being divided by r^{5}.0495

*Let us see if we can simplify some of those fractions.*0512

*How may times this 4 go into 28? 7 times and 7 will go into 28 four times.*0515

*This is (r*^{7} y^{4})/r^{5}0527

*We can go ahead and reduce our y.*0534

*5 of them on the bottom and with 5 of them on top that will leave us with an r*^{2} and y^{4}.0536

*Let us try another one.*0548

*This one has a lot of fractions and a lot of negative signs.*0549

*(P*^{-1}/5 q^{-5}/2)/(4^{-1} p^{-2} q^{-1}/5)^{-2}0553

*I’m going to use my rule to apply this -2 to all of my exponents.*0565

*I’m sure that will help get rid of a bunch of different negatives.*0572

*-2 × -1/5 = 2/5, -2 × -5/2 = q*^{5}0576

*Then on to the bottom, -1 × -2 = 2 and p*^{4} = 4, q^{2}/5.0593

*That does simplify it quite a bit.*0610

*At least I can see that this guy right here will be 16 but we will also have to reduce these a little bit more. *0611

*I'm going to take care of these ones I want to think what is 2/5 – 4.*0620

*If we can find a common denominator it will helps out with that ones.*0626

*2/5 - 20/5 we will call that one -18/5, so I know that I will have 18/5 on the bottom.*0629

*Let us try it out.*0646

*I have a 16 on the bottom and now I discovered I have a p2*^{18}/5 on the bottom as well.0648

*These ones we can reduce. *0656

*I want think of 5-2/5.*0659

*The common denominator there will be 5, 25/5 - 2/5 = 23/5 we will put that on top q*^{23}/5.0663

*We have our final simplified expression. *0679

*In this next two I have some radicals and we will go ahead and write them as our rational and see what we can do from there.*0686

*The top, this would be y*^{2}/3 ÷ y^{2}/50696

*If I’m going to end up simplifying using our quotient rule, we will look at this as 2/3 – 2/5.*0705

*We need a common denominator on those fractions to put them together.*0713

*Let us look at this as over 15.*0721

*10/15 – 6/15*0726

*This would give us y*^{4}/15.0731

*Then we can continue writing this as a radical if I want it to be the 15th root y*^{4}.0735

*That is what I was talking about earlier about being able to combine these radicals into a single radical.*0744

*Let us try this other one.*0752

*I have z and I’m looking for the 5th root of it, I will write that as z*^{1}/5.0753

*When I’m taking the 3rd root of all of that, that is like to the 1/3.*0761

*My rule for combining exponents in this way says I need to multiply the two together.*0769

*Z*^{1}/150774

*If I will write this one as a radical them I’m looking at the 15th root of z.*0779

*There are many of the different examples that we can get more familiar with using these radicals and rational exponents.*0786

*Let us do one where we work on writing it using these radicals.*0796

*I’m looking at the 8th root of the entire 6z*^{5} – 7th root 5m^{4}0805

*Be careful when it comes to trying to combine things any further from there.*0824

*We have not covered yet on what to do with addition.*0828

*Most of our rules cover our rules for exponents you got to be careful on what you do with addition.*0834

*I’m going to leave this one just as it is and not work on combining.*0843

*One problem that I will have is that my bases are not the same in anyway.*0849

*This is good as it is.*0855

*Be very more familiar with taking the radicals and turning them into rational exponents.*0858

*Remember that the top will represent the power and the bottom of those fractions will represent the index of the radical.*0864

*Thank you for watching www.educator.com.*0871

*Welcome back to www.educator.com.*0000

*In this lesson we are going to look at simplifying rational exponents.*0003

*The things we will look at are our product rule, quotient rule and what to do when they involve things like radicals.*0009

*We will also look at a few radicals that involve variables and how can we get into simplifying much higher roots.*0018

*If you have two non negative real numbers call them A and B.*0028

*And what you can do with them when they are being multiplied underneath a radical is put them under the same radical.*0033

*If those are already underneath the same radical and you are also free to go and separate them back out. *0041

*They are each underneath their own radical.*0047

*This rule right here is good for either simplifying or combining things.*0050

*You can use it in both direction, combine or break apart. *0055

*One very important thing to note is that this rule only works for multiplication, *0060

*do not try and use any type of splitting thing if you have addition or subtraction.*0065

*Where the simplifying part comes into play is by taking a very large number, large expression and breaking it up into those individual parts.*0075

*We can consider a radical simplified when there is no perfect square under the radical anymore.*0085

*Look at these different radicals, find the perfect squares that are in there and split up the roots over those perfect squares.*0093

*In a similar rule, we have that we can either combine things under radical when we are dealing with division *0106

*or if they are both under division, we can split up over the top and over the bottom.*0112

*This works as long as the value on the bottom is not equal to 0. *0118

*Of course that is because we cannot divide by 0. *0122

*Feel free to use this rule in both directions either combine or separate them.*0127

*Note that this rule only works for division.*0131

*Again, do not try and use this for addition or subtraction. *0133

*If we happen to have a variable underneath the root and it is raised to the power, *0140

*we can assume that variable represents a non negative number.*0143

*That will make sure that we are dealing with real numbers and not imaginary numbers.*0149

*The product in the quotient rule that we have just covered apply when all the variables up here under a radical sign as well. *0155

*We are dealing with non negative real numbers with these various different rules.*0162

*You can even use these rules for some much higher roots and when you are looking to simplify those it depends on the index. *0171

*For example, if you are looking to simplify a cubed root, then what we are doing is *0183

*we are trying to take out all the cubes that are underneath that root. *0187

*Here is a quick example. *0190

*Maybe we are looking at the 3√a*^{3} without simplifying it to just an (a).0192

*In cube roots this happens to be true whether (a) is positive or negative.*0198

*As long as you are dealing with square roots, the square root of the square number is always non negative. *0204

*The product and quotient will be applied to both our higher roots as well as just square roots.*0212

*Let us practice a few of these rules with some example questions. *0221

*If you want to simplify some numbers like the √4/49 feel free to split it up over the top and or the bottom.*0225

*And then you can take each of these separately. *0234

*This would be 2/7.*0237

*Here us one where they already have a root over the top and the bottom but I cannot simplify them individually. *0245

*I’m going to use that the quotient rule in the other direction and we will put them underneath 1 root.*0251

*From there I can figure out how many times 3 goes into 48.*0259

*I have that 3 goes in the 48 at least 10 times. *0269

*I have 18 left over, that would be 3 × 6.*0277

*By combining them underneath one root it allowed to go ahead and simplify the numbers a little bit better.*0291

*I just have to look at √16 is just 4.*0296

*Whether you are putting things underneath the same root or separating them out sometimes they cannot be simplified.*0302

*In this next one I will look at the root of the top and the root of the bottom.*0311

*With the bottom I can take the √36 is just 6 but the √ 5 has to stay as it is because it is not perfect square.*0316

*Let us look at one more.*0331

*You want to look at the √3/8 × √7/2*0335

*With this one I have two different radicals.*0347

*Do not let those fractions distract you we are looking to combine them underneath the same radical.*0351

*I will just combine them together using multiplication.*0359

*Once I do that I could go ahead and continue simplifying from here. *0365

*This will be √21/60 and on the bottom 16 is a square number so I can even go further.*0368

*√21 / √16 will be √21 / 4*0379

*All of these examples are designed to get you more familiar with either taking that root *0386

*and putting over both the parts or combining them into a single root.*0390

*Let us try a few that have some variables in them and notice how the same process works out. *0397

*What I want to do is simplify the √x*^{6} and I could imagine splitting up my x^{6} into lots of x^{2}.0404

*If I use my rule to split up the root over each of those I will have √x*^{2}, √x^{2}, I will have that three different times.0419

*Each of these would simplify to just an x.*0429

*I can go ahead and package that altogether as x*^{3}.0433

*Notice how that fits with some of our other rules such as rewriting this as x*^{6} / 2 which is x^{3}. 0436

*All our rule stays nice and consistent with one another, they both agree.*0448

*Let us try another one.*0454

*We will slip this one up over 100 and over p*^{8}.0457

*I could think of the √100 that is 10 and then for the p*^{8} that could be (p^{4})^{2}.0465

*That square and square root can take care of each other 10p*^{4}.0475

*That one will be good.*0481

*Let us try another one. This one is √7/y*^{4}0485

*I will put the square root on the top and in the bottom.*0492

*We want to look at that y*^{4} as (y^{2})^{2}.0498

*In that way we can see that this square and square root will take care of each other.*0505

*I’m left with the √7/y*^{2}0511

*In this higher root the same rules apply.*0523

*When it gets down to breaking them up and simplifying them, nothing changes.*0526

*We just have to be worried about looking for our cubed numbers or in some of these other examples a number raised to the fourth power.*0531

*Starting with 108, I want to think of this one, I think about a cubed number.*0539

*108 is the same as 4 × 27.*0553

*That is important because 27 is one of my cubic numbers.*0558

*I will split it up over the 4 and 27.*0565

*We can go ahead and simplify that 3rd root of 27.*0570

*That one will be just 3.*0577

*We like to put our numbers first, let me write this one as 3 × 3rd root of 4*0581

*Note how we do not simplify the 4 because it is not a cubic number, it is a square number.*0589

*Let us try the next one, the 4th root of 160.*0597

*I want to think of numbers raised to the fourth power.*0603

*If you want you can even make a list.*0606

*1*^{4} = 1, 2^{4} = 16, this is 4 16 × 10, the 4th root of 16, 4th root of 10.0607

*We will go ahead and simplify the 4th root of 16 is 2, the 4th root of 10*0627

*That one is done.*0633

*Let us use the quotient rule on this last problem.*0640

*On top I will have the 4th root of 16 and in the bottom 4th root of 625*0644

*We saw earlier that the 4th root of 16 we will go ahead and simplify that will be just 2.*0652

*With 625 that is 5*^{4}, 2/5.0658

*Let us do the same thing with higher roots but we will deal with some variables underneath the roots.*0671

*The first one I have 3rd root of z*^{9}.0679

*I could look at this as (z*^{3})^{3}.0685

*What I want to do in this is highlight this root and that 3rd root end up getting rid of each other.*0693

*What is left over is z*^{3}.0700

*Note that this meshes with some of our earlier work and I could write this as z*^{9}/3 and I will still get z^{3}.0705

*All of our rules are staying nice and consistent with one another.*0715

*In here the3rd root of 8x*^{6}.0719

*I’m looking at 3rd root of 8 and 3rd root of x*^{6}.0724

*That will reduce to 2 and this would be like (x*^{2})^{3}.0730

*2 × x*^{2} that will be the final reduced expression.0745

*I have the 3rd root of 54 t*^{5}0753

*In this one I will try to break down as much as possible but remember if we have things that are still not cubic, *0757

*we have to leave them underneath the root.*0763

*Let us see.*0766

*What cubic numbers can I find in 54?*0767

*That will be the same as 27 ×2 and what cubic numbers can I grab from t*^{5}?0772

*That is a t*^{3} and t^{2}.0785

*Notice here I am thinking of my product rule for exponents and how I have to add those exponents together to get 5.*0790

*I’m splitting them up just this way so that I would have one of them as cubed.*0797

*We could take the cubed of everything in here.*0805

*Cube of 27, 2 and t*^{3}, t^{2} 0808

*Some of these will simplify and some of them would not.*0817

*3rd root of 27 is 3, 3rd root of 2 has to stay, 3rd root of t*^{3} is t, and the 3rd root of t^{2} has to stay.0821

*Gathering up what was able to be taken out I will have 3t × 3rd root of 2t*^{2}0837

*Simplify and bring out as much as you can, and if you cannot go ahead and leave them under the root.*0846

*One more, I will first split up the root over the top and on the bottom.*0853

*A*^{15} / 3rd root of 640860

*On the top of this that is (a*^{5})^{3} underneath the cube root.0868

*For the 64 on the bottom that is something that I could have just take the cube root of.*0879

*This would simplify into a*^{5} and 3rd root of 64 is 4.0888

*That one is simplified.*0895

*Be familiar with your rules for these radicals especially when it gets to being able to split things up *0900

*and simplify over each of the individual components.*0907

*Thank you for watching www.educator.com.*0910

*Welcome back to educator.com.*0000

*In this lesson we are going to take a look at adding and subtracting radicals.*0003

*You may notice in a lot of my other lessons I avoid trying to add or subtract radicals as much as possible.*0009

*It is because there is a few problems that you will run into when you try and these radical expressions.*0015

*Look at some of the problems that we want to avoid. *0020

*Do not get into the actual rule for adding and subtracting that way you will know *0023

*what situations that you can add and subtract and which situations you cannot.*0027

*A lot of the other rules that we have picked up for radicals so far we have been doing lots of multiplication and division.*0035

*You will notice that in those rules they follow pretty much exactly from the rules of exponents.*0042

*There is a nice product rule, quotient rule, and they seem to mimic one another.*0048

*To understand some of the difficult things that we run into with adding, subtracting we have to look at what happens *0053

*when some of our rules for adding, subtracting when you have things with exponents. *0060

*For example, I want to put together x*^{2} + 3x what problems when I run into? can I put those two things together or not?0064

*You will notice that you will run into quite a bit of a problem.*0075

*These ones do not have the same exponents, I cannot put them together.*0079

*Since the terms are not like terms and they have different exponents maybe we can change it around and try a different situation. *0084

*Let us go ahead and try ab*^{2} + r^{2}.0100

*Can we put those together? After all the exponents here are exactly the same. *0104

*Can we combine those like terms?*0110

*These one are not like terms.*0114

*We ran into a very similar problem the bases are not the same with these two.*0115

*Those ones are not the same, they do not have the same base then we cannot put them together. *0124

*We usually run into one of those two problems where we are dealing with radicals.*0131

*Either they do not have the same base or they do not have the same exponents. *0135

*Here is a quick example involving radicals so we can see what I’m talking about.*0139

*Here I have 3rd root of x + √3*0142

*If I write those as exponents, then it is like x*^{1}/3 + 3^½.0147

*These have different bases and they have different exponents, there is no reason why you should be able to put these together.*0153

*After seeing many of these different examples, you might that be under the false assumption that we cannot add any radicals together.*0163

*In some instances you will be able to put these radicals together you have to be very careful on certain conditions. *0173

*One, when putting radicals together you have to make sure that their powers, those would be the indexes of each *0182

*of the radicals are the same and you have to make sure that the radicand or the bases are exactly the same.*0187

*I can put together the 3rd root of 5x with the other 3rd root of x.*0194

*It is completely valid because when written as exponents I have the bases the same and they are both raised to the power of 1/3.*0200

*How would I go about actually putting them together?*0210

*I will treat them just almost like an entire variable.*0213

*If I was adding u + u I will get two u.*0216

*Notice how this common pieces here.*0221

*Another way of saying that is, we simply add together our coefficients out front so one of those should equal two of them.*0224

*Make sure that you keep in mind that if you are going to add and subtract radicals *0237

*you must have the same index on those radicals and you must have the same radicand.*0242

*That is the part underneath radicals.*0246

*Once you get to the addition or subtraction process, look at your coefficients out front, so (5 × √x) + (3 × √x) – (6 × √x) .*0248

*I’m looking at 5 + 3 - 6. *0260

*That will give me a result of 2.*0265

*This is exactly the same process that you would go through if you are just adding like terms.*0270

*That will be u*^{2} + 7u^{2} - u^{2} then you are just looking at these initial coefficients like 2 + 7 - 1, and that would give you the 8.0275

*Let us see if we can take a look at some examples on when we can add and subtract these radicals.*0288

*In the first one I'm looking at 3 × 4th root of 17 – 4th root of 17*0294

*Let us check, the indexes are exactly the same we are looking at the 4th root and our radicands. *0300

*That is the part underneath, they are both 17.*0308

*We are going to look at the coefficients, 3 - 1 = 2.*0312

*I have 2 4th root of 17 and that one is good.*0318

*Let us look at this other one.*0326

*21√a + 4 3rd root of a*0328

*It is tempting to want to put these ones together but we cannot do it. *0335

*This one is the square root and this one is a cubed root.*0340

*It must have the same index if you have any hope of getting those together.*0353

*Let us look at some others.*0357

*On this one we want to add or subtract if possible.*0361

*I have 3 + √xy + 2 × √xy*0365

*Both of these are dealing with square roots and that is good.*0373

*Both of these are with an xy underneath that root.*0376

*We will simply add together their coefficients.*0381

*This will give us a 5√xy.*0386

*Let us see how that works for the next one.*0392

*7 × 5th root of u*^{3} - 3 × 5th root of y^{3}.0393

*That is so close.*0400

*Both of them have a 5th root and things are being raised to the third power, all of that is matching up but the variables are completely different. *0402

*One is a u and one is a y.*0412

*Make sure you check your indexes and you check your radicand before you ever put them together using addition or subtraction.*0420

*There are a few instances where the indexes are the same, but it looks like that radicand on the part underneath is completely different.*0431

*It is tempting to write those off and say okay, I probably cannot put those together using addition or subtraction.*0440

*Sometimes if you can do a little bit of simplification and get them the same then you can go ahead and put those together.*0446

*Let me show you an example of numbers. *0453

*Suppose I wanted to put together √2 + √8 and just looking at them I will say that wait 8 and 2 they are not the same, I cannot put them together.*0455

*The √8 over here that is the same as 4 × 2 and I can take √4 and that would leave me with 2 × √2 .*0466

*I can simply rewrite the next one as 2 × √2 .*0480

*In doing so now my radicals are exactly the same and I can simply focus on these coefficients out front.*0485

*I can see that 1 + 2 does equal my 3. *0493

*Do not be afraid to try and simplify these a little bit before you get into the addition or subtraction process.*0497

*Let us try that and keep it in line with these ones.*0504

*We want to rewrite the expressions and then try and add or subtract them if possible.*0507

*The first one I have a -√ 5 + 2 × √125.*0513

*I have -√ 5, 2 and 125 if I want to end up rewriting that, that is a 5 × 25 so that one reduces.*0520

*I have the √5 × 5.*0542

*Let us write that as √ 5 + 10√ 5 and now that I have my radicals the same now just focus on this coefficients -1 + 10 would be 9√ 5.*0548

*The next one I chose a big number but no worries, we can take care of this one.*0568

*We are looking for the 4th root of 3888 + 7 × 4th root of 3*0575

*If I have any hope of putting these together I want to match this 4th root of 3 over here.*0584

*As I go searching for ways to break down that very large number, the very first thing I'm going to try and break it down with is 3.*0591

*Let us see if I can.*0600

*It is the same as 1296 × 3 that is good because if I look at 1296 I can take the 4th root of that and I will get 6.*0608

*I simply have to add together these other radicals here by looking at their coefficients.*0631

*6 + 7 = 13 4th root of 3*0637

*Let us try one more, √72x - √32x *0648

*Let us try and simplify this as much as possible.*0655

*With the first one, looking at 72 is the same as 36 × 2 and with 32 that 16 × 2.*0659

*Notice how I have the square numbers underneath here, but I can go ahead and simplify.*0674

*√36 that will be 6 and I still have that 2 underneath there, √16 will be 4 and there is the 2x for that one.*0679

*I’m looking at 6√2x -4 × √2x or 2 √2x*0691

*These ones are a little bit larger involving some much higher roots, but the same process applies.*0708

*We must get the part underneath the roots the same if we are going to be able to put these two together.*0714

*10 × 4th root of m*^{3} is already broken down as far as it will go.0724

*The next one I could look at the 6561 and try and take its 4th root and break it apart from its m*^{3}.0736

*The good news is that one does break down, you will get 9.*0751

*4th root of m*^{3}0763

*10 4th root of m*^{3} + 9d4th root of m^{3} and we could put those together 10 and 90 is 1004th root of m^{3} .0767

*Let us try this next one here, this one is the 3rd root of 63xy*^{2} – 3rd root of 125x^{4}y^{5}0789

*In this one we will not only need to simplify those numbers but also take care of the variables like the x and y.*0799

*First, the numbers the 3rd root of 64 what does that break down into?*0811

*That will go in there 4 times and I have a 3rd root of xy*^{2}, 0820

*that one does not break down any further because both of those powers are smaller than the index of 3. *0827

*The 3rd root of 125 would be 5 and let us see what we can do with those variables.*0834

*X*^{3} × x that will be x^{4} and y^{3} × y^{2} that would give me y^{5}.0843

*This right here I can go ahead and take out of radicals. *0855

*Okay, taking out the x, taking out the y, and I still have xy*^{2}.0865

*Things are looking good and the part underneath the radical is now exactly the same, we will worry about our coefficients. *0873

*Notice how our coefficients are not like terms, I will be able to write them simply as 4 - 5xy package them together and then write my radical.*0880

*It is like we are just factoring out this common piece and writing it outside here.*0895

*In all cases, make sure you get those radicals exactly the same and combine their coefficients.*0904

*One last example that we can see many of our different rules in action, we will try and combine 3rd root of 2 / x*^{12} - 3 × 3rd root of 3 / x^{15}.0912

*Starting off, I'm going to use my quotient rule to break that up over the top and over the bottom.*0927

*We will break this one up over the top and over the bottom as well, looks pretty good.*0938

*I will go ahead and simplify the roots on the bottom.*0948

*2*^{3}rd root of 2 / x^{4} - 3 × 3rd root of 3 / x^{5}.0953

*Since we are looking to combine things as much as possible, *0967

*I will get a common denominator by putting an x on the bottom and on the top for the left fraction.*0969

*2x 3rd root of 2/x*^{5} - 3 3rd root of 3 /x^{5}.0979

*As I continue trying to put them together here is one of those situations where we are stuck, we cannot move any farther from there.*0993

*Since this is 3rd root of 2 and this is the 3rd root of 3 and those are different.*1000

*You will know it is tempting but we cannot put them together anymore. *1006

*We will leave this as 2x3rd root of 2 - 33rd root of 3/x*^{5}.1011

*A lot of different rules to keep track of our radicals but as long as you remember the rules follow directly from the rules for exponents you should be okay. *1021

*Be very careful in adding and subtracting those radicals and make sure everything is satisfied before you even attempt to put them together. *1031

*Thank you for watching educator.com*1040

*Welcome back to www.educator.com.*0000

*In this lesson let us take a look at how you can multiply and divide radicals.*0003

*We will first cover some rules for multiplying and dividing radicals and then get into that division process *0011

*and see how we want to rationalize the denominator that involves a radical expression.*0018

*There are only a few rules that you have to remember when working with radicals and the good news is we have seen many of these already. *0027

*For example, you can change radicals into fractional exponents.*0033

*If you want to combine them for division you can separate those out if you want to separate them out over multiplication.*0039

*You can also add and subtract radicals as long as we make sure that the radicands and the indexes are exactly the same.*0050

*Radicals follow our properties for all the other types of numbers so we can also use *0059

*the commutative property, associative property, even on these radicals.*0064

*The reason why that is important is because there is a few situations where you often try to apply some rules that do not work.*0072

*Treat these radicals like they are any other numbers. *0080

*Let us see this as we walked through the following problem. *0084

*I want to multiply the √6 + 2 × √6-3*0087

*For this one treat it like binomials and use foil to multiply everything out.*0093

*Every terms multiplied by every other term. *0099

*Let us see our first terms would have √6 × √6, outside terms -3 √6, inside terms 2√6 and our last terms 2 × -3 = 6.*0103

*Then we can use our other rules to go ahead and simplify this further. *0119

*I see I have two roots here, I can put those underneath the same root and that is the same as the √36 or 6.*0125

*-3 × √6 + 2 × √6 can I add them together or not?*0135

*Yes I can because they have the same radical I will just do the coefficients.*0142

*-3 + 2 - √6 *0147

*- 6 is still there. *0157

*You can continue simplifying your like terms by combining the 6 and -6 giving -√6.*0159

*We are using many of our tools that we have learned up to this point, in order to handle these radicals.*0168

*Watch out for lots of situations where you need to use foil.*0176

*In this example I have 4 + √7 *^{2} it is tempting to try and take that 2 and distribute it over the parts in between.0181

*However, do not do that. *0191

*Do not even attempt any type of distribution with this one. *0194

*What you should do with it is foil because as we learned in our exponent section, this stands for 4 + √7 and 4 + √7 .*0197

*Those two things are being multiplied by each other.*0208

*I can see where foil comes in to play.*0212

*I will take my first terms 16, outside terms, inside terms, and last terms.*0216

*Unlike before I could use some other rules to go ahead and combine things.*0227

*I can go ahead and add these two since they have the same radical and get 8√7.*0232

*I can combine these under the same radical and get the √49 and that is simply 7.*0239

*We can go ahead and finish off this problem, 16 + 7 would be 23 or 23 + 8 × √7.*0249

*Let us take a close look at division.*0266

*When dividing a radical expression we go ahead and rewrite it by rationalizing the denominator. *0269

*If you have never heard that term before rationalizing the denominator, *0278

*it is a way of rewriting it so that there is no longer a radical in the bottom.*0281

*That seems a little weird. *0286

*I mean, if we are interested in dividing by radical why are we writing it that there is no radical in the bottom. *0287

*It seems like we are side stepping the problem like we are not ending up dividing by the radical.*0294

*It is just a simpler way of looking at the whole division process and it is something that you have done before with fractions. *0299

*For example, when we have 1/3 ÷ 2/5 you have been taught that you should always flip the second and then multiply.*0307

*Why are we doing that? Why do not we just go ahead and divide by 2/5?*0314

*What is the big deal with a turn it into multiplication problem?*0317

*One, we will show you how accomplish the same thing, but by flipping the second one multiplying it does it in a much simpler way.*0321

*Also supposed I write this problem as 1/3 ÷ 2/5, you recognize that this is one of our complex fractions. *0330

*I can simplify a complex fraction by multiplying the top and bottom by the same thing.*0341

*I will multiply the top and bottom by 5/2, that should be able to get rid of our common denominator.*0348

*On top I would have 1/3 × 5/2 and on the bottoms 2/5 × 5/2 would be 1.*0359

*What we are sitting on the top there is the 1/3 and there is the 5/2 which comes from that rule we learned.*0369

*That we should take the second one, flip it and multiply. *0379

*But notice how we are doing that by simplifying the bottom now we are dividing by 1.*0382

*It is a great way to end up rewriting the problem, and taking care of it in a much simpler way.*0390

*That is exactly what we want to do when we are rationalizing.*0396

*The actual steps for rationalizing the denominator look a lot like this. *0402

*First, we will end up rewriting the rational expression, so that we will end up with no root in the bottom.*0407

*When we rationalize we try to get rid of that root.*0413

*We will do this by multiplying the top and bottom by the smallest number that gets rid of that radical expression.*0416

*That sometimes you can use some larger numbers, but it is best to use the smallest thing that will get rid of that radical. *0422

*It saves you from doing some extra simplifying in the end.*0429

*If we are dealing with square roots I recommend try and make that perfect square *0434

*in the denominator and that should be able to rationalize it just fine.*0437

*Let us see one of this division by radical in process and this is also known as rationalizing the denominator.*0444

*I have 2 ÷ √(2 )*0451

*I’m going to end up rewriting this so that there is no longer √2 in the bottom.*0456

*We will do this by multiplying the top and bottom by another root so that we will have a squared number in the root for the bottom.*0462

*On top I have 2√(2 ) and in the bottom I can go ahead and put these together and get √2 × 2, which is the √4.*0471

*I have created that square number on the bottom and now we can go ahead and simplify it.*0485

*That will be 2.*0493

*If you can simplify from there go ahead and do so, you will see that this one turns into the √2.*0496

*When I take 2 ÷√2 like the original problem says the √2 is my answer.*0503

*Let us try another one.*0513

*This one is 12 ÷ √5*0515

*We are looking to multiply the top and bottom by something to get through that √5 in the bottom we will use another√5.*0519

*12 √5 for the top and bottom √5 × √5 would be the same as √25.*0529

*The good news is that one reduces and becomes just 5.*0541

*If you are doing a division, you are dividing by the √5 even though it looks like you are changing into multiplication problem.*0549

*This is also known as rationalizing the denominator.*0558

*You can use these tools like rationalizing the denominator for some much higher roots as well. *0565

*The thing to remember by is that you want to multiply by the smallest root actually complete set root.*0571

*When we are dealing with the square roots it look like always multiplying by the same thing on the bottom.*0577

*With high roots, sometimes that might not be the case.*0583

*Let us look at this one.*0586

*1 / 3rd root of 2 if I try and multiply the top and bottom by 3rd root of 2 it is not going to get rid of that radical.*0587

*We will be left with the 3rd root of 2 / 3rd root of 4 and 4 is not a cubic number.*0598

*It is like it did not have enough of the number on the bottom to go ahead and simplify it completely.*0604

*Which should we multiply?*0612

*If I want a cubic number in the bottom but I'm going to use 3rd root of 4 .*0615

*When I take that onto the top and bottom you will see that indeed we do get that cubic number that we need.*0625

*From the bottom this would be 3rd root of 8 and on the top I will just have 3rd root of 4 . *0632

*The bottom simplifies becomes 2 now my answer is 3rd root of 4 ÷ 2.*0640

*Let us try another one of those higher roots. *0652

*This one is 3 ÷ 4th root of 9*0654

*Let us see, what would I have to use with a 4th root of 9?*0659

*9 is the same as 3 and 3, it will be nice if I had even more 3’s underneath there.*0664

*Let us say a couple of more.*0671

*I will accomplish that by doing the top and bottom by 4th root of 9 .*0674

*3 × 4th root of 9 for the top and 4th root of 81 on the bottom.*0682

*Now it is looking much better. *0693

*34th root of 9 on the top, the bottom simplifies to just the 3 and now we will cancel out these extra 3s. *0696

*I have 4 4th root of 9 .*0706

*This one we want to rationalize the denominator and if you look at it you must think what denominator are we trying to rationalize?*0715

*Is there a root in the bottom?*0722

*Because of our rules that allow us to break up the root over the top and bottom, there is.*0725

*In this one we have the 3rd root of 2y / 3rd root of z.*0731

*We can see we have 3rd root of z and it definitely needs to be simplified.*0744

*How are we going to do that?*0750

*Since I'm dealing with a cubed root I will need an additional 2 z's for that bottom.*0752

*Let us use the cube root of z*^{2}.0758

*Watch what that would give us here on the bottom.*0765

*3rd root of z*^{3} on top, the 3rd root of 2y z^{2}.0768

*The bottom simplifies and there will no longer be any roots in the bottom.*0779

*3rd root of 2y z*^{2} / z and I know that this one is done.0784

*One thing that can make the rationalization a more difficult process since we have more than one term in the bottom.*0797

*Our main goal is to end up rewriting the bottom so that there is no longer a root. *0805

*If we have more than one term we are going to end up using something known as *0812

*the conjugate of the expression to go ahead and get rid of it. *0815

*What the conjugate is, it is the same as our original expression, but it has a different sign connecting them.*0820

*That will allow us to get rid of that root.*0827

*To show you why we get through the root we will use an example. *0830

*Here I have 4 + √3, the conjugate of this one would be the same I have a 4√3 it will have a different sign connecting them 4 - √3.*0835

*Watch what happens when I foil these two together.*0852

*The 4 + √3 and its conjugate.*0855

*Starting with the first terms I have 4 × 4, which would be 16. *0863

*My outside terms would be -4√3, my inside terms will be 4√3.*0868

*I can move on to my last terms -√3 × √3.*0877

*A lot of things are happening when you multiply by its conjugate.*0882

*One, notice how our outside and inside terms where the same but one was positive and one was negative.*0886

*When you are dealing with conjugates that should always happen.*0892

*Those two things are gone.*0897

*We will focus was going on down here on the end.*0899

*-√3 × √3, -3 × 3 which be 16 - √9 which would be 9. *0903

*The numbers may change to make it different but at this point, there is no more radical.*0918

*Because those two radicals multiply and I get that perfect square number, there is no more radicals to deal with.*0925

*I can just take 16 - 9 and get a result of 7.*0932

*By multiplying by that conjugate, I got rid of all instances of all radicals.*0940

*This is why it will be important to use it when getting rid of our radical on the bottom.*0946

*Let us see for these examples.*0953

*Notice we need that conjugate because we have two terms in the denominator. *0955

*I'm going to multiply the top and bottom of this one by the conjugate.*0962

*√5 – 2 and √5 - 2 *0968

*When dealing with more than one term, remember that you will have to foil out the bottom.*0976

*Also remember that though you will have to possibly foil or even distribute the top.*0984

*The top 9√5 – 18 and working with the bottom and foiling that out my first terms would be √5 × √5 =5.*0991

*Outside terms and inside terms they are going to cancel out I know I’m on the right track.*1006

*2 × -2 -4*1015

*We will go ahead and do some canceling and let us see what we have left over.*1020

*9√5 - 18 / 5 – 4 = 1 and this reduces to 9√5 - 18. *1024

*Notice how we have divided by that radical because we have gotten rid of that radical in the bottom.*1037

*In this last example, let us look at rationalizing the following denominator.*1046

*We have 7 ÷ 3 - √x*1050

*With this one I’m going to have to multiply the top and the bottom by its conjugate.*1055

*You know what is next in there, this will be 3 + √x we will do that on the bottom and on the top. *1064

*Let us see what this does to the top as we distribute and remember that on the bottom we will foil.*1073

*I get 21 + 7√x for the top.*1082

*On the bottom we have 9 + 3 × √x - 3 × √x and then - √x × √x.*1089

*If we do things correctly we should get rid of all those radicals in the bottom.*1108

*+ square root - square root those two will take care of each other and then my √x× √x = x.*1114

*This will leave us with 21 + 7 √x / 9 – x.*1123

*We have got rid of all those radicals in the bottom you can say that our denominator is rationalized.*1133

*You can see that when you are working with dividing radicals you always have to keep in mind *1139

*what you would put on the bottom in order to get rid of all those radicals. *1144

*If you only have a single term feel free to multiply by what would complete whatever that radical is.*1148

*Complete the square or complete the cube.*1154

*If you have more than one term use the conjugate to go ahead and rationalize the denominator.*1157

*Thank you for watching www.educator.com.*1162

*Welcome back to www.educator.com.*0000

*In this lesson we are going to go ahead and solve some radical equations. *0003

*We only have one thing to pickup in this lesson it is just all the nuts and bolts on how you solve on these radical equations.*0009

*A radical equation is an equation in which we have variables present in our radical.*0018

*That will be something like this, I have √x-5 - √x-3 = 1.*0026

*To solve these we essentially want to end up isolating these roots then raise each side of our equation to a power in order to get rid of them. *0033

*It has to be the appropriate power. *0044

*For dealing with square roots we will square both sides.*0045

*If we have say 4th root then we will raise both sides to the 4th power.*0048

*With these ones it is extremely important that you go ahead and you check your solutions when you are done.*0054

*In the solving process and you raise both sides to a power we may introduce solutions or possible solutions that do not work in the original.*0060

*Always check your solutions for these types of equations.*0069

*Let us get an idea of what I’m talking about with √3x + 1 - 4 = 0. *0075

*The very first thing you want to do when solving something like this is get that root all by itself. *0082

*Let us go to both sides of the equation so we can do that. *0088

*√3x+ 1 = 4*0092

*We have isolated the root we are going to get rid of it by raising both sides to the power of two.*0097

*The reason why we are doing this is because we have a square root.*0104

*3x + 1 = 4*^{2} which is 16 0109

*Now they got rid of your roots, we simply solve directly.*0117

*The type of equation that we might end up with could be quadratic, could be linear, but we use all of our other tools from here on out.*0122

*-1 from both sides we will get 15 and divide both sides by 3 will give us 5.*0129

*You will realize that this looks like a possible solution always go back to your original for these ones and check to see if it does work. *0136

*I’m going to write this out here -4 does that equals 0 or not.*0146

*Let us put in 5, 3 × 5 would be 15 and all of that is underneath the root and then 15 + 1 would be 16.*0156

*√16=4, 4 -4 does equal 0, I know that this checks out and x equals 5 is our solution.*0172

*This one, we will go ahead and try and solve it is 5 + √x+ 7 = x.*0191

*Isolate our radical by subtracting 5 from both sides of the equation √ x + 7 = x - 5.*0200

*To get rid of that radical let us go ahead and raise both sides to the power of 2.*0213

*On the left side we will just have x + 7.*0222

*Be very careful on what is going on over here, keep your eyes peeled because notice how we have a binomial and we are squaring it.*0226

*In fact, you should look at it like this x - 5 × x – 5.*0238

*That way you can remember that this is how it should be foiled instead of trying to do some weird distribution with the two.*0244

* It does not work like that.*0250

*Let us see what we have when we foil.*0253

*Our first terms are x*^{2}, outside terms - 5x, inside terms - 5x, and our last terms 25.0256

*We can go ahead and combine to see what we will get x*^{2} - 10x + 25.0270

*You can see that this one is turning into a quadratic equation because we have x*^{2} term.0285

*Since it is quadratic we will get all of our x on to the same side and see if we can use some of our quadratic tools in order to solve it.*0291

*0 = x*^{2} - 11x + 180301

*How shall we solve this quadratic? *0312

*It is not too bad, it looks like we can use reverse foil to go ahead and break it up and see what our parts are.*0315

*x × x = x*^{2} then 2 × 9 = 180322

*I know that 2 and 9 are good candidate because If I add 2 and 9 I will get that 11.*0330

*Let us make these both negative.*0335

*Have it x = 2 or x = 9.*0337

*Two possible solutions and they are possible because they might not work.*0342

*Let us check in the original 5 + √ 2 + 7 does it equal 2?*0347

*Let us find out, 2 + √7 = √ 9, √9 is 3 so I get 8.*0359

*Unfortunately that looks like it is not the same as the other side.*0372

*This one does not work and we can mark it off our list.*0377

*Let us try the other one 5 + the square roots and we are testing out a 9 so we will put that in 9 and 9.*0383

*Underneath the square root, 9 + 7 would give us a 16 and √16 =4, I will get 9 which is the same thing as the other side. *0396

*That one looks good. *0410

*You will keep that one as your solution. *0413

*Be very careful and always check these types of ones back in the original, you will know which ones you should throw out.*0416

*If there happens to be more than one radical in your equation then we can solve these by isolating the radicals one at a time.*0427

*Do not try and take care of them both at once. *0434

*Just focus on one, get rid of that radical and focus on the other one and then get rid of that radical. *0436

*It does not matter which radical you choose first.*0442

*If you have a bunch of them just go ahead choose one and go after it. *0446

*Be very careful as you raise both sides to a power since you have more than one radical in there, *0450

*when you raise both sides to power it will often end having to get foiled.*0456

*Let us see exactly what I'm talking about with this step, but will have to be extremely careful to make sure we multiply correctly.*0460

*Always make sure you check your solutions to see that they work in the original.*0468

*But this one has two radicals in it. *0474

*I have √x - 3 + √x + 5 = 4*0477

*It does not matter which radical you choose at the very beginning and I'm going to choose this one.*0483

*I’m going to work to isolate it and get all by itself x + 5 = 4 – and I have subtracted the other radical to the other side.*0490

*Since √x+ 5 is the one I’m trying to get rid of, I’m going to square both sides to get rid of it. *0505

*That will leave me with just x + 5 on that left side.*0514

*Be very careful what happens over here on the other side, it is tempting to try and distribute the 2 but that is not how those work.*0520

*In fact if we are going to take look at this as two binomial so we can go ahead and distribute.*0528

*Let us go ahead and take care of this very carefully. *0542

*Our first terms 4 × 4 = 16, our outside terms we are taking 4 × -√x – 3, -4 × √x - 3.*0545

*Inside terms would be the same thing -4 √x - 3 and now we have our last terms.*0560

*negative × negative = positive and then we have √x -3 × √x -3.*0570

*That will give us √x - 3*^{2} since it will be multiplied by themselves. 0581

*When you do that step the first time, it usually looks like you have made things way more complicated. *0590

*I mean, we are trying to get rid of our root but now it looks like I have 3 of them running around the page. *0596

*It is okay we will be able to simplify and in the end we will end up with just one root, *0600

*which is good because originally we started with two and if I get it down to one we are moving forward with this problem.*0605

*Let us see what we can do.*0612

*On that left side I have 16 these radicals are exactly the same, I will just put their coefficients together -8 √x -3 *0614

*The last one square of square root would be x -3.*0626

*Now you can start to combine a few other things. *0634

*I will drop these parentheses here, if I subtract x from both sides that will take care of both of those x.*0639

*Let us see I can go ahead and subtract my 3 from the 16, 5 =13 - 8√x -3.*0650

*Let us go ahead and subtract a 13 from both sides, -13 – 13, -8 = -8√x -3.*0665

*It looks like we can divide both sides by -8, 1 =√x -3.*0682

*That is quite a bit of work, but all of that work was just to get rid of one of those radicals and we have accomplished that.*0689

*We got rid of that one. *0695

*But we still have the other radical here to get rid of.*0696

*We go through the same process of getting it all alone on one side, isolating it, squaring both sides get rid of it, and solving the resulting equation.*0700

*This one is already isolated we are good there.*0709

*I will simply move forward by squaring both sides. *0711

*1 = x -3 let us get some space. *0716

*If I add 3 both sides this will be 4 = x.*0724

*Even after going through all of that work and just when you think you have found a solution, we got to check these things. *0729

*We have to make sure that it will work in the original. *0735

*4 - 3 + √4 + 5 we are checking does that equal 4.*0741

*Let us see what we have.*0756

*4 - 3 would give us √1 , 4 + 5 would be 9, so√4 =1 and √9 = 3 and fortunately 4 does equal 4.*0758

*We know that this solution checks out, 4 =x.*0774

*If you have more than one of those radicals just try and get rid of them one at a time.*0781

*You might be faced with some higher roots and that is okay.*0788

*You will end up simply using a higher power on both sides of your equation in order to get rid of them.*0791

*In this one I have the 3rd root of 7x - 8 = 3rd root of 8x+ 2.*0796

*If I'm going to get rid of these cube roots I will raise both sides of it to the power of 3.*0802

*Leaving me with 7x - 8 = 8x + 2*0810

*I can work on just giving my x together.*0818

*-7x from both sides, -8 = x + 2.*0822

*We will subtract 2 from both sides and this will give us -10 = x.*0833

*Let us quickly jump back up here to the top and see if that checks out.*0841

*3rd root of 7 × putting that -10 - 8 and 8 × -10 + 2*0845

*Let us see if they are equal.*0866

*This time I’m dealing with 3rd root of (-70-8) to be the 3rd root of -78.*0868

*The other side I have -80 + 2 which is the 3rd root of -78.*0880

*It looks like the two agree. *0887

*I know that the -10 is my solution.*0890

*With those radicals try and isolate them, and then get rid of them by raising them to a power.*0894

*Always check your solutions with these ones to make sure they work in the original.*0899

*Thanks for watching www.educator.com.*0904

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at complex number. *0002

*I held off on this very special type of number so you could build a lot of things about them that you will see in the section *0007

*such as multiplying them together which will look like multiplying polynomials and rationalizing denominator will look a lot like the division process.*0013

*Here are some other things that we will cover.*0022

*First we will get into a little bit about that the vocabulary from the complex numbers.*0024

*You have heard about imaginary numbers and you will see how that works with complex numbers. *0028

*We will learn about the real and the imaginary part of complex numbers. *0033

*Many of the properties involving these complex numbers are the same properties that you have seen before for real numbers. *0039

*Then we will get into the nuts and bolts on how you can start combining these things. *0046

*That means adding, subtracting, multiplying and dividing our complex numbers.*0051

*Near the very end we will also see some nice handy ways that we can simplify powers of (i) *0056

*so you can always bring them down to the most simplest form.*0061

*An imaginary number (i) is defined as the √-1.*0067

*It seems like a small definition and it does not seem to be very useful, especially if we want talk about lots of different numbers.*0075

*By defining it in this way, the √-1 it can actually express the root of any negative number using this imaginary number (i).*0082

*Here is a quick example to show how it works. *0091

*Suppose I’m looking at the √-7 using some of our properties we can go ahead and break it up into it -1 × 7.*0093

*And then break up the root over each of those parts.*0102

*You will see the definition shows up right there. *0105

*I'm taking the √ -1.*0108

*It is that piece that gets turned into an (i) and now I'm representing this number as (i) × √7.*0111

*Think of doing this for many other numbers where you have a negative underneath the square root.*0119

* If I had -23 underneath the square root that will be (i)√23 and a negative comes out as the (i).*0125

*√ -16 would be (i) × 4 or just 4i.*0137

*How those imaginary numbers relate to our complex numbers?*0150

*Complex number is a number of the form A + B(i).*0154

*You can see that the imaginary number is sitting right there, right next to that B.*0158

*Those other values A and B are both real numbers.*0164

*We call A the real part of the complex number and B the imaginary part, since it is right next to the imaginary number.*0168

*You can represent a lot of different numbers using this nice complex form.*0175

*Maybe I will have a number like 3 – 5i that would be in a nice complex form.*0180

*I can read off the real and imaginary part.*0186

*We could throw in some fractions and in here as well, so maybe ½ × 2/3(i).*0189

*We can also have complex numbers like 0 + 3(i).*0196

*We would probably would not leave it like that, something like this would probably we will write as 3(i).*0202

*Notice how we can still say it is in a complex form where the real part is 0.*0207

*The good part about these complex numbers is they will obey most of the usual laws that you are familiar with.*0217

*Our commutative, associative, and distribution, all of those will work with our imaginary numbers as well. *0223

*The things that we could do with real numbers, we can do with imaginary numbers as well. *0230

*Some of the things that we will have to watch out for is when we get down in the simplification process, so keep a close eye on that.*0236

*Let us get into how you can start combining these numbers together.*0247

*To add or subtract complex numbers this is a lot like having like terms.*0251

*Be very careful to watch the signs when combining these numbers together.*0257

*To show you how this works I have 2 – 5(i) that complex number + 1 + 6(i).*0262

*Since I'm adding I'm going to go ahead and drop my parentheses here and just figure out which terms I should add together.*0269

*Notice how all of our real parts I can go ahead and put those together and I can go ahead and collect together the imaginary parts.*0280

*That is what I'm talking about when I say adding like terms. *0290

*The real ones 2 + 1 would be 3 and the imaginary I have -5(i) -6(i) that will give me -11(i).*0294

*I will write those as 3 -11(i).*0309

*When you are working with subtraction we have to be very careful with your signs.*0314

*Let me show you, suppose I have 3 + 2(i) and now I'm subtracting a second number 4 + 7(i).*0319

*Parentheses will come in handy because I need to distribute this negative sign to both parts of that second complex number.*0334

*I will look at this as 3 + 2(i) – 4 – 7(i) and now I can see the parts and the like terms that I need to go ahead and put together.*0343

*3 – 4 = -1 and 2 – 7(i) = -5(i)*0357

*This would be written as -1 – 5(i)*0369

*Again, combine those like terms.*0373

*To multiply complex numbers think of multiplying together two binomials.*0379

*It looks a lot like the same process. *0384

*When we are multiplying together two binomials we have great way of remembering that we can use the method of foil to accomplish that.*0387

*There is one additional thing you have to remember, when you have (i)*^{2}, we can go ahead and reduce that to -1,0394

*that seems a little odd but let us see why that works.*0401

*If I'm looking at the √-1 and that is our (i) and suppose I take that and I square it.*0404

*I want to square that square root I will get -1 on the left.*0413

*That reveals that (i)*^{2} = -1.0419

*Watch for that to show up in the simplification process.*0424

*Let us go ahead and foil out these two complex numbers.*0428

*2 – 5(i) multiplied by 1 – 6(i)*0432

*We will begin by multiplying the first terms 2 × 1 = 2.*0436

*We will move on to those outside terms 2 × -6(i) = -12(i).*0441

*On the inside terms -5(i) and now let us go ahead and do the last terms -5(i) × -6(i), *0453

*minus × minus would be +, 5 × 6 is 30 and now there is my (i)*^{2}.0465

*We will go along and we will start crunching things down like we normally would do if they were a couple binomials.*0473

*2 – 17(i) now comes this interesting part this (i)*^{2} here is the same as -1.0480

*We will go ahead and rewrite it as a -1. *0491

*You can see there is more than we can do to simplify this.*0498

*I’m not only multiplying the 30 × -1, but then I can go ahead and combine it with the two out front.*0501

*2 – 17(i) - 30 and then let us combine these guys that will give us -28 – 17(i).*0508

*This one is good.*0524

*With many of these problems where combining complex numbers you know that you are done *0527

*when we finally get into that nice complex form.*0532

*You can easily see the real and imaginary parts.*0535

*To go ahead and divide complex numbers, we want to multiply the top and the bottom by the complex conjugate.*0542

*The complex conjugate of a number will look pretty much the same but it will be different in sign.*0550

*If I'm looking at the complex conjugate of A + B(i) that will be –B(i).*0557

*Let us just do some quick example. *0563

*Let us see if I have 3 + 2(i) its complex conjugate would be 3 – 2(i). *0564

*If I had -7 – 3(i), the complex conjugate would be -3 + 3(i).*0573

*The only thing that is changing is that negative sign or that sign on the imaginary part.*0583

*This will have the feel of rationalizing the denominator that is why we had to cover a lot of that information before working on these complex numbers.*0591

* I’m going to look at the bottom of this particular division problem 4 + 2(i) ÷ 3 + 5(i).*0599

*I'm going to find the complex conjugate of 3 + 5(i), that would be 3 – 5(i).*0606

*Once we find that complex conjugate we will multiply on the top and on the bottom of our complex fraction.*0616

*Multiply on top, multiply on the bottom.*0628

*This one involves quite a bit of work.*0631

*I have two terms in my complex number, I will have to foil the top and we will also have to foil out the bottom. *0634

*This will make things look a lot more complicated at first but if you just go through the problem carefully you should do fine.*0644

*Let us start with the top.*0653

*Taking our first terms I have 4 × 3 and that would give me 12.*0656

*The outside terms would be 4 × -5 (i) = -20(i).*0662

*On the inside 2(i) + 3(i) then would be 6(i) and our last terms 2 × -5 = -10(i)*^{2}.0668

*Do not forget to multiply those (i) together as well.*0680

*On the button 3 × 3 = 9, outside will be -15(i), inside would be 15(i).*0684

*And then we have 5(i) × -5(i) = -25(i)*^{2}.0695

*Some things you want to notice after you go through that forming process.*0703

*We are using that complex conjugate on the bottom, the outside and inside terms will end up canceling. *0707

*If they do not cancel, make sure you have chosen the correct complex conjugate.*0715

*Notice that we have a couple of (i)*^{2} showing up on the top and bottom. 0720

*That comes from the (i) being multiplied together. *0725

*Each of these will be need to change into -1.*0728

*Let us go through and do those two things and clean this problem a little bit.*0733

*I have 12 – 20 + 6(i) = -14(i), -10 × -1 ÷ 9 my outside and inside terms on the bottom cancel -25 × -1.*0737

*We can see the beauty of using that complex conjugate.*0759

*On the bottom we no longer have any more imaginary numbers. *0762

*That one I was talking about how I have the feel of rationalizing the denominator and we are getting rid of those roots in the bottom.*0766

*Here we got rid of the imaginary numbers in the bottom.*0772

*Let us continue to simplify and see where we can take this problem.*0776

*12 – 14(i) + 10 ÷ (9 + 25) *0781

*12 + 10 = 22 – 14(i) ÷ 34 *0790

*It is tempting to try and stop right there but do not do it, we want to continue doing this entire problem *0801

*and try to get it in that nice complex form where I can see the real and imaginary part.*0806

*To get in that form I’m going to use the part where I breakup the 34 and 22 and under the 14.*0811

*It looks a lot like when we are taking polynomials when we are dividing by a monomial.*0818

*This will be 22 ÷ 34 -14(i) ÷ 34.*0824

*This particular one, both of those fractions can be reduced.*0833

*I will divide the top and the bottom by 2, 11/17 – 7(i) ÷ 17.*0837

*This one is actually completely done.*0846

*I can easily see my real part over here and my imaginary part -7/17.*0849

*That is I know that this one is done.*0855

*Division is quite a tricky process remember to multiply by the complex conjugate of the bottom *0859

*and then work very carefully to simplify the problem from there.*0864

*It should be in its complex form when you are all done so, you can see the real and imaginary parts.*0868

*After working with doing some combining things with these various complex imaginary numbers, *0876

*you may have noticed that you rarely see any higher powers of (i).*0881

*In fact the largest part of (i) that we came across was i*^{2} and we turned it immediately into a -1. 0885

*The interesting part is you can usually take much higher powers of (i) and end up simplifying them down.*0892

*To see out why this works for some much higher powers, let us just take a bunch and start picking them apart.*0898

*We see many instances while we are working with (i) and (i) was just equal √1 which of course we call (i).*0905

*When we have i*^{2} and then we thought about squaring the √-1, so we got -1 as its most simplified result.0912

*We will see what will happen with i*^{3}.0922

*One way that you could interpret this would be i*^{2} × (i).0927

*The reason for doing that is because you can then take the i*^{2} sitting right here and simplify that.0934

*That will give us -1 × (i) the final result of that would be –i as it is simplest form.*0942

*Let us choose that same idea and see if we can go ahead and simplify i*^{4}.0955

*That will be i*^{2} × i^{2}.0959

*Both of the i*^{2} = -10964

*I have -1 × -1 = 1*0970

*Take note how all of these are simplifying to something else that does not have any more powers on them.*0975

*Let us do a few more that hopefully we can develop a much easier or a shortcut way of doing this process.*0983

*Let us go on to i*^{5}.0990

*This can be thought of as i*^{4} × (i).0994

*We have already done i*^{4}, it is way back here, that is equal to just 1.0999

*1 × (i) I know this simplifies to (i).*1003

*On the next one i*^{6}, i^{4} × i^{2}.1010

*We have already done i*^{4} that is just 1.1020

*I will put in i*^{2}, I will just write across from it -1 so 1 × -1 = -1, very interesting.1024

*I think we are starting to see a pattern of some of these (i).*1035

*Notice how they are looking exactly the same as the ones on the other side here.*1038

*Let us see if that continues.*1044

*I*^{7} will be i^{4} × i^{3}, 1 × i^{3} = -i.1046

*Maybe the last one, i*^{4} × i^{4}, both of those are equal to one, this is equal to 1 as well.1058

*Notice how we did starting to see same numbers.*1069

*There are some good patterns we can see, I mean we got past i*^{4} and all of them contained an i^{4} 1072

*so we are getting this similar pieces right here.*1079

*If I was even to continue on to i*^{9} then I will have even more groups of i^{4} and I could see that it would simplified down.1083

*All of those much higher powers will end up simplifying down to something which does not have any more powers whatsoever.*1097

*We can take as much farther and develop a nice shortcut formula so we do not always have to string out into a bunch (i).*1104

*Let us see how to do that.*1111

*Looking at all of these examples you can see that there is a pattern on how they simplify.*1115

*In fact they repeats in blocks of 4 and you start with (i) then it goes -1, then –I, then 1, it just repeats over and over again.*1120

*To figure out where in that pattern you are at, you can take the exponent and simply divide it by 4 and observe what the remainder is.*1129

*From the remainder you will know exactly what it will simplify to.*1138

*I have made a handy table to figure out what that looks like *1142

*If I’m looking at (i) to some large power, I will take the power and divide it by 4 *1146

*and the remainder happens to be 3, according to my chart here it will simplify to –i.*1152

*Let us do a quick example to see how this works and why it works.*1160

*Let us take i*^{13}.1166

*If I wanted to I could imagine a whole bunch of i*^{4} in here.1173

*i*^{4}, i^{4}, i^{4} and then i^{1}.1178

*If we add up all those exponents they add up to 13.*1186

*You can see that most of them are gone since i*^{4} = 1.1191

*1 × 1 × 1 × (i) =(i)*1196

*Let us use our shortcut formula to figure out the same thing. *1207

*In the shortcut we take the exponent and divide that by 4.*1210

*13 ÷ 4, 4 goes in the 13 exactly 3 times.*1216

*Notice why that is important, that is exactly how many bunches of i*^{4} I have.1227

*When you are going through that division by 4 process you are counting up all of the i*^{4} that will simply break down and simplify into 1.1233

*It goes in there 3 times, 3 × 4 = 12 and I will subtract that away getting my remainder of 1.*1243

*That remainder tells me how many extra (i) I still have left over.*1253

*From there I can see I have only one (i) and I know it simplifies down to (i).*1259

*I get the same exact answer as the 4, i*^{13} = (i).1267

*Now that we know a lot more about these complex numbers let us go through some various examples *1275

*of putting them together and see how we can simplify powers of (i).*1279

*This first one we want to do a subtraction problem 4 + 5(i) – 6 – 3(i).*1284

*With the addition and subtraction we are looking to add like terms.*1290

*Subtraction is a tricky one, you have to remember to distribute our sign onto both parts of that second complex number. *1295

*I’m looking at 4 + 5(i) -6 + 3(i)*1302

*I can see my like terms, let us put together 4 and -6, and 5 and 3.*1311

*4 - 6 = -2, 5(i) + 3(i) = 8(i)*1320

*We can see that it is in that final complex form and we know that this one is done.*1329

*In this next example, we want to multiply two complex numbers. *1337

*Think foil.*1342

*Be on the watch out for additional things that we will need to simplify such as the i*^{2}. 1346

*Okay, starting off, multiplying our first terms together we will get 1.*1351

*Our outside terms we will multiply that will be 3(i) then we can take our inside terms 2(i) and then finally take our last terms 2(i) × 3(i) + 6i*^{2}.1357

*We go through and start combining everything we can.*1377

*1 + I have my like terms here, so 3 + 2 = 5(i) + 6 and take the i*^{2} and turn it into -1.1380

*Let us continue 1 + 5(i) - 6 now we can see we have just a couple of more things that we can go ahead and combine.*1397

*This will be -5 + 5(i) and now I can easily see my real and imaginary part, so I know that one is done.*1408

*Let us divide these two complex numbers. *1420

*I have 2 – 5(i) ÷ 1 – 6(i)*1422

*This is the lengthy process where we find the complex conjugate of the bottom of if we multiply the top and bottom of our fraction by that.*1426

*Let us first find that complex conjugate.*1434

*That will be 1 + 6(i).*1437

*We are going to use that on the top and on the bottom.*1440

*We have to remember that since we are multiplying we will have to foil out the top and the bottom.*1445

*Foil top and foil bottom.*1452

*Let us see what the result of that one looks like.*1458

*I’m looking at the top, first terms 2 × 1 =2, outside terms would give me 12(i), inside terms -5(i), and last terms -30(i)*^{2}.1461

*Lots of terms in there, be very careful and keep track of them all.*1479

*On the bottom, first terms 1, outside 6i, inside -6i, and our last terms -36i*^{2}.1483

*It is looking much better, looking pretty good. *1496

*If we do this correctly, we should not have any more powers of (i) in the bottom.*1498

*Let us see what else we can simplify.*1502

*We will go ahead and combine these 2i on the top and we will combine these 2i of the bottom *1505

*and then we will put these i*^{2} to make them both -1.1510

*2 and then we will have 7(i) - 30 × -1.*1518

*On the bottom 1 – 36 -1.*1527

*It looks like we have indeed accomplished our goal there is no longer imaginary numbers on the bottom. *1533

*We continue putting things together 2 + 7(i) + 30 since the negative × negative is a positive and 1 + 36.*1538

*Just a few more things to put together 32 + 7(i) ÷ 37.*1553

*This point it is tempting to stop it but keep going until you get it into its nice complex form, we can see that real and imaginary part.*1562

*I’m going to give 37*^{32} and 37^{7}.1570

*7 /37(i) this one looks much better.*1579

*Sometimes you can go ahead and reduce those fractions but this one the way it is.*1587

*In this last problem we will go ahead and see if we can simplify some very large powers of (i).*1595

*Since they are very large powers we will go ahead and use our shortcut to take care of that process.*1602

*In the first one I have i*^{107}. 1607

*Let us start off by taking that exponent, 107 and we will divide it by 4.*1611

*This will figure out all the bunches of i*^{4} that we have hiding in there.1618

*4 of those in the 10, twice and we get 8.*1623

*Subtract them away I will end up with 27.*1629

*That goes in there 6 times and 6 × 4 = 24 and then we can subtract that away. *1635

*I'm down to 3 and that is smaller than 4 so I know that 3 is my remainder.*1645

*Think back, if my remainder is 3 then what does this entire thing simplify to?*1653

*It simplifies to –i.*1660

*If you ever forget the way you use that table, you can always build it very quickly by doing just the first few values of (i).*1664

*(i) = (i), i*^{2} = -1, and we have –I and 1.1674

*You quickly have built your table.*1682

*On to the next one i*^{2013} something very large.1686

*We will grab the exponent and we will divide it by 4.*1692

*4 goes into 20 5 times and I will get 20 exactly.*1701

*Let us put in our 0 placeholder over the 1 and then bring down the 13.*1707

*4 goes into 13 3 times and we will get 12.*1713

*This one shows that my remainder is 1.*1720

*What does this entire thing simplify down to?*1725

*If I get a remainder of 1, 2, 3 or 0, now I know what it will simplify down into.*1729

*It simplifies down just to (i).*1736

*There are many things you can do to work with these complex numbers but they are often mere things that you have learned before.*1740

*Keep track of this handy method for simplifying powers of (i) that way you can take those much higher powers *1745

*and bring them down to something nice and compact.*1751

*Thank you for watching www.educator.com.*1754

*Welcome back to www.educator.com.*0000

*In this lesson we are going to start looking more at linear equation starting off with the vocabulary of linear equations.*0003

*There will be lots of new terms in here, it will definitely take some time to look at them all and what they mean and play around with them a little bit.*0010

*Some of the terms that will definitely get more familiar with are variable, term.*0018

*We will look at coefficients and we will definitely see how we can combine like terms.*0026

*We will be able to tell the difference between equations and expressions and get into a linear equation.*0032

*What we want to solve later on and solutions.*0038

*When looking at an equation, we often see lots of letters in there, those are our variables.*0046

*What they do is they represent our unknowns.*0054

*One favorite thing to use in a lot of equations is (x), but potentially we could use any letter.*0058

*You could use a, b, it does not really matter but most the time our unknown is (x).*0064

*A term is a little bit more than just that variable.*0070

*It is a number of variables or sometimes the product or quotient of those things put together.*0074

*To make it a little bit more clear, I have different examples of what I mean by a term.*0079

*All of these things down here are types of terms.*0085

*The first one is the product of an actual number and a variable.*0088

*Down here with the (k) it is simply just a variable all by itself.*0093

*A coefficient of a term is a number associated with that term.*0102

*If I'm looking at a term say 2m, the coefficient is the number right out front that is associated with that term.*0109

*I'm looking at another one like 5mq, but again the 5 would be the coefficient of that term.*0121

*Terms with the exactly the same variables that have the same exponents those are known as like terms.*0133

*There are two conditions in there you want to be familiar with.*0140

*It must have exactly the same variables and it must also have exactly the same exponents.*0143

*Let us say I have both of those and you can not consider them like terms.*0150

*Let us take a look at some real quick.*0156

*I'm looking at 5x and I'm looking at 5y, these are not like terms.*0158

*Why, you ask? They do not have the same variable.*0169

*One has (x) and the other one has (y).*0172

*How about 3x*^{2} and 4x^{2}, these are like terms.0176

*These ones are definitely good because notice they have exactly the same variable and they have the same exponent.*0188

*They have both of those conditions.*0203

*This one is little bit trickier so be careful, they both have an (x), that looks good.*0218

*They both have a (y), that seems good but they have different exponents.*0224

*This one has the y*^{2} and this one has nothing on its y.0230

*I would say that these are not like terms.*0238

*An expression is the statement written using a combination of these numbers, operations, and variables.*0244

*This is when we start stringing things together so I might have a term 2x and then I decide use may be addition and put together a 4xy.*0251

*That entire thing would be my expression.*0262

*In the equation, we take a statement that two algebraic expressions are actually equal.*0265

*I can even borrow my previous expression to make an equation.*0271

*I simply have to set it equal to another expression, maybe 2x*^{2}.0276

*What a lot of students like to recognize in these two cases is that in an equation there better be an equal sign somewhere in there.*0285

*With your expression there is not an equal sign because it is just a whole bunch of string of terms and coefficients, numbers, operations.*0292

*Since we are interested in linear equations and eventually getting solved for those, what exactly is a linear equation?*0306

*It is any equation that can be written in the form, ax + b = c.*0314

*There is some conditions on those a, b, and c.*0321

*Here we want a, b, and c to be real numbers, we would not have to deal with any of those imaginary guys.*0324

*We want to make sure that (a) is not 0.*0330

*The reason why we are throwing that condition in there is we do not want to get rid of our variable.*0335

*If (a) was 0, you would have 0 × x and then we would have a rare variable whatsoever.*0340

*It is an equation, not necessarily look like that but it almost can be written in that form, it is a linear equation.*0347

*A number is a solution of that equation if after substituting it in for the value the statement is true.*0354

*That means if you actually take out your variable and replace it with a number that is the solution.*0361

*Then it is going to balance out, it is going to be true with that number in there.*0367

*This first part we just want to identify the different parts of the equation so we can better feel of what we are looking at.*0379

*First of all I know that this is an equation because notice how we have an equal sign right there.*0386

*I have the expression 7x + 8 that is one expression one side and expression 15 on the other.*0394

*What else do I have here? I have my 7x and I have the 8 both of these are terms.*0400

*If I pick apart that one term on the left, I can say that the 7 is a coefficient and that the x is my variable.*0410

*There are many different parts of the equation and you want be able to keep track of it.*0428

*And probably terms are one of the most important for now.*0432

*Let us look at this one, let us see 30x = (4 × X) - 3 + (3 × 3) + (x + 2).*0438

*Again identify the parts of the equation, let us see.*0444

*It has the equal sign, I know it is an equation, it is important to recognize.*0449

*Let us see, over on this side I have a term 30x, I have this term, I have this term.*0454

*I have a bunch of different terms.*0461

*Terms are always combined using addition or subtraction.*0465

*That is how we can usually recognize them.*0468

*In my terms I have some coefficients but you know it might be easier to use my distributive properties to see even more of those coefficients.*0471

*Let us use that distributive property, let us take the 4 multiplied by the x and 3 and do the same thing with 3.*0481

*4x -12 + 3x + 6, not bad.*0489

*Looking at that I have even more terms, I got my 30x, I get 4x, 12, 3x, 6 and lots of different terms now.*0497

*Into those terms I can identify what its coefficient is and I can identify the variable, the x.*0508

*This one says simplify it by combining like terms.*0528

*Remember our like terms are terms that have exactly the same variable and they have exactly the same exponent.*0532

*We have to be careful on which things we can actually combine here.*0539

*Let us see I have 12w and 10w those are like terms, they both have a w to the first.*0543

*Over here is 8- 2w, all three of those are like terms, we only write them next to each other.*0550

*I know that I need to combine them.*0558

*The 9 and the 3 I would also consider those like terms because both of them do not have a variable associated with them.*0567

*I will combine those together.*0574

*Let us take care of everything with the (w), 12w + 10 would be 22w.*0576

*That would give me a 20w when combining those, now we will take care of 3 and 9, -3 + 9 = - 6.*0591

*It is important to recognize that you should not go any further than here because the 20w and 6, those are not like terms.*0609

*We are not going to put those together.*0616

*On to example 4, this one we want to simplify by combining like terms.*0622

*I have lots of grouping symbols in here so it is hard to pick out what my like terms are.*0630

*I think we can do it though but we might have to borrow our distributive property first.*0637

*I'm going to take this negative sign and distribute it inside my parentheses here.*0641

*Now that would give me (2y - 3y) - 4 + (y*^{2} + 6y^{2}), not bad.0648

*I can see there is a few things I can combine, let us see.*0665

*Specifically I can put these y's together since both of them are single (y).*0670

*I can combine these ones together over here because both of them are y*^{2}.0675

*Let us put those together, 1y - 3y would be -2y, I have a 4 hanging by itself, -4.*0683

*y*^{2} + 6y^{2} would be a 7y^{2}, that looks good.0695

*Remember, I have not distributed my 2, it is not going to help me combine anymore like terms.*0703

*It will definitely help me see my final results, -4y - 8 + 14y*^{2}.0708

*My final answer would be -4y - 8 + 14y*^{2}.0722

*I would not combine those anymore together because none of those are like terms.*0726

*I have a single (y), I have an 8 that does not have any variables whatsoever.*0731

*I have that y*^{2}, definitely not like terms.0735

*Alright, thanks for watching www.educator.com.*0739

*Welcome back to www.educator.com.*0000

*In this lesson we will look at more of the nuts and bolts of solving linear equations.*0002

*Specifically some of the things that we will actually do is we will look at types of solutions.*0010

*What does it mean to actually have a solution versus when it is consistent or inconsistent?*0015

*The things that will make this handy is how you can also deal with fractions when they are in the solving process.*0023

*It is something that my students always ask me.*0031

*What a solution means is a value that when you substitute it in for the variable, it makes the equation true.*0039

*We are looking for that magic value that would make the whole thing true.*0046

*In some cases, only one value would end up making the whole thing true.*0051

*If you only have one value that will do that or sometimes only two or three, we call this conditional.*0057

*The condition is that this variable must be that value.*0064

*There are other cases where we actually end up where any value will do.*0069

*It does not really matter what the value is as soon as you substitute it in for the variable it will make it a true statement.*0075

*If any value will do we say that the equation is an identity.*0083

*Lastly, there are very few types of equations that no matter what you try and substitute in for that variable, nothing seems to work.*0089

*No matter what that value is if nothing works then we say that that equation is a contradiction.*0100

*I have hidden a couple of those into the examples later on so watch for the one that is an identity and one that is a contradiction*0106

*and how we actually pick that out of just one that actually have a solution.*0116

*How do you exactly solve a linear equation?*0123

*You will use a few properties to help you out in the solving process.*0126

*One of them is your addition property of equality.*0130

*It is a handy property but what it says is that you can add the same amount to both sides of the equation.*0133

*In fact, many people like to look at this as almost like a balance scale.*0140

*Whatever you add to one side of your equation then you better be sure to add it to the other side as well to make sure that it all balances out.*0146

*If you are going to add + 4 over here, make sure you also add + 4 to the other side.*0155

*There is also the multiplication property of equality.*0163

*For this one you are allowed to multiply both sides by the same value and it will keep the equation the same.*0168

*You do have to be a little bit more careful with that property because it works as long as you do not multiply both sides by 0.*0175

*But you can use any other number you want.*0184

*You can multiply both sides by 3, multiply both sides by (x) as long as (x) is not 0, it will keep that equation the same.*0186

*One thing that I like to do when solving equation is to think of this process, *0197

*it often help with fractions as well as to get the variable that we are looking for all alone.*0203

*The very first thing that I like to do is scan it over and see if it does contain any fractions*0209

*and immediately clear them out if I can using some sort of common denominator.*0213

*After I have done that, I like to simplify each side of the equation as much as possible before shifting things to one side or the next.*0217

*We will simplify each side as much as possible first.*0225

*After we simplify both sides, we will try and isolate the variable on one side of the equation.*0230

*It may be difficult to do especially if there is more than one copy of that variable but if we can get them together usually we can isolate it just fine.*0236

*Lastly, it is always a good idea to check the solution just to make sure it actually works in the original problem.*0244

*That is always a good thing to do in case we make a mistake in one of these earlier steps.*0251

*We will definitely be able to find that out by the four steps when we check it by substituting it back into the equation.*0256

*Let us get into the examples and see how the solving process actually works.*0264

*I want solve (2 × x) +3 = 4x – 8, our goal is to figure out what the value for x is that would make this entire thing true.*0270

*In order to help this process out, I can see that it does not have any fractions, I’m going to try and simplify both sides and make it as compact as possible.*0281

*This means I'm going to use my distributive property on the left so they do not have my x inside parentheses 2x + 6.*0291

*Looking at both sides of the equation now, there is not much I can simplify if I was looking at just left or if I was looking at just the right.*0310

*What I want to do is to try and isolate my variable, in other words try and get it all alone.*0320

*Since I have the next on the left and on the right, I have to work on getting these together first.*0326

*In order to get them together, I will use my addition property of equality to add the same thing to both sides.*0333

*What I will add is -2x, we will put that on both sides and see the results.*0340

*2x and -2x, those two would cancel each other out and get rid of each other and just be left with 6.*0351

*On the right side, I have 4x - 2x so only 2x left – 8.*0358

*That is good, I worked out that way because now I only have one x to deal with and I can work on isolating that and getting it all by itself.*0368

*How am I going to do that? I better move this into the other side by adding 8 to both sides.*0377

*6 + 8 that will give me 14 and -8 + 8 will cancel each other out and be gone.*0384

*I have 14 = 2x, we just divide both sides by 2, I have 7 = x.*0400

*It looks like we have found a solution something that will make our equation true.*0415

*My question is does it actually work or not?*0420

*On to that last step, the one we actually see if this is the solution and we do that by substituting it back into the original problem.*0424

*I’m going to write the original problem, I’m going to write it all except for those x’s.*0431

*I’m going to take that value and put it in wherever I saw an x, let us see.*0443

*I had an x + 3 put 7 there and 4 × x now 4 × 7.*0451

*I'm doing here is simplify both sides of our equation now, let us see if it balances out.*0458

*2 × 7 + 3 is 10, 2 ×10 that would be 20, I’m not sure if these are equal, if we continue we find out.*0465

*4 × 7 is 28, 28 - 8 is 20, sure enough it looks like the value of 7 does make this equation true.*0479

*I know that 7 = x is my solution.*0494

*Let us try another this one, 3/4x – 1 = 7/5.*0498

*This one contains fractions so remember how I suggested taking care of them and you do not have to worry about quite as much.*0507

*We are going to try and find a common denominator that we can just multiply it by and get rid of our fractions.*0515

*Looking at our denominator, I have 4 and 5.*0522

*A common denominator in this case would be 20.*0529

*I’m going to multiply both sides by that 20, 20 and 20.*0541

*Let us write down the rest of our equation as well, 3/4x -1 and 7/5.*0555

*Sometimes it is no fun to deal with numbers like 20 but watch what it does when I start distributing it on the left side here,*0567

*and we will multiply the 20 and 7/5 together.*0573

*20 × 3/4 would be a 15x and 20 × 1 being – 20.*0578

*We do not have to deal with fractions on the outside of the equation anymore.*0591

*Perfect, I like it.*0593

*20 × 7/5 if we want to cancel out 5 from there, when you are looking at 4 × 7 = 28.*0595

*I just have to solve 15x – 20=28, let us work on getting an x all by itself by isolating it.*0608

*I will add 20 to both sides giving us 15x equal to 48 and then we will divide both sides by 15.*0622

*This will give us that x is equal to 48/15 but if we want we can even reduce that a little bit further.*0650

*I think 3 goes into the top and into the bottom.*0657

*This will use 16/5, not bad. *0665

*You have x equals 16/5, now get in front of that last step.*0672

*Does it equal to 16/5? Does it not? Let us find out by playing it back into the original equation.*0676

*I’m just testing it out and see if it holds true.*0682

*The original equation is ¾ and I have that x, that is where we will put our number -1 and we will see if it is equals 7/5.*0686

*Let us see our number was 16/5, check and see what happens.*0700

*I can take out a 4 from the 16 so this would be 12/5 – 1, it will give me a common denominator of 5.*0707

*I would have 12/5 – 5/5 and how do we know? That is equal to 7/5.*0734

*This one definitely checks out, I do know that 16/5 is my solution, it looks good.*0743

*We will have something like this one, more fractions here and that is good.*0755

*4 + 5x = (5 x 3) + x and see what we can do this one.*0758

*Let us simplify both sides of the equation first and then go from there.*0765

*I'm going to distribute my 5 into the parentheses.*0769

*4 + 5x = 15 + 5x, it looks pretty good.*0775

*I want to get my x’s together and hopefully isolate it.*0786

*We will subtract 5x from both sides, 4 = 15.*0791

*Look at what happened there, when I subtract 5x from both sides, I lost all of my x’s completely.*0805

*Even worse than that than I thought I have left over this 4 -15 that is not true.*0812

*That does not make any sense, 4 does not equal 15.*0819

*What I was left here is known as a false statement.*0823

*What this is telling me is that since I did not make any mistakes that you do not know the value for x is going to work.*0828

*This is an example of one of those situations where we have a contradiction.*0837

*No matter what value for x you are trying use, it simply not going to work in this equation.*0849

*In fact, no solution exists for it.*0854

*Let us look at another one and see if we have any better luck.*0860

*In this one is -x + 3 = (1 + 4) - 2x/2.*0863

*We have a fraction, I'm going to try and get rid of those fractions first by multiplying everything through by a 2.*0869

*I will multiply it on the left side and I will multiply it on the right side, both by 2.*0877

*Just to keep things nice and balanced.*0886

*-x + 3, 1 + 4 - 2x / 2.*0890

*We multiplied both sides by 2, we can go ahead and distribute it.*0901

*Let us see what the result will be.*0907

*-2x + 6 equals, I have 2 + and when it distributes on this second part here, it is going to get rid of that 2 in the bottom, 4 - 2x.*0911

*Looking good, let us continue trying to combine our terms specifically those x’s and the numbers.*0933

*I will add 2x to both sides and you will get 6 = 2 + 4.*0943

*You will notice in this one that we lost our x’s but this one is a little different.*0958

*Instead of being left with other nonsense, this one we are left with 6 = 6 which is actually something that is true.*0963

*Before you get too worried, of course check all the steps and make sure they are all correct, which they are.*0977

*What this is trying to indicate here is that we have an identity.*0982

*It does not matter the value of x, you can use any value and the equation is still going to ring true.*0985

*Let us mark this one as an identity.*0992

*In short, that is a great way that we can identify both of those situations.*0999

*If it has an actual solution, it is a conditional equation.*1003

*You usually go through the solving process and you will be able to figure out what that value is and able to test it and see if it actually works.*1007

*If it is a contradiction, so nothing works, then you will go through the solving process and usually your variable drops away entirely,*1014

*and what you are left with is a false statement, it does not make sense.*1022

*That is even after all of your steps are completely correct.*1026

*If you have an identity like this one then you will go through that solving process and you will be left with a true statement*1029

*which indicates that you could use any value want for x.*1036

*Let us try some more examples.*1042

*In this next one I have 1/3x – 5/12 = ¾ +1/2 x.*1046

*There is a lot of fractions in here, let us see if we can take care of them all by getting some sort of common denominator.*1053

*What would 3/12, 4, 2 go into?*1062

*I think our common denominator would have to be a 12.*1067

*We are going to multiply both sides by 12.*1072

*Let us write everything we got here, 1/3x – 5/12 and I have ¾ + 1/2x.*1084

*Then we are going to distribute through on both sides and see what we get.*1100

*1/3 of 12 would be 4x, 12 × 5/12 would give us 5, then I will have 12 × 3/4.*1108

*That goes in there 9 times, then (12 × ½) + 6x.*1124

*Notice that common denominator, we clear out all those fractions then you do not have to worry about them.*1134

*What we have left here is 4x - 5 = 9 + 6x.*1139

*If you want to continue trying to simplify this by getting the x’s together and getting them isolated.*1144

*Our x over here and x over here let us get them together by subtracting 4x from both sides, -5 = 9.*1153

*6x – 4x that would be 2x, that looks good.*1170

*I only have one x to deal with.*1177

*Let us go ahead and subtract 9 from both sides, -9 -9 -14 = 2x.*1180

*There is only one last thing to do, divide both sides by 2.*1194

*It looks like our solution is -7 or we are not too sure until we check it.*1200

*Let us substitute it back into the equation and see what happens.*1211

*I have 1/3 - 5/12 = ¾ + ½ and into all of those blank spots where we are used to have the x, let us go ahead and put that -7 in there.*1215

*Let us see if these things are true or not, also -7 × 1/3, that is -7/3 – 5/12, is that equal to ¾ - 87/2?*1237

*I do not know, we got to do a lot more simplifying before we can figure that out.*1259

*We will use our common denominator to help us out, the common denominator of 12.*1263

*Let us see if I need a common denominator over there that would be -28/12.*1270

*A common denominator of 12 on the other side would be 9/12 minus, top and bottom by 640, 2/12.*1279

*Let us see what we can do, -28 - 5 would be -33/12.*1294

*What is going on the other side? 9/12 – 42.*1306

*It looks like I have another -33/12.*1310

*They are exactly the same and what that means is that x does equal in -7.*1314

*You can see when you go through that process of checking it, *1322

*it can be a lot of work especially if you have lots of fractions and stuff that you are dealing with.*1324

*But it is a good idea just to make sure that the solution you come up with is the solution.*1329

*Let us do one more.*1338

*Here I have (3x -15)/4 + (x + 47)/7 = 11.*1339

*This one involves fractions, let us take care of them out of the gate by multiplying it by a common denominator, let us use 28.*1347

*A lot to keep track of in here, (3x -15)/4 + (x + 47)/7 = 28 × 11.*1364

*We will take that 28, we will actually distribute it to both parts on this left side.*1379

*We will take it to this giant fraction here and I will take it to this giant fraction over here.*1384

*What this is going to do is, it is going to allow us to cancel out those fractions in the bottom.*1388

*28/4 would give us 7, that is still going to be multiplied by that 3x -15 part.*1394

*When it distributes on the second piece, 28/7 reduces and that goes in there four times. *1404

*We still have a 4 multiplied by x + 47.*1411

*Over on the side we have 28 × 11 that is 308, some fairly big numbers.*1417

*Let us continue distributing and see what we can do.*1426

*Let us take this 3 × 7, we will also take the 15 × 7 and do the same thing with 4 on this side.*1429

*Hopefully free of those x’s and get them out of parentheses.*1440

*21x - (7 × 15) = 105 + 4x.*1444

*Now I have 4 × 47 = 188, again, some big numbers but just have to push on through.*1461

*I have more than one copy of x here let us get those guys together.*1476

*I also have some things that do not have an x, let us get those together as well.*1480

*21 + 4x would be 25x - 105 and 188, I think that would give us 83.*1487

*I only have a single x to deal with so I will simply work on getting that isolated.*1508

*Let us subtract 83 from both sides so that the 25x is the only thing on the left side.*1516

*308 – 83 will give us 225, almost done.*1529

*Let us go ahead and divide both sides by 25 now.*1539

*This will give us x = 9.*1549

*This one is a lengthy process to get all the way down to just x = 9.*1554

*Even when you get that far, go ahead and double check it just to make sure that it actually works out.*1558

*I have my (3 × x -15)/4 on one side and both in those blank spots, let us go ahead and put that 9.*1571

*Let us start to simplify and see what we will get.*1589

*3 × 9 = 27/4, I see a 9 + 47 = 56.*1592

*We are hoping that this will equal 11.*1610

*27 - 15 that is 12/4 that is 56/7, we are getting closer.*1614

*12/4 is 3, 56/7 let us see that goes in their 8 ×, 3 + 8 =11.*1625

*Sure enough this one checks out.*1639

*I know that x does equal 9.*1642

*When going through the solving process, you do have a lot to keep track of.*1648

*The most important thing is truly working getting those x’s all by themselves and isolate it so that they are the only thing on one side of the equation.*1652

*If you have to deal with fractions, use a common denominator to clear them all out.*1660

*And also definitely go back and check your solution to make sure it is a solution.*1664

*Thank you for watching www.educator.com*1671

*Welcome back to www.educator.com.*0000

*In this lesson we will take a look at solving formulas and that is our only objective for our list of things to do.*0002

*But one thing I want you to know while going through this one is how similar formulas to solve the equations.*0011

*If you are curious what exactly is a formula and how they differ from those linear equations that we saw earlier.*0020

*A formula is a type of the equation and usually conveys some sort of fundamental principle.*0025

*Below, I have examples of all kinds of different formulas, D = RT and I = P × R × T.*0031

*What these represents are usually something else.*0039

*For example, D = RT we will look at that a little bit later on because it stands for distance = rate × time.*0042

*I = PRT, that is an interesting formula and it stands for interest = principal, rate, time.*0058

*It is not that these equations are too unusual but they actually have lots of good applications behind them.*0075

*One thing that is a little bit scarier with these formulas is you will notice how they have a lot more variables than what we saw earlier.*0081

*They have a lot of L, W and other could be multiple variables for just a single formula.*0089

*You know we have those multiple variables present in there and usually we are only interested in solving for a single variable in the entire thing.*0097

*The way we go about solving for that variable is we use exactly the same tools that we used for solving several linear equations.*0105

*You have the addition property for equality and you have the multiplication property for equality.*0114

*Both of these say exactly the same thing that they did before.*0120

*One, you can add the same amount to both sides of an equation, keep it nice and balanced.*0124

*And that you can multiply both sides of an equation by the same value and again keep it nice and balanced.*0129

*The only difference here is that we will be usually adding and subtracting or multiplying, dividing by a variable of some sort.*0135

*One thing that we usually thrown in as an assumption is that when we do multiply by something then we make sure that it is not 0.*0142

*If I do multiply it by a variable, multiply by both sides by an X, then I will make the assumption that X is not 0.*0151

*Just to make sure I do not violate the multiplication property of 0.*0158

*You have lots of variables and the very first thing is probably identify what variable you are solving for.*0167

*I usually like to underline it, highlight it in some sort of way just to keep my eye on it.*0174

*If you have multiple copies of this variable, try to work on getting them together so that you can eventually isolate it.*0180

*Some of these formulas do involve some fractions.*0186

*Use the common denominator techniques so you can clear them out and not have to worry about them.*0189

*Simplify each side of your equation as much as possible then isolate that variable that you are looking for.*0196

*Try and get rid of all the rest of the things that are around you by using the addition and multiplication property.*0203

*You can check to make sure that your solution works out by taking your solution for that variable and playing it back into the original.*0212

*When you do this for your formula, what you often see is that a lot of stuff will cancel out and you will be left with a very simple statement.*0221

*Usually we do not worry about that one too much unless we have some more numbers present in there but you can do it.*0229

*Let us actually look at a formula and watch the solving process.*0238

*You will see that it actually flows pretty easily.*0242

*In this first one I have P=2L + 2W.*0245

*This formula stands for the perimeter of a rectangle.*0248

*It goes through and adds up both of the lengths and both of the widths.*0251

*What we want to do with this one is we want to solve it for L.*0257

*Let us identify where L is in our formula right there.*0261

*What I’m going to do is I’m trying to get rid of everything else around it.*0265

*The 2W is not part of what I'm interested in so I will subtract that from both sides.*0271

*P and W are not like terms so even though it will end up on the other side of the equation, I will not be able to combine them any further.*0280

*Now you have 2 - 2 or P - 2W = 2L, continuing on.*0291

*I almost have that L completely isolated, let us get rid of that 2 by dividing.*0298

*Notice how we are dividing the entire left side of this equation by 2 so that we can get that L isolated.*0310

*What I have here is (P -2W)/2 = L.*0322

*One weird part about solving a formula is sometimes it is tough to tell when you are done *0328

*because usually with an equation when you are done you will have a number like L = 5 or L = 2.*0334

*Since you have many variables in these formulas that when we get to the end what we developed is another formula right here.*0340

*It is just in a different order or different way of looking at things.*0348

*This formula that we have created could be useful if we were looking for L many times in a row and we had information about the perimeter and W.*0351

*It is solved, it is a good the way that it is and the way we know it is solved is because L is completely isolated.*0362

*Let us look at another one, in this one we want to solve for R.*0371

*I have Q= 76R + 37, a lots of odd ball numbers but let us go ahead and underline what we would be looking for.*0375

*We want to know about that R.*0382

*I think we can get rid of a few fractions and we would have to multiply everything through by 6.*0387

*Let me do that first, I will multiply the left side by 6, multiply the right side by 6.*0395

*Q = 76R + 37, on the right side we better distribute that 6 and then we will see that it actually does take care of our fraction like it should.*0403

*6Q = 7R, I know I have to take care of 37 × 6.*0422

*I guess I better do some scratch work.*0434

*37 × 6 I have 42, 3 × 6 is 18 + 4 = 22, a lot of 2.*0436

*6Q = 7R + 222, I do not have to deal with any more fractions at this point so let us continue isolating the R and get it all by itself.*0452

*I will subtract 222 from both sides, awesome.*0463

*One final step to get R all by itself, we will divide it by 7.*0482

*(6Q – 222)/7 = R and I will consider this one as solved.*0490

*The reason why we can consider this one done is because we have isolated R completely.*0503

*Even though we do have a Q still floating around in there, it is solved.*0508

*Do not get too comfortable with that, it is not equal to just a single number, very nice.*0513

*This one is a little different, we want to solve the following for V.*0521

*The reason why this one is a little bit different is I have a V over here but I also have another V sitting over there.*0525

*When you have more than one copy of the variable like this, you have to work on getting them together before you can get into the isolating process.*0534

*Let us take care of our fractions and see if we can actually get those V’s together and work on isolating it.*0543

*We only isolate one of them then it is not solved, we still have a V in there.*0551

*To take care of our fraction I will multiply both sides of this one by W, it is my common denominator.*0556

*(RV + Q) / W = W × 5V, on the left side those W’s would take care of each other.*0569

*I will be left with RV + Q = W × 5V.*0585

*In this point I do not have to deal with the fractions but notice we have not got those V’s any closer together so let us keep working on that.*0594

*If I'm going to get them together, I at least better get them on the same side of the equation.*0602

*I'm going to subtract say an RV from both sides right.*0609

*Now comes the fun part, I still have two V’s, I’m going to make them into one.*0627

*The way I’m going to do this is I’m going to think of my distributive property but I’m going to think of it in the other direction.*0633

*If both of these have a V then I will pull it out front and I will be left with W5 – R.*0641

*This step is a little bit tough when you see it at first but notice how it does work is that I'm taking both of these and moving them out front into a single V.*0653

*I'm sure if it is valid, go ahead and take the V and put it back in using our distributive property and you will see that you be right back at this step.*0664

*It is valid.*0673

*The important part of why we are using it though is now we only have a single V and we can work further by isolating it.*0675

*How do we get it all by itself, these V’s be multiplied by 5W – R, that entire thing inside the parentheses.*0682

*We will divide both sides by that and then it should be all alone 5W – R = V.*0690

*I literally just took this entire thing right here and divided it on both sides.*0702

*To your left, now I can call this one done because V is completely isolated, it is all alone.*0709

*There is no other V’s running around in there.*0717

*I have worked hard to get them together and I know it is completely solved.*0719

*Welcome back to www.educator.com.*0000

*In this lesson we will look at applications of linear equations.*0002

*Think of those linear equations but more of a word problem sense.*0006

*Some of the things we will cover is the six step method that you can use to approach some of these word problems and to use that method.*0012

*We will have to look at how you can start translating these math things into actual equations.*0020

*We will also loot at very specific types of word problems that you might encounter such as when you just have some unknown quantities involving numbers.*0028

*Some of that involve some trigonometry and ones that might just involve some consecutive integers.*0036

*How should you approach some of these word problems?*0046

*In the six step method, we like to try and understand as much about the problem as we can.*0049

*That is why in the first step we are looking at reading the problem as much as possible and gathering up information and trying to understand it.*0055

*Once we think we have enough information it is a good idea to assign some sort of variable to the unknown.*0063

*In some of these problems, it might look like there is more than one unknown *0069

*and in cases like that you should assign a variable to the one that you know the least amount about.*0072

*From there we will be able to actually start writing equation from the information of the problem.*0080

*This is probably the most difficult step and sometimes you might get stuck on that one.*0084

*Once we have an equation when you go ahead and move on to solving that equation and see if we can actually get a solution.*0091

*We are not done even when we do get a solution it is a good idea to state the solution in the context of the actual problem.*0098

*That way we know if we get done we will say x = 3.*0104

*What exactly does x represent? Is it the number of miles driven?*0107

*Is it the number of beans in our basket? What exactly is the variable?*0111

*It is also important that we identify it early because it will make step five much easier.*0117

*When we are all done, we want to make sure that we check our solution to make sure that it is reasonable.*0125

*Especially with a lot of word problem, sometimes you might get a solution that just does not make sense in the context of the problem.*0131

*For example maybe I'm going through on solving something and I actually get time being -3 seconds.*0137

*It might be a problem because we can have negative seconds being time.*0143

*It is definitely a good idea to check your problem, check your solution and make sure that they make sense.*0147

*One thing I will note is that these steps often make it look like a nice linear process like you have simply moved through 1, 2, 3, 4, 5, 6. *0154

*In practice that may not always happen.*0163

*Do not be too surprised if you end up going through step 1, step 2, and you try to build that equation in step 3*0165

*and you realize you might not have all the information you need.*0171

*In cases like that, it is good to go right back to step 1 and see if we can gather up even more information.*0175

*When you are completely done with solving the problem, you will eventually move through all 6 of these steps*0180

*but you may end up bouncing back a few times if it looks like you do not have quite enough information.*0185

*If I’m taking a lot of word problems and being able to create equations then we want to be on the lookout for some keywords that are in those word problems.*0196

*We will use these keywords, these nice common terms to help us translate some of what we see in the problem into the actual formula.*0204

*Here are some terms that you should be familiar with.*0212

*When we are looking at addition, you want to look for words such as sum, together, total, more than or added to, *0215

*that are usually your clues that something will be added.*0224

*Think of a number and 3, that would say I want x + 3.*0227

*For subtraction we are looking for words like difference, minus, less than and decreased by,*0234

*all of those usually signify subtraction.*0239

*You have to be a little bit careful on the order of subtraction, for example you might see something like 3 less than a number*0242

*and that would say you have like x – 3 because that would be 3 less than that number x.*0250

*The order is important and sometimes it seems a little backwards.*0257

*For multiplication, look for product, times, twice or three times they might say something like that and percent of.*0261

*Those all indicate that something should be multiplied together.*0270

*Division has some good terms, you are looking for quotient, divided by, or ratio.*0276

*In all of our equations, we will need an equal sign there somewhere so we will be looking for equals is or will be.*0283

*That would be a good way to signify where the equal will go in the equation.*0292

*We will definitely come over our problems and look for some of these key terms so we can build our equation and solve from there.*0296

*Let us go ahead and jump into some of the examples and we will see what we have.*0305

*Let us first just start off by reading it and seeing what information we can dry out.*0309

*This one says the quotient of a number and 6 is added to twice the number and the result is 8 less than the number, find the number.*0314

*It looks like I see lots of things packaged up in here and just coming over.*0324

*I'm dealing with the quotient, remember that is division.*0330

*Added to means there are some addition.*0334

*Twice the number means we have some multiplication.*0337

*The result is there is our equal sign.*0340

*8 less than the number, this will represent subtraction.*0344

*There are lots of good pieces that are flying around in there.*0349

*I think the very thing I need to do now that I got some good information is.*0352

*Let us go ahead and identify our unknown.*0356

*If I’m going to be using an x, also that x is the number and that is the one we are looking for.*0360

*We will use that in our equation to package up all the rest of this information.*0371

*We have identified are variable, let us see if we can come over, and start putting it together.*0376

*I have the quotient of a number and 6.*0382

*The quotient of our number and 6 would be division or x ÷ 6.*0386

*That entire piece is added to twice the number, notice how I’m using multiplication.*0393

*The result is equals 8 less than the number, this is the one that seems a little backwards.*0406

*But we will take our number and subtract 8 so that we can get 8 less than our actual number.*0415

*We have read it over, identified our variable, we set up our equation, now we have to move through the actual solving process.*0420

*In this process we use a lot of our tools as before.*0428

*We try and clear out some of our fractions with a common denominator and get those x’s isolated onto one side.*0431

*With this one, I’m going to multiply everything through by 6.*0437

*This will definitely help us take care of that fraction that x ÷ 6.*0453

*We will use our distributive property, that will give us x + 12x = 6x - 48. *0461

*Now we do not have to deal with our fractions and we can just work on simplifying each side of the equation and getting our x’s alone.*0477

*I see some like terms on the left, 1x and 12 x = 13x, 6x – 48.*0484

*I simplified each side now let us go ahead and move the 6x to the other side.*0495

*It looks like we are almost done.*0514

*Finally, we will divide by 7 the x will be completely isolated.*0517

*It looks like our number, the one we are looking for, is a -48/7.*0532

*In terms of does this fit the actual context of the problem?*0539

*This one is less of a real word problem that is why I do not have to deal with time or distance. *0542

*It seems a perfect reason if that this is the number we are looking for.*0547

*We can always check the number to make sure that it works by taking it, and putting it back into the original.*0552

*If I want to take this and put it all the way back into all of the x here it should work out just fine.*0559

*That is the process in a nutshell to get your solution.*0568

*Let us look at some ones that are little bit more real world-ish and see how we can move through those.*0573

*This one says the perimeter of a rectangle is 16 times the width and length is 12 cm more than the width.*0583

*Find the length and width of the rectangle.*0590

*It seems like a pretty good problem I want to get as much information on this as I can.*0594

*I’m going to start off by trying to draw a picture of what we are dealing with here.*0599

*It sounds like we have some sort of rectangle and the perimeter of the rectangle is 16 times the width.*0604

*It may be my unknown here should be the width.*0613

*Let us mark that w is the width.*0618

*The length is 12 cm more than the width and I can mark out and then find the length and width of the rectangle.*0630

*I draw a little picture to give myself a sense of what is going on.*0645

*I have labeled my unknown w being the width and now I think we have enough information to start packaging this together into an equation.*0649

*Let us see what we can do.*0657

*The perimeter would be the sum of all the sides.*0659

*w + w that would take care of these two sides right here, + w + 12 + w +12 and that will take care of both of my lengths.*0664

*That entire thing is the perimeter right there and it is equal to 16 times the width.*0687

*It looks like a pretty good equation.*0696

*We just have to work on solving to make it all makes sense.*0698

*Let us see on the left side I have lots of w's and we can go ahead and combine all of them together into just 4w since all of those are like terms.*0703

*Then I have a 12 + 12 so we will say that is 24, all equal to 16w.*0715

*Simplify each side, let us go ahead and put our w's together, 24 = 12w.*0727

*One last step, divide both sides by 12 and looks like we get that 2=w.*0740

*It starts the work where we interpreted in the context of the problem.*0753

*Since we identify this earlier that w is the width I know that this represents the width of the rectangle 2.*0756

*Since our units in here are in centimeters and let us say that our width is 2 cm.*0764

*That is all we wrote we would actually be in a lot of trouble because we have not answered the entire problem.*0778

*In the entire problem we want to know the length and the width of this rectangle.*0785

*We only have the width so let us use this width to see if we can find the other part.*0789

*Since it says that the length is 12 cm more than the width then we can simply add 12 to this 2 and we will get the other.*0796

*Length is 14 cm.*0807

*We have answered both parts of this question I know that the width is 2 cm and the length is 14 cm.*0819

*Let us look at another one, this one says during a sporting match the US team won 6 more medals than Norway and the two countries won a total of 44 medals.*0831

*How many did each country win?*0840

*This is one of those examples where I said it looks like there is more than one variable in here.*0842

*After all, we have no idea how many medals that US won and I have no idea how many medals Norway won.*0847

*Let us see if we can pick apart and see which one we know the least about.*0853

*During a sporting match the US team won 6 more medals than Norway, I know a little bit about the US, *0856

*they won 6 more than Norway and the two countries won a total of 44.*0862

*It looks like the one I know the least amount about is actually Norway.*0867

*I have no idea how many medals they won.*0871

*Let us set that up as our variable so x is the number of medals won by Norway.*0872

*If the US team won 6 more medals than Norway, I could represent that using addition, x + 6 that would represent the number of medals that the US won.*0898

*I want to look at their total wins and it should be 44.*0913

*I will add Norway's medals to that should equal 44.*0918

*Let me highlight the parts of this equation so you can see exactly what I'm doing here.*0923

*This x right here is the number of medals from Norway and this over here is the number of medals from the US.*0927

*You can see I'm taking both of those quantities and put it together and look at the total number of medals from both one.*0938

*Now that we have our equation, again let us work to solve it.*0946

*My two like terms I got x and x that would give me a 2x then I can work to get my x’s all by themselves.*0950

*Let us go ahead and subtract 6 from both sides 2x = 38.*0964

*Now we will divide by 2, this will give us x = 19.*0972

*We have got to the problem and we figured out what x is but again interpret it in the context of the problem.*0993

*x = 19 but what is x?*1000

*Earlier we said that x is the number of medals won by Norway so I know that Norway won 19 medals.*1003

*Let us write that down.*1011

*We are not done yet, we still have to say how much each country won.*1025

*I also need to know how many the US won.*1030

*Since we know that the US won exactly 6 more than Norway and I can just take 6 and add it to the 19, the US won 25 medals.*1033

*I just have to double check this quickly, add these together and make sure we get 44.*1050

*19 + 25 = 44 total, it looks like we are doing okay.*1054

*Let us look at another example and see how well that one turns out.*1074

*In this example, we are going to look at some angles and a little bit of trigonometry.*1080

*It is okay if you do not know anything about trigonometry, I will give you all the background information you need for this one*1084

*You can see how you need to put it together.*1089

*This one says find the measure of an angle such that its complementary angle and its supplementary angle both added to be 174°.*1092

*After looking at these terms in here and saying wait a minute what is supplementary and what is complementary?*1101

*Let me tell you.*1107

*Two angles are said to be complementary if you can add them together and they add to be 90°.*1108

*90° is complementary.*1114

*Two angles are said to be supplementary if you can add those together and you get 180°.*1117

*180° is supplementary.*1122

*What we are looking to do is you have some unknown angle and it has a complement and a supplement and we are looking at those two to get that 174°.*1126

*Let us first identify our unknown in this problem and that would be our angle, x is the unknown angle.*1137

*We have to use this unknown angle in order to build what its complementary angle would be and what its supplementary angle would be.*1154

*That is going to be a little bit tricky, because we want to know what would you have to add to this unknown one in order to get an x? *1162

*In order to get 90 and since we do not know, how could we figured that out.*1168

*Let us do a quick example.*1174

*What if we knew that our angle was 70°, what we have to add to that in order to get 90?*1175

*It would not take that long to figure out, you should just take another 20° and sure enough you get 90°.*1183

*If you think about how you came up with that, you will be amazed that all you have to do is take 90 and subtract the angle that you knew about.*1189

*We will say that 90 - x that right there will be our complementary angle.*1199

*In a similar way you can also build your supplementary angle.*1216

*You can say that it will be 180° minus whatever angle you started with.*1220

*We have our complementary angle and we have our supplementary angle.*1239

*In the context of the problem you want to be able to look at these two added together so that would be 174°.*1242

*Let us just take both of these expressions here, put them together.*1249

*(90 – x) + (180 – x) there is our complementary and supplementary must be equal to 174°.*1253

*Let us work on combining our like terms and see what we can get.*1266

*90 + 180 = 270, -x, -x, -2x = 174.*1270

*We will subtract 270 from both sides, -2x = -96 and then let us go ahead and divide by -2, x = 48.*1286

*It looks like our unknown angle is 48°.*1314

*We can check to make sure that this is our angle by using it back into the context of the problem.*1332

*If this angle is 48 then what do I have to add in order to get 90°?*1338

*In other words, what would be its complementary angle?*1343

*If I take 48 and I add 42 that will give me 90, so 42 is the complementary angle.*1353

*Let us do the similar process, what would I have to add to 48 in order to get 180° or even if you take 180 - 48 to see what its supplementary angle would be.*1365

*Looks like 132.*1388

*If we take both of these and add them together, sure enough we get 174° like we should, so we know that our angle is 48°.*1392

*One last example and in this one we will deal with looking at a lot of different numbers but there is only one unknown*1409

*and we will look at it as being the one we know the least amount about.*1416

*This one says we have a piece of wire and it is 80 feet long.*1421

*We are going to cut this into 3 different pieces with the longest piece being 10 feet more than the middle sized piece *1425

*and a short piece being 5 feet less than the middle sized piece.*1431

*Find the length of the three pieces.*1434

*It looks like we do not know anything about the short, the middle or the long one, but we will write them all just using one variable.*1438

*I know a little bit about all of them put together since I know that my total is 80 feet long.*1447

*I'm going to come in three pieces and the longest piece is 10 feet more than the middle.*1454

*I know a lot about the long one.*1460

*The middle sized piece and the shorter piece is 5 feet less than the middle sized piece.*1463

*I know a little bit about the middle one, its 5 feet more than the shorter one since the shorter one was 5 feet less.*1471

*I think what I need to set up as my unknown and we will say that x is the length of the short piece.*1479

*I will try and describe all the other pieces using just that short length right there.*1499

*The short piece is 5 feet less than the middle piece.*1505

*If I take x and I add 5 to it that should give me my middle piece just fine.*1514

*The long one is 10 feet more than the middle, then I can start with the middle and add another 10 and this would represent the long piece.*1528

*I need to take my short piece, my middle piece, and my long piece to be able to put all those together and represent the 80 total feet of wire.*1543

*Let us set up this equation.*1552

*short piece = x + our middle piece x + 5 + our long piece x + 15 all of this should equally total of 80 feet.*1554

*We got our equation so let us work to solve it.*1572

*Adding together our like pieces, I have 3 x’s but I can put together the 5 + 15 = 20.*1575

*I have 3x + 20 = 80.*1588

*Let us subtract 20 from both sides, giving us 3x = 60.*1593

*We will divide both sides by 3 and we will see what x needs to be, x = 20.*1604

*It looks like our shorter piece of wire is going to be 20 feet long.*1613

*We also want to identify what all the other ones need to be.*1636

*We can use these smaller expressions over here to figure out what they need to be.*1639

*Since the shorter piece is 5 feet less than the middle piece, you can simply add 5 to this and get that middle.*1646

*The middle piece is 25 feet long.*1654

*Our long piece since it is 10 more than middle one, we will add 10 + 25 = 35 feet long.*1668

*Now I have the information of all three bits of wire.*1685

*If we add all of these up, we should get a total of 80.*1691

*20 + 25= 45 + 35 sure enough adds up to a total of 80 feet.*1695

*You can see that the process of interpreting these word problems can be a little tricky *1703

*But if you look for those keywords and try and hunt down your unknowns as much as possible, it can work out pretty well.*1708

*You can always look at the context of the problem to make sure your answers make sense.*1714

*Thanks for watching www.educator.com.*1719

*Welcome back to www.educator.com.*0000

*In this lesson we are going to continue on with our examples of applications of linear equations.*0003

*In this we are going to look at some trickier applications, the ones involving lot of motion and mixtures.*0008

*It usually gives students a lot of headaches so pay close attention on how we approach this with some nice methods that will organizer our information.*0013

*Our two big goals, motion and mixture problems is what we want work at.*0026

*And the two methods we will use to attack these will be using the table method and the beaker method.*0031

*You will see that the table method usually works pretty good for a lot of motion problems *0037

*and the beaker method works pretty good when you are dealing with those mixtures.*0040

*As I said earlier, motion and mixture problems tend to be a weak area for many people.*0048

*The problem comes from being able to take all the information and organize it in some sort of way so they can actually build your equation.*0053

*In terms of equation, there is only two equations that work in the background.*0061

*For a lot of motion problems, it is the distance, the rate at the time, that all need to be related together.*0066

*The equation for that is distance = rate × time.*0073

*Let us go ahead and put it in there.*0077

*We have our distance, we have our rate, and we have our time.*0078

*With our mixture problems, there is a very similar formula that we are after and that is the one that deals with the pure substance in the mixture.*0086

*For that one we are looking at the amount of pure substance, multiply it by some sort of percent and the amount in the total mixture.*0094

*This guy over here is just the pure amount.*0112

*To help us organize these we will begin by using what is known as the table method.*0120

*In the table method, you build a small table usually has about three rows and about four columns in it*0125

*and we end up putting into this table is information about your distance, rate, and time.*0132

*The information about different things that is moving in the problem.*0137

*Every row in this table here usually represents a different object.*0142

*If I’m talking about to bicyclists, I might talk about bike number 1 and bike number 2.*0147

*That way I can keep track of each of their motions separately.*0157

*In each of the columns, we will put information about the rate, distance, and the time.*0160

*The way we usually do that is we have information about the rate here, the time here, and the distance over here.*0166

*What we are looking at is that equation of distance = rate × time, in fact it is hidden in the background.*0179

*rate × time = distance.*0186

*For that reason it will be important on how we pull the information out of this table.*0191

*We want to remember that we want to multiply across.*0196

*In this direction we will multiply the rate and time in order to get our distance and what we want to do from there is add down to get our total.*0200

*It will help us build that equation and even if it does involve a couple of objects moving.*0211

*The other method which is good for mixtures is known as the beaker method.*0219

*The way this method works is you end up creating a beaker for each of the mixtures you are putting together.*0224

*Maybe this one over here is my 30% solution, this guy over here is my 90% solution and I'm looking to create a new solution over here on this side.*0230

*You will notice I have already put in the addition and equals signs so I can see that I'm mixing two beakers together to get a third one over here.*0241

*There are two things that we want to record in each of these beakers.*0250

*First we will record the amount, how much is in this beaker.*0254

*Is it 50 L? Is it 40 L?*0260

*We want to know what the exact amount in each of these will be.*0263

*We will also want to know what percentage is each of the solution. 30%, 20%, we will definitely keep track of that for each of these beakers.*0269

*In order to create the equation from these we will multiply the amount and the percentage together from each of the beakers.*0279

*Watch for me to use this method when we get to those mixture problems.*0300

*Let us go ahead and see this method in action as we are looking at example 1.*0307

*This says that building an equation to represent the following problem.*0313

*We have two airplanes and they are leaving a city at the same exact time and they are flying in opposite directions.*0317

*If one flies at 410 mph and the other one 105 mph faster, how long will it take them to be a total of 3290 miles apart.*0323

*You can see what I'm saying, there is a lot of information in this problem that is very confusing to try and keep it all straight.*0333

*It will help us out, let us keep track of both of these planes.*0339

*We have plane 1 and plane 2, what do we want to know about these planes?*0343

*We want to know how fast they are going so we will keep track of their rate.*0357

*We want to know how long they have been traveling, their time.*0361

*Of course we want to know how far they have traveled so we can get their distance.*0364

*Let us see what we can now figure out about these planes and put it into the table.*0372

*They are leaving a city at the same time and fly in opposite direction and one flies at 410 mph.*0378

*Let us say that is plane 1, they are traveling at 410 mph and the other one is 120 mph faster, if I was to add 120 to 410 that will give me 530 mph.*0384

*I know how fast each of these planes are going, that is definitely good.*0401

*How long have they been traveling now?*0404

*I’m not sure I got a lot of information about that because that is what I'm looking for in the problem.*0408

*It says how long will it take them to be 3290 miles apart?*0413

*My time here is completely unknown.*0418

*I do not how you know long each of them have been travelling.*0424

*To think about how the distance is working to all of this, remember the distance is equal to the rate multiplied by time.*0428

*I have multiplied across to get my distance is 410x for plane 1.*0436

*My distance for the other one is 530x.*0445

*I get a sense of how far each of them had gone and I want to know when they will reach a total of 3290 miles.*0452

*I’m looking at their total distance.*0460

*We will add this last column down.*0462

*The distance from plane 1+ the distance from plane 2 must equal a total distance of 3290 miles.*0466

*You can see that the table helps out so we can build this equation.*0479

*It organizes that information and we can crunch it down.*0485

*Now that we have this equation, we can not necessarily stop.*0490

*We have to move forward and figure out what the solution is.*0492

*Let us see if we can solve this.*0495

*3290 miles and we also have the 410x + 530x = 3290.*0503

*The first thing one I want to do is get my x’s together.*0518

*Let us see what that will be, 940x = 3290.*0521

*We will divide both sides by 940 and this will give me x = 3 ½.*0533

*In the context of the problem, what was that mean?*0548

*x was our time so this tells us that the planes traveled for 3 hours and ½ of an hour or 30 minutes.*0551

*When we use this table again to set up another problem and you will see how it helps us organize that information.*0577

*In this example, we are dealing with two trains and they are leaving the downtown station at the same time and they are traveling in opposite directions.*0587

*One had an average speed of 10 mph more than the other one and at exactly 1/5 of an hour they were 12 miles apart.*0595

*What are their two speeds?*0603

*We are going to start this one often much the same way we did the last one.*0605

*We have two trains, we will label our rows.*0607

*I have train 1 and train 2.*0611

*I want to keep track of how fast they are going, their rates.*0618

*How long they have been traveling, their time?*0623

*How far they have gone, their distance?*0626

*Let us go to the problem over here to even pull out the information and put in the right spot in the table.*0631

*One have an average speed of 10 mph more than the other one.*0636

*That does not give me a definite speed for any of these trains.*0642

*I just know that one is traveling faster than the other one.*0644

*Let us say we have no idea how fast train 1 is going we will use our unknown for that.*0650

*Since the other one was traveling 10 mph faster, I can take its rate and add 10.*0655

*Onto time, it says at exactly 1/5 of an hour and that is pretty good.*0666

*At least I know how long each of them have been traveling so 1/5 of an hour for each of them in terms of the time.*0671

*Let us put these together and see if we can get our distance.*0683

*Multiply rate and time, I will get 1/5x, multiplying rate and time here 1/5x + 10.*0686

*After 1/5 an hour they were a total of 12 miles apart.*0702

*I'm looking at both of these distances added together so 1/5x + 1/5 x + 10 so when will their distances be a total of 12 miles apart?*0706

*There is my equation and I'm going to use and try and solve from here 1/5x + 1/5x + 10 = 12.*0727

*We will use our techniques for solving linear equations and see if we can pick this apart.*0748

*The one is coming to mine right now since I'm staring at those fractions is we will multiply by a common denominator to clear at those fractions.*0753

*Let us go ahead and multiply everything through by 5.*0760

*I will do that on the left side and I will do it on the right side x + x + 10 and 5 × 12 = 60.*0764

*We can start combining our like terms 2x + 10 = 60.*0781

*I will go ahead and subtract 10 from both sides and finally we will divide both sides by 2.*0791

*I will get that x = 25 but again what exactly does this represent?*0810

*If you remember when we hunted down our variables and we said what x was, this is actually the speed of our first train.*0815

*Let us write that down, train 1 is moving at 25 mph and since the other train is going 10 mph faster, we will say that train 2 is moving at 35 mph.*0822

*Now we have the speed of both trains.*0856

*Now that we have seen a couple of those the table methods, let us get into using the beaker method with some good mixture problems.*0861

*In this problem we want to know how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution*0868

* in order to get a new acid solution that is only 50%.*0875

*A lot of information and it is tough to visualize this one unlike the motion problem where you can actually see two trains moving.*0881

*This is why I use the beaker method so I have something a little bit more visual that I can grab onto each of our beakers.*0887

*Let us go ahead and start recording the amount and the percentage for each of these things.*0894

*Let us start with the amount so it says how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution?*0902

*It looks like it does not tell me how much of the 40% I have, so we will leave that as an unknown amount.*0912

*We will be mixing it with 80 mL of this other one.*0922

*The percentage of this beaker is our 40% and the other beaker is our 70%.*0928

*This new beaker on the, we are trying to mix the two together, how much will it have in terms of their amount?*0942

*I’m mixing some unknown amount + some 80 mL so this new beaker at the end will have a total of x + 80 mL.*0949

*It is important that the last beaker should have a total of both the amounts.*0960

*What is its percentage? This will have a 50% solution.*0965

*We have recorded the amounts, we have recorded the percentages, and we will multiply those two quantities together and actually get our equation.*0971

*(40% is the same as .4 multiplied by x) + (70% is the same as .7 multiplied by 80) all of this must be equal to .5 multiplied by x + 80.*0981

*By multiplying those two together and putting in the appropriate addition and equals I have now set up my equation and I can solve from here.*1000

*This one is like our equation that has fractions only this one has in decimals.*1008

*I’m going to multiply through by 10 to get rid of a lot of those decimals.*1013

*.4 multiplied by 10 = 4x, .7 multiplied by 10 will be 7 and .5 multiplied by 10 would be 5.*1024

*This is the equation that I need to solve in order to figure out the amount of that 40% solution.*1041

*Let me just copy this onto the next page and we will go from there.*1049

*I’m going to start combining together each side of this equation using my distributive property and then I can work on getting those x’s isolated.*1072

*4x + 560 = 5 × 8 another 0.*1082

*We will subtract 4x from both sides 560 = 400 + x subtract the 400 from both sides, so 160 = x.*1100

*What this is telling me is since x represents the amount of 40% solution, I know that we need 160 mL of the 40% solution.*1127

*It is always a good idea to keep track of what your unknown represents.*1148

*I have one last example and this one is using the beaker method again but this one involves a lot of coins.*1154

*This is to demonstrate that these two methods are flexible and you do not always have to use them *1162

*with just a pure mixture or even just with distance, motion and travel.*1167

*They still can apply to a lot of other types of problems.*1171

*This one says a man has $2.55 in quarters and nickels.*1176

*He has 9 more nickels than he does quarters.*1180

*How many nickels and quarters does he have?*1183

*We have to figure out what the two are mixed together.*1185

*It is like a mixture problem that is why we are going to use the beaker method to attack it.*1190

*The two things that I will try to keep track of is how much of quarters and nickels do I have and what would the amounts of each of these be.*1194

*I better label my beakers to make it a little bit easier.*1205

*We will say that this first one contains just nickels and the second one contains quarters.*1208

*Let us see what we know about the nickels.*1219

*He has 9 more nickels than quarters.*1221

*We do not know a lot about those quarters, do we?*1223

*The amount of quarters we will leave that as x and the amount of nickels there is 9 more than however many quarters he had, x + 9.*1226

*I can write down what each of them represent.*1239

*A quarter is the .25 and nickel.05.*1245

*What we are looking to do is to combine these together, the amount and how much each of them are to get a total amount in that last beaker.*1253

*He has a total of $2.55 and the reason why I’m not writing down a specific amount or specific amount per coin in there*1267

* is because this one represents the two already put together just the $2.05.*1277

*Let us go ahead and build our equation.*1283

*.05 multiplied by the amounts x + 9.*1286

*.25x = $2.55 this one is similar to the previous one.*1292

*It has lots of decimals running around and I want to get rid of a lot of those decimals.*1302

*I’m going to multiply this one by hundreds and then we can do it just some nice numbers and no decimals.*1307

*This would make it 5 × x + 9 and 25x = 2.55 that is the equation that we will solve and figure out how many quarters and nickels we actually have.*1316

*5x + 9 + 25x = 2.55 working to simplify this equation I will distribute through on the left side of the equation giving me 5x + 45.*1342

*I will go ahead and combine my like terms 5x + 25x that will give 30x.*1376

*Then subtract 45 from both sides and you can get that x a little bit isolated.*1390

*Finally let us go head and divide it by 30 to see if we can figure out what that x is, so x = 7.*1404

*In the context of this problem, x represents the number of quarters we have.*1416

*We have 7/4 if you remember we have exactly 9 more nickels than quarters so we can add 9 to this number and figure out the total nickels.*1423

*We have 16 nickels.*1440

*We have figured out the total amount of both coins.*1447

*In each of these types of methods, they are great ways that you can get your information organized and into a good spot.*1453

*It will be handy with motion and with mixtures.*1461

*Thanks for watching www.educator.com.*1465

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at the rectangular coordinates system.*0002

*It will help us out as we get into more of the graphing process with some of equations later.*0006

*Some of the specific things that we will look at as we are looking at our Cartesian coordinate system are just some of the parts of it.*0012

*We will learn all about the x axis, y axis, origin, quadrants and how we can plot an ordered pair.*0018

*When we get into plotting more things such as the graph we will look at looking at a table to do that graph.*0026

*And again how you can find the x and y intercepts of that line once on the graph.*0032

*The Cartesian coordinate system is formed by taking 2 number lines and putting them together after 0.*0041

*You will notice that one is in the up and down direction and one is completely horizontal.*0047

*Now the horizontal one, the one that goes completely flat, this guy right here, we will call this one our x axis.*0053

*The one that go straight up and down that will be known as our y axis.*0064

*The 0s of each of them when it both connects, we will also give that one a special name, we will call this point right here the origin.*0071

*The little tick marks that you see usually represent how far you have to go on each of these number lines.*0087

*You can see that it breaks down into these number lines into so much larger parts of this graph.*0095

*We will give each of these squares a name, we will call them quadrants.*0103

*We usually number these quadrants starting in the upper right corner, say quadrant 1 and move in a counterclockwise direction.*0110

*This entire space here would be quadrant 2 but I have quadrant 3 and quadrant 4.*0118

*Now in each of these quadrants we can plot a point.*0126

* A point is an ordered pair and we look at it as a pair of the x and y put together.*0131

*If I'm looking at a point say out here, I have identify what it is x and y value are *0138

*by looking at how may tick marks I have to go over on the x axis and this would be 2.*0148

*How may tick marks I have to go in the y direction, 1, 2, 3, 4, I can plot out that point.*0154

*Keep in mind that with all of these points that you see, the first one is the x and the second one is y.*0163

*Let us just do a real quick example of how we know little bit more about the parts of the graph and see if we can plot out some of these ordered pairs.*0177

*In all of these, the first one will represent an x value and the second will represent our y value.*0185

*We will start at the origin with each of these and moving the x and then the y direction.*0191

*Starting with first one, we need to plot 2, 4.*0198

*I’m going to start at the origin and I'm going to move 2 units to the right and 4 units up.*0201

*This point right here will be the point 2, 4.*0213

*The next point has a negative value in it and it is a negative in the x.*0222

*I actually move left 1, 2, 3 and then I will go up 7, that point is located right here, 4, -3, 7.*0227

*As long as you can keep track of which one was x and which ones what is y, not a bad task at all.*0244

*The next one is -1 and -5, at we will move in the x direction -1 and then down 5.*0251

*1, 2, 3, 4, 5, so there is our point -1, -5.*0258

*Just a couple of more to do, the next one is 3, -2, that is in the positive direction 1, 2, 3 and then down 1, 2.*0269

*And one last one to go, this one is 0, 4.*0286

*What should you do if 0, should you go right or should you go left?*0293

*Since it is at 0, we do not go right or left in the x direction, we actually stay at the origin and then we will move in the y direction 4.*0297

*Up 1, 2, 3, 4 that point is right here 0, 4.*0304

*Now that we have identified some points on here, we can actually say which quadrants each of these are in.*0313

*If we look at our first point up here, our 2, 4 it is in quadrant number 1 because it is in that first square.*0319

*The next point -3, 7 is in quadrant number 2.*0331

*-1, -5 quadrant number 3.*0338

*3, -2 that one is in quadrant number 4.*0345

*The last point 0, 4 as the tricky one is not in quadrant 1, it is not in quadrant 2, it is actually right on the y axis.*0351

*Some more things about this rectangular coordinate system, when we get into actually trying to graph an equation on here*0370

*we can look at it as a special relation between its x values and its y values.*0377

*There are a lot of good things we can say about that equation.*0383

*To actually visually see what that relationship is and of course put it on the graph, *0386

*we will go ahead and create a table of values and see how they are related.*0390

*Usually this is done by picking a lot of different values for one of the variables and see what the other corresponding other values need to be.*0395

*Here are very important terms you want to keep in mind.*0402

*Where the graph of that equation crosses the x axis that will be known as its x intercept.*0404

*In a similar fashion, if it crosses the y axis, we will call that the y intercept.*0415

*Just draw a general graph so this is not a line since it is all curvy but it still crosses the x axis in some spot so I will call this the x intercept over there.*0422

*It still crosses the y axis so it has a y intercept.*0440

*You want to be able to identify both of these points.*0446

*A little bit more about generating those points from the table.*0453

*I said that a line is a special relationship between its x values and its y values and to see that relationship, *0457

*we want to be able to create a little chart and put in some values for either x or y.*0463

*Let us look at one I have 5x - 3y =12.*0476

*If I want to go ahead and try and visualize what this equation looks like, I need to know what values go on my graph.*0486

*What point should be there?*0493

*A good way to figure out if the point is on the graph or not, is to see if it simply satisfies the equation.*0495

*If I’m just picking some point like 3, 1, I can substitute the 3 in for x and 1 for y and see if it does satisfy it.*0502

*And see if 3, 1 is on the graph or not.*0513

*(5 × 3) – (3 × 1) does that equal 12, let us find out.*0516

*That will be 15 - 3 = 12 or 12 = 12.*0524

*What that tells me is that 3, 1 is one of the points on my graph, 1, 2, 3 on the x axis up one.*0532

*Let us pick another point, 0, -4.*0546

*But when I plugged this one in, I get 5 × 0 which is 0 - 3 × - 4 = 12.*0554

*I know that one is on the graph as well, 0, -1, 2, 3, 4.*0570

*In practice, we usually do not pick up points out of thin air and then test to see if they work or not.*0583

*Usually we end up picking a lot of different things for one variable then see what it has to be for the other variable.*0588

*Let us get a little bit of space in here and see how we would do this more in practice rather than just picking things out of thin air.*0595

*Let us pick out some more values for x.*0603

*For example what if x was 2, what with that 4y to be, we can find out by putting in the 2 for x and actually solving for y.*0606

*That will be 10 - 3y = 12, I will subtract the 10 from both sides and get y = -2/3.*0620

*I know sure enough that is another point that I can put on my graph, 2 in the x direction then down 2/3.*0632

*What you may notice is I start to do more and more points over here on the graph.*0641

*You will get a better sense of what the entire graph looks like.*0645

*I have done about 3 points here but I’m going to draw lines for all of them to say that if I had even more points they would all line up along there.*0650

*Let us get into some more examples and see what other parts of the graph we can identify*0665

*and get a little bit more into the nuts and bolts of using that chart to graph out an entire line.*0669

*In this example we just simply want to find the x intercepts and the y intercept of the line.*0676

*Remember, this is where it crosses the x axis and where crosses the y axis.*0681

*My x axis is horizontal and I can see it crosses right at that point.*0686

*That would be at -1, 2, 3, 4, 5 in the x direction and 0 in the y direction.*0691

*x intercepts is at -5, 0 it crosses the y axis as well right up here at 4.*0701

*For that one we would not go in the x direction whatsoever but we would go out 4, so its y intercept is at 0, 4.*0717

*With these ones we want to determine whether the point is on the line or not.*0735

*We can figure this out by substituting the values in for x and y.*0739

*The first one will always be an x and the next one a y.*0746

*-2 + 3y = 21, let us put in the -3 for x and 5 for y.*0753

*-2 × -3 = 6, 3 × 5 =15, for a total of 21, 6 + 15 = 21.*0768

*It looks like this one checks out since it makes the equation true, we can say that -3, 5 is on the line.*0783

*Let us do the same thing for the next one, my x is 1 and my y is 7.*0794

*Let see if this one rings true.*0811

*Plugging in 1 for x, 7 for y and simplify 3 + 4 is 7 and 5 × 7 is 35, these things are definitely not equal.*0815

*What is that tells us that 1, 7 is not on the line.*0836

*You can take a point, plug it into the equation and see if it satisfies it and makes it true.*0845

*Let us get into more of the graphing process.*0852

*Here is an entire equation and I want to know what its entire graph looks like.*0854

*Rather than just picking points out of thin air and testing them, we are going to try and create a few values of our own by generating them.*0858

*We will generate them using a nice little table.*0867

*I will pick some values for x and we will see what makes y.*0870

*To make this processing a little bit easier, we will take our equation and we will solve for y first.*0875

*Our equation is 2x + 3y =12, to solve this for y I would move the 2x to the other side by subtracting a 2x and then divide it by 3, I will get -2/3x + 4.*0881

*This is the same equation, I just manipulated it a little bit so I can work with it a little bit easier.*0905

*We are going to pick some values for x, plug them in and see what they make y.*0912

*It does not matter what values you pick but I do recommend choosing some negative values or may be throw in 0 and also choose some positive values.*0918

*Let us choose -3, -2, -1, 0, 1, 2 and 3.*0927

*With each of these, imagine what happens if you take -3 plug it in into the equation, what result do you get.*0938

*If you do some scratch work that is okay, y = -2/3 we multiply that by -3, add 4 to it and see what the result is.*0944

*-2/3 × -3 that would give us 2, 2 + 4 =6.*0955

*You put that on the other side, I know that -3, 6 is one of the points on our graph.*0966

* -3, 1, 2, 3, 4, 5, 6 there is a point right up there.*0972

*Let us take up the -3 and try this again, this time we will plug in -2.*0982

*This would give us 4/3 + 4 which should be the same as 5 and 1/3.*0991

*It is okay if you get fractions with these just the way it turns out since you are choosing all kinds of different values for x.*1001

*-2 up 1, 2, 3, 4, 5 and just a third more there is that point right there.*1009

*Let us take out the -2 and put in -1 and let us see if this works out.*1019

*negative × negative would be a positive, 2/3 + 4 = 4 2/3.*1032

*-1 up 1, 2, 3, 4 and 2/3 right above there.*1043

*I have not even graphed all of the points yet and you can see that they all seem to be following along in a straight line.*1049

*We will do one more then will go ahead and connect the dots.*1058

*Let us see what happens if x =0, 0 × -2/3 would be 0, 0 + 4 = 4.*1066

*We will put that point on there as well right up here at 4 and now we will just connect all of our dots and this would be our graph of the equation.*1078

*Using this table we can definitely see a lot of points even sometimes we do not need to graph all the points,*1094

*we should graph enough of them that you have a good sense of what it looks like.*1100

*Also if your graphing a line and you are going through this table and you got one that is a little off from all the rest, it is okay.*1104

*But if it is off from all the rest then we will check your work on now to make sure that you have done it correctly.*1110

*In this next one, we want to graph the equation again using a table of values.*1121

*Like last time, I will go ahead and solve this for y so we have a little bit easier of the time working with it.*1126

*I will move that 5x together just by adding 5x to both sides then I will go ahead and divide by 2.*1133

*This is the equation that I will be using it is the same as the original.*1149

*I have just manipulated it a little bit so I will have an easier time working with it.*1153

*Let us pick some values to put in there for x so we can see what y needs to be.*1158

*We will pick some nice ones like -4, -2, 0, 2 and 4.*1165

*This one I’m picking values that are multiples of 2 since all of them we have to multiply by 5/2 first.*1172

*That will make our lives a little bit easier.*1179

*y =5/2, let us plug back in -4 then we will add 10.*1184

*-4 × 5/2 =-10 when we add 10 to that we get 0 so our first point is -4, 0.*1194

*Let us try another one, this one is -2, I will plug that one in there -2 × 5/2 =-5.*1212

*I will add 10 to that and get a total of 5 so this point -2, 5.*1229

*What happens if we go ahead and put in a 0, 0 × 5/2 = 0 + 10 = 10.*1250

*I have another point for my chart here and this one is off the chart 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 way up here.*1263

*We can already see that after doing a bunch of these, the rest of my graph is going to go completely off of my graph paper.*1275

*We will go ahead and do the other 2 just so you can see what their value will be but we already have plenty of points so we go ahead and graph these line for 1, 2.*1286

*If I put in 2 I will have 5 +10 = 15.*1293

*If we put in 4, 4 × 5/2 = 10 + 10 =20.*1305

*Now we have plenty of points, let us go ahead and graph this out.*1325

*I will go ahead and connect the dots.*1328

*The graph of this equation looks like this and with this one we have taken it definitely farther and mark out its x intercept.*1332

*This one is at -4, 0 and its y intercept way up here will be at 0, 10 and see they show up in the chart here and here when one of the variables is 0.*1342

*We will look at more next time with being able to do some shortcuts for graphing these lines *1362

*But with many types of equations you can often use a table to generate some values and put on some points.*1367

*Thank you for watching www.educator.com*1373

*Welcome back to www.educator.com.*0000

*In this lesson we will be taking a look at more about the slopes of lines and how we can use that to better graph.*0002

*Specifically some of the information that we went up looking at is how we can first determine the slope of line*0012

*whether we are just given a couple of points or whether we have the entire graph of that line.*0017

*Once we know more about slope we will be able to learn how to graph the entire thing using its slope and its y-intercept.*0026

*This will bring about many different forms that you can represent a line.*0032

*We will learn about switching back and forth between these many forms.*0035

*We will learn about some very special lines, the ones that are vertical and horizontal.*0040

*Look for those equations so you can easily recognize them in the future.*0043

*When it comes to a line, there is actually many ways that you can go about graphing or representing that line.*0051

*You can write it in standard form, slope intercept form and point slope form.*0056

*For now I will be mainly concerned with these first 2 forms, the standard form and the slope intercept form.*0062

*We will get more into the point slope form in another lesson.*0068

*When it comes to standard form, it looks a lot like this ax + by = c.*0074

*The way you want to recognize a standard form if you ever come across it, is that both of your x’s and y’s will be on the same side of the equation.*0082

*We like to usually put them on the left side.*0089

*There are no fractions or decimals present in the equation so the a, b, and c, those are nice whole numbers in there.*0093

*The x term here will be positive that is the a value.*0102

*We do not want to be a -6 or -7, we usually like to be 3, a nice positive value. *0106

*The reason why this form of the line will be so important is that if you have a line in standard form, it usually easy to graph.*0114

*The way we go about graphing something in standard form is we use its intercepts.*0122

*That is where it crosses the x and y axis.*0127

*The reason why the intercepts are so nice for our standard form is because when it does cross one of those axis, one of the values either x or y will be 0.*0130

*It will be making a table, but it would not be that big.*0140

*Let us do a quick example of something in standard form to see how easy it is to graph.*0145

*In this one I have 7x + 2y = 14.*0151

*You can see that it is in a standard form because I have both of my x’s and y’s on the same side.*0154

*I do not see any fractions, no decimals and the coefficient of x here is positive, a nice 7.*0160

*In order to graph this, I will find its x and y intercepts.*0169

*I will make my chart here rather than picking a lot of different points, I will look at where x = 0 and where y = 0.*0174

*Watch what this does to the equation.*0184

*If I use 0 for x, then it is going to get rid of the entire x term.*0187

*Since 7 × 0 right here, all of that is just going to go away.*0196

*I just have to solve the nice simple equation 2y = 14 which I can do by dividing both sides by 2.*0201

*I know that one of my intercepts, my y-intercept is at 0, 7, nice and simple.*0212

*Let us do the same thing by putting in a 0 for y.*0219

*7x + 2 × 0 = 14.*0222

*You will see that 0 is in there and again it is going to get rid of this term entirely since 2 × 0 is 0.*0229

*Then I have 7x = 14, divide both sides by 7 and we will get x = 2.*0237

*Now that I have both of these points, we will put them on our graph.*0249

*0-7,0, 1, 2, 3, 4, 5, 6, 7 way up here and the other 1, 2, 0 over here.*0253

*Connect those intercepts and I will get the graph of the entire line.*0273

*One disadvantage with using just the intercepts to graph a line is if you make a mistake on one of them, it is often hard to catch.*0277

*If you want to get around that problem, it might not be a bad idea to actually put in an additional point to see what happens with the graph.*0284

*That way if you do make a mistake with one of them you be able to quickly see that they are not all in a straight line.*0292

*To understand some of the other forms like slope intercept form, you have to understand a lot more about slope.*0301

*What exactly is this slope?*0307

*If I had to describe it, it is a measure of the steepness of a line, how steep is a line, if it is shallows or more steep?*0310

*Can you attach a number to that steepness?*0319

*What we do and we call it the slope.*0322

*In many equations we will use the letter m to represent that slope.*0324

*How do we attach a number to the steepness?*0329

*What we will do is we take 2 points from the line and we end up looking at the difference in the y values over the difference in the x values.*0331

*This gives us a nice equation for figuring out the slope of a line.*0340

*You will see these whole numbers in here and you can interpret that as each of these points.*0345

*These x’s and y’s come from point number 1 and this other one, these values come from point number 2.*0351

*In our work later, it is often a good idea to label one of your points as .2, and one of this is .1, just we do not mix up things in the slope formula.*0360

*You may have also heard of other ways to describe slope.*0373

*One of the most common is to call slope the rise of the line divided by the run of the line or simply rise over run.*0376

*That is actually a good way to remember how it is related to its steepness.*0384

*Its change in the rise would be the y direction and this change in the run would be the x direction.*0388

*One thing that may throw you off is that sometimes there is a negative sign in the slope.*0401

*You can interpret that negative line sign as whether you are going up in your rise or down, or do you need to go left or right in your run.*0405

*Let me go ahead and pick this is apart.*0414

*If you see a positive sign in the top then think of going up on your rise.*0417

*If you see a negative sign then think of going down.*0424

*With the run if it is positive, you will end up going right and if it is negative, then go left.*0431

*To make it a little bit more sense once we start seeing some more lines.*0439

*Maybe I can give me a quick example right here.*0445

*A line like this, I can say that maybe the rise is 5 and run over here is 4 so the slope will be equal 5/4.*0450

*If I had a different line like this, I can see them going down and then I have to go right.*0466

*Since I’m going down, I will mark this one as -3 maybe this one over here as 5, then I will have a slope of -3/5.*0477

*In practice 1, if I do have a negative sign in my slope, I usually just give it to the top part of the fraction.*0488

*That way I only have to remember about going up and down.*0492

*I do not even have to worry about left and right since the bottom is positive.*0496

*Another word of warning, be careful not to mix up all your points.*0503

*If you have your y values from point 2 being first then take the x values from point 2 being first as well.*0507

*I also use the sign of this to give you a little bit more intuition as to which direction that line should be facing.*0516

*There are only a few instances of what your line can look like.*0523

*It could be going from the lower left to the upper right and that is an indication that your slope is going to be positive,*0527

*since both your rise and your run are going to be positive numbers.*0534

*If your line is going from the upper left to the lower right and you are going to have a negative slope.*0543

*This will be because your rise is negative, but your run is positive.*0551

*The other 2 special cases that you have to come and watch out for is what happens when your slope is 0 *0557

*and what happens when it is undefined.*0562

*In these 2 instances, you either have the alignment as completely horizontal.*0565

*This is when you have a 0 slope and it is completely up and down if you have an undefined slope.*0570

*Let us get into our examples and see how we can start finding our slope just from a couple of points.*0580

*With these ones we will use the formula for the slope of the line to pick it apart.*0587

*I’m going to try and keep things a little bit together by marking these out as point number 2 and I will mark the other one out as point number 1.*0593

*In the formula, here is what we are looking at.*0610

*Our slope should take our y values and subtract them all over our x values and subtract those.*0613

*Notice how I’m keeping things all lined up, I have both of my x and y from point 2 over here and both my x and y from point 1over here.*0622

*Let us put in some values.*0636

*I need the y value from point number 2, that is a 4 then we will minus our y value from point number 1 that is -1.*0638

*All over x value from point number 2, -2 and x value from point 1, -1.*0654

*Notice I’m subtracting these even though they already have a negative sign in them.*0668

*Be very careful on your signs with that.*0672

*Looking at the top, 4 - -1 is the same as 4 +1 = 5.*0675

*-2 - -1is the same as -2 + 1= -1.*0683

*It looks like this slope for the first one 5/-1 I will just get a slope of -5 between these 2 points.*0690

*It does not matter which one you will label as point number 2 or point number 1 just as long as you keep them straight.*0702

*I’m going to switch which ones I’m calling point number 2 and point number 1 just to highlight this, but you will get the same either way.*0712

*Let us start off, I want to subtract my y2 from my y1.*0721

*Y2 is 6 – y1 6, my x2 is -5 and my x1 is 3.*0727

*On the top of this, I have 6 - 6 giving us 0.*0752

*On the bottom I have -5 - 3 - 8 and 0 divide by anything is 0.*0757

*This indicates that our slope is 0.*0764

*This is one of our special cases where we have a horizontal line, it is completely horizontal.*0767

*Let us do one more, point number 2 and point number 1.*0772

*The slope for this I will take my y value from the second point I will subtract the y value from the first one.*0789

*Then to our x’s, x from our second point minus x from our first point.*0804

*Now we will work to simplify, 5 - -4 is the same as 5 + 4 =9, 3 – 3 =0.*0814

*Be very careful with this one, we can not divide by 0.*0826

*Whatever that mean, it will get us around the bottom.*0839

*This is indicating that our slope is undefined.*0841

*It is not that there is not a line there, there is a line.*0845

*The line is just completely straight up and- own, it is our vertical line.*0847

*Just to make this a little more clear, I will say slope is undefined or sometimes we might say that there is no slope.*0851

*In order to know a lot more about slope, we will get a little bit more into the slope intercept form.*0868

*Slope intercept form looks like this, y = mx + b.*0875

*The way you can recognize this form is that the y will be completely alone on one side of the equation.*0880

*Usually I like to put it on the left side.*0886

*Our slope will usually be represented by that m and we will put it right next to the x.*0889

*The b in this equation stands for our y-intercept of the line.*0896

*That is where it crosses the y axis on our Cartesian coordinate system.*0900

*The reason why that this form is usually everyone's favorite is because it makes graphing a nice and simple process.*0906

*The way it makes a graphing so nice for us, is we start at the y intercept.*0913

*Just rely from the graph by whatever that d value is.*0919

*What we do is we use the slope as directions on how to get to another point on our graph.*0922

*Let me give you a real quick example of how this would work in practice.*0929

*I have 1/7x + 3.*0934

*The very first thing that I would do is I would look over here at the y intercept and I would take its value.*0945

*I know that this particular line crosses the y axis 3 and I would put a point on the y axis right at 3.*0952

*Starting at that y-intercept, I would use the slope as directions to get to another point.*0967

*Keep in mind that it is the same as rise over run.*0976

*Starting at that y-intercept, I will go up one into the right 7, up 1 to the right 1, 2, 3, 4, 5, 6, 7.*0980

*Now that I have 2 points, I can go ahead and connect them and make the entire graph of this equation.*0995

*One you start with the y-intercept and two you use the slope as directions to get to a second point.*1006

*Let us see this in action by actually graphing out some linear equations.*1016

*They said before let us start here on the nth with our y-intercept and make that our first point that we put on the graph.*1022

*This one is -5, it crosses the y axis down here at -5, it looks good.*1029

*In terms of our slope, we want to think of this as rise over run so starting at that y intercept, we go up 3 and to the right 4.*1038

*1, 2, 3 to the right, 1, 2, 3, 4 and now we have a second point.*1050

*Now that we have 2 points, simply connect them to a nice solid line and there is your entire graph.*1063

*It makes the graphing process much easier.*1073

*You do not even have to worry about the table and doing all of those values.*1075

*Let us try this one.*1082

*Graph the line using the y intercept and the slope.*1083

*On the back in here, I see that my y-intercept is 3 and that will be the first point on my graph.*1087

*My slope is -2, how will that work with rise over run?*1099

*It does not look like a fraction like in some of my other examples.*1107

*It does not feel free to turn it into a fraction by simply putting it over 1.*1111

*This tells me that I need to go down 2 since that is negative and to the right 1, down -2 and right 1, now I have a second point right there.*1116

*I can draw the entire graph, very nice.*1132

*We have both of these forms under our standard form and our slope intercept form,*1142

*you may be curious which one should you be using most of the time.*1152

*Both of them are good for graphing.*1159

*And what I often recommend is if you have to graph something and it is already in standard form, just go ahead and use the intercept.*1163

*It is usually one of the quickest ways to do it.*1169

*If it is already in slope intercept form then use its y intercept and slope as a direction to the second point and graph it that way.*1171

*That is usually the quickest.*1178

*If you do have to switch back and forth between these 2 maybe you are more comfortable with slope intercept form, then feel free to do so.*1179

*If you have something that is in some other form and you want to get in the slope intercept form then the process is pretty quick.*1188

*What you should do is simply solve for y and get it all by itself on one side of the equation.*1195

*In doing so, you will be able to better see what it slope is and the y-intercept.*1200

*Not very many people go the other direction, but potentially you could end up rewriting something into standard form.*1206

*There is a lot more criteria that go in there.*1212

*One of the first things is you should get both of your x’s and y’s on the same side.*1215

*Then you want to make sure that your constant is on the other side.*1221

*Only x’s and y’s on one, constant on the other.*1226

*Try and clear out your fractions by multiplying by a common denominator.*1231

*a, b, and c should not be fractions.*1233

*Then look at the coefficient in front of x and it should be positive.*1237

*If that is not positive, then multiply it by -1 and make it positive.*1243

*More practice switching back and forth so we can see how this process works.*1250

*Even though we have our standard form and our slope intercept form.*1259

*You want to be aware that there are 2 special cases and we seen them come up but once before.*1262

*We have some lines that are completely vertical and some that are horizontal.*1268

*The vertical ones are straight up and down, and the horizontal ones are left and right.*1275

*The way you can recognize their equations are they are simply x equals a number or y equals a number.*1281

*You may see something like x = 2, maybe this is like y = 15.*1289

*For the one it says x equals these are your vertical lines.*1296

*For the one that says y equals these are your horizontal lines.*1304

*The way that they can keep in track of which one should be horizontal and which one should be vertical is the way you interpret them. *1312

*If you have an equation like x = 2, what that is trying to tell you is that the x value no matter where you are in that line is always 2.*1318

*If my line looks something like that and I decide to pick up some individual points,*1330

*no matter what point I pick out I can be sure that the x value will be 2, no matter where I am on that line.*1335

*In a similar fashion, if I’m looking at y = 15, no matter where I am on that line I should end up with the y value being 15.*1346

*Watch for these 2 special cases to come up in my examples.*1364

*Let us first work on switching back and forth between these 2 different forms.*1371

*What I have here is a line in standard form and we want to put it into slope intercept form and we want to put it into slope intercept form.*1375

*I want to actually go through the process of graphing it and I’m more familiar with slope intercept.*1383

*When I already did take this and put it in to that other form, I need to solve for y.*1388

*Let us start by moving the 7x to the other side, 2y = -7x + 14.*1395

*And then we will divide everything by 2 and that should get our y completely by itself.*1407

*Notice on the right side there, I have to divide both of those terms by 2, -7/2 x + 7.*1416

*The most important part about writing it in this new form is now I can easily identify what my y intercept is.*1427

*It looks like it 7 and I can more easily identify what my slope is.*1435

*It has a slope of -7/2 and I know it is facing down from the upper left to the lower right.*1441

*Let us go to the other direction.*1450

*Let us take a line that is written in slope intercept form and put it into standard form.*1452

*It requires a little bit more work but we can do it.*1458

*The first thing I’m going to do is try and work to get my x’s and y’s on the same side.*1461

*I will subtract 1/2x on both sides, -1/2x + y = 3.*1466

*I want to make sure that my constant is on the other side, that is the 3 and it is already there.*1481

*I want to get rid of all fractions so I need to get rid of that ½.*1487

*I can do this if I multiply both sides of the equation by a common denominator and in this case that would be 2.*1492

*-x + 2y = 6, we are almost there.*1500

*You can see that it certainly look a lot more like that standard.*1507

*The last thing we need to make sure is that the coefficient in front of x is positive.*1510

*Right now looks like it is negative, I already fixed that.*1515

*I will multiply everything through by -1.*1518

*-1 × -x would be x and -1 × 2y = -2y, equals -6.*1528

*This form is in standard form.*1544

*They might be looking at it and say what good is that? Why would you want it in standard form?*1547

*Remember that you can graph it in standard form now by simply looking at its intercepts plugging in at 0 for x and 0 for y.*1551

*Let us get into some very special cases.*1560

*We want to graph the line y =5.*1563

*There is not much of the equation to look at.*1568

*What should be the slope? What should be the y intercept?*1570

*This is one of our special cases, y equals a number.*1574

*Since it is y over here, this is going to help me indicate that this is going to be a horizontal line.*1578

*1, 2, 3, 4, 5 would be one point.*1584

*I will just make a giant horizontal line with all points y = 5 and double checking this make sense.*1588

*If I was to pick a point on a line at random, in this case it is 1, 2, 3, 4, 5, 6 its y value is 5.*1596

*If I pick something over here, its y value is still 5 no matter where you go on this line its y value will always be 5.*1606

*One last special case, this is x = -2.*1618

*This will be a vertical line straight up and down because we are dealing with x over here.*1623

*I’m at x = -2 and we will make it straight up and down.*1631

*The reason why this makes sense is no matter what point you choose on line, as a way up here at 1, 2, 3, 4, 5, 6, 7 the x value will always be 2.*1641

*With the slope of this one, remember that its slope is undefined.*1663

*There you have it some have very nice techniques you can use for graphing lines and now 2 forms that you can use to represent lines.*1668

*Thanks for watching www.educator.com.*1676

*Welcome back to www.educator.com.*0000

*In this lesson we are going to continue on with our linear equations and look at how we can build these linear equations for ourselves.*0002

*Specifically, some of things that we will look at is we will look at the point slope form on a line.*0014

*It will help us when building those equations.*0020

*For some of our examples, we will also have to know a little bit more about parallel and perpendicular lines.*0023

*Watch out for my explanation on what these are and how you can tell if two lines are parallel or perpendicular.*0029

*Earlier, we looked at two forms of a line.*0038

*We looked at standard form and we looked at slope intercept form.*0041

*Remember both of these forms are important because they were good for graphing.*0045

*When we want to build our own line, sometimes we can do this as long as we have enough information.*0057

*To form that we use for building our line is known as point slope form.*0064

*The reason why it is called that is because those are the two bits of information that we need. *0068

*We need to know at least one point on a line and we need to know what the slope of that line is.*0072

*Once we have both those bits of information and we can drop it into point slope form.*0077

*This form here requires a little bit of explanation.*0086

*You will see that there is an x and y that has some subscripts on it and that is where the point goes in terms of its x and y.*0090

*The slope is the n here so we can just put that in and then we also have a couple of other x’s and y’s which do not have subscripts.*0098

*With the ones that do not have any subscripts whatsoever, you will not be putting any values in for those.*0108

*Those simply stay in our equation. *0114

*One last thing to note is even though we will use point slope to actually build the equation on a line.*0116

*Usually you take point slope and end up converting it into slope intercept form because you are looking to do some other things without a line,*0122

*other than just than build it or maybe you want to eventually graph it.*0129

*It is good to go ahead and move it in to one of those other forms.*0132

*To build lines using point slope form, we want to substitute in our point and we will go ahead and substitute in our slope.*0143

*I already have point slope form, I will put it over here and I have a point and I have a slope.*0151

*Let us go ahead and just substitute these in and see how it works.*0157

*I will have y - -3 then the m represent the slope, I will put in ¾, x – xy.*0161

*Notice how the x and y which did not have any subscripts are still in there.*0176

*Once we have everything substitute in here, we want to go ahead and start rewriting this*0181

*either into standard form or slope intercept form so we can do some other stuff with it.*0186

*I’m going to go ahead and put this into slope intercept form by just getting y all by itself and maybe cleaning up a few other terms.*0190

*I'm subtracting a negative on the left side and that will be the same as adding 3.*0198

*I will go ahead and distribute my ¾ here.*0203

*I have ¾ x and -3, looking much better.*0206

*I will go ahead and subtract 3 from both sides.*0216

*Y = 3/4x – 3 – 3 is 6.*0219

*Our point slope gets our foot in the door so we can actually build the equation and then put it into a different form, maybe a slope intercept down here.*0227

*We can go ahead and graph it and do some other things with it.*0241

*Sometimes we will be given information about another line in order to build the one we want.*0247

*And information about that other line, we may know that it is parallel or perpendicular to the one we want.*0253

*What exactly does that mean? *0258

*We can say that two lines are parallel if they had exactly the same slope.*0261

*In my little picture here, you can see what that does.*0266

*You will end up with two lines and they usually have a little bit of a gap or space in between them, but they have exactly the same slope.*0269

*They are going in the same direction.*0275

*You might also have lines that are perpendicular.*0279

*With this one, they end up meeting at a right angle.*0282

*We do not talk much about angles in this course, another way that you can say two lines or perpendicular is *0286

*if their slopes are negative reciprocals of one another.*0291

*Let me show you what that means.*0295

*I suppose this blue line had a slope of 4/5, if the red one was perpendicular then I wanted to be the negative reciprocal of the blue line.*0296

*I made it negative and I flipped it over.*0312

*With perpendicular lines, they will always be different in sign, one will be positive and one will be negative and they will be reciprocal when flipped over.*0314

*Another way that you can test if two lines are perpendicular or not is actually you take both of their slopes and just multiply them together.*0325

*Two things could happen if you multiply them together and their slope is negative or their combined value is -1 *0333

*then you will know for sure that they are perpendicular.*0342

*If you multiply them together and you get anything else other than -1 then you will know that they are not perpendicular.*0346

*It is a nice and easy task you can use to know how the two lines are related.*0352

*We will definitely know this as we get into more of those examples.*0358

*Let us start off with example 1, in this one we want to write the equation on a line using the given point and the slope.*0363

*Since the two bits of information that I need for point slope, I will just go ahead and nearly drop it into the formula.*0372

*y -y1 equals slope x - x1.*0378

*Let us put in the y value first, y - 3 equals my slope is -2/3x - -6.*0386

*You will know the formula has a minus sign in there since the x value is a negative, go ahead and put that negative sign in there as well.*0399

*Now all we got to do is clean it up a little bit and maybe turned it into a different form.*0408

*I will see what we can do.*0412

*y – 3 = -2/3, when I subtract the negative that is the same as addition.*0413

* x + 6 and I think I will go ahead and distribute this -2/3 in there.*0422

*-2/3x – 12 ÷ 3 and a -12 ÷ 3 = -4, it looks pretty good. *0430

*Let us add 3 to both sides and now we have that same line written into slope intercept form.*0449

*Again we use point slope form to go ahead and create the equation of the line and then probably put it into some other form.*0462

*This one is a little different, in this one we simply want to determine whether the two lines are parallel*0472

*or maybe the perpendicular or maybe they do not fall into either of those two cases.*0477

*With these ones, notice how many of them are written not in one of the forms that we have covered. *0482

*In other words they are not in slope intercept form but the second one is in standard form but we have to be able to figure out what their slope is.*0489

* A way to determine what the slope of the line is to go ahead and rewrite it into our slope intercept form and we can just read it out of the equation.*0501

*Let us do that first before we actually compare what these slopes are.*0509

*Starting with this first one here, I can move the 2 to the other side by subtracting 2 so y =5x – 2.*0512

*Let us see what the second one, let us go ahead and move the x to the other side, -x – 15 and then we will divide it by 5.*0521

*Here is our first line, here is our second one.*0539

*Now they are both in slope intercept form we can say that the first one has a slope of 5 *0545

*since it is right next to x and the other one has a slope of -1/5.*0550

*Looking at the two slopes, they are definitely not the same, they are not parallel.*0555

*It looks like one is positive and one is negative and they are reciprocals.*0562

*I think these two are perpendicular.*0567

*If you want to test out feel free to just take the two slopes and multiply them together and see that when you do this you get a -1.*0570

*You will know that these two lines are perpendicular.*0579

*Let us try this process with the next pair of lines.*0590

*I will begin by just putting this into slope intercept form.*0595

*This one is almost in slope intercept form.*0599

*I just have to reverse the y and putting it first.*0602

*This one I will go ahead and move the 3x to the other side, so now I have my two lines right here.*0606

*y = 2x + 1 and y = -3x + 4, in this form, I can read off what both of their slopes are.*0616

*Looking at their slopes I can see that they are not the same, they are definitely not parallel.*0627

*One is positive and one is negative, but they are not reciprocal so they are not perpendicular either.*0633

*They are not parallel or not perpendicular, I will put this in the neither categories.*0638

*They are just two lines hanging around on the graph.*0643

*Let us try that same process with another pair of lines.*0650

*Let us see if we can find a couple of that which actually are parallel.*0652

*We will begin by putting these into a better form so we can read off that slope.*0656

*I'm moving the 4x to the other side, this one the y is still not completely all by itself.*0662

*Let us go ahead and divide everything through by 2, now we have one of our lines.*0668

*The second one let us go ahead and rearrange things.*0678

*The y is on the left and let us get it completely isolated by multiplying everything through by -1.*0683

*Now we have both of our lines.*0693

*Now that they are in this form, the slope of the first one is -2 and the slope of the second one is -2.*0702

*I can see that they are exactly the same and we will call these pair of lines parallel.*0709

*Onto one more pair of lines, let us see what they are, parallel, perpendicular, or neither, we will find out.*0719

*I need to get my y isolated I will start off by moving the 4x to the other side.*0726

*I will divide both sides by 3, y =4/3x +2.*0734

*The second one, let us go ahead and move the 2 over.*0745

*I have 3x + 2 and I will divide everything by the 4, ¾ x + ½.*0749

*Here is equation 1 and here is equation 2.*0760

*If you look at them in this form, we can pick out what each of their slopes are, I have 4/3 and 3/4.*0767

*These ones are pretty close, they are definitely reciprocals of one another since you would take 4/3 and flip it over and get ¾.*0775

*When those have been both positive so we can not say that these are perpendicular.*0783

*Even though they are reciprocals they are not negative reciprocals of one another.*0788

*They are definitely so they are not parallel.*0792

*Even though they are close, they do end up in the neither category.*0795

*They are not parallel or not perpendicular, they are simply neither.*0799

*They are just two lines.*0802

*With this one we want to write the equation of a line that happens to be parallel to the given line and also goes through the given point.*0809

*This one is a little different, notice how we need to know what the slope of the line is*0818

*and we need to know what the point is but they have not quite given us what the slope is.*0822

*We are going to have to extract that out of this other line that they have given us by knowing that it is parallel to the one we want to build.*0827

*Let us hunt down what the slope of this other line is.*0834

*We will do that by dividing everything through by 4.*0837

*It looks good so y = ¼ x + 5.*0845

*Let us see, I know that the slope is ¼.*0850

*Now that I have a point, I have a slope, I can use point slope form and build our line.*0855

*Y - -3 = ¼ x – xy and now let us clean it up and see what line we have.*0861

*y + 3 = ¼ x, ¼ of 2 would be -1/2.*0872

*Now we will subtract 3 from both sides.*0890

*y = ¼ x – ½ - 3, get a common denominator over there.*0893

*Our line is y = ¼ x – 7/2.*0908

*The way we built this line, I know for sure that it goes through the given point 2, -3.*0917

*I already explored what the slope of the other line was so I know it has a slope of 1/4.*0923

*Let us try this one, write the equation of a line which is perpendicular to the given line and it goes through the given point.*0932

*Similar to the other one, but it have not given us the slope directly.*0939

*It just told us it is perpendicular to this other line over here.*0943

*Let us figure out what slope is, y equals, I will move the 8 to the other side, -8x + 3.*0948

*The slope of this line is -8 which is good but that is not exactly the slope we want to use.*0958

*Our line is perpendicular to this line, so we need to use the slope -1, 1/8. *0966

*That way it is the negative reciprocal of the other.*0979

*Now we will drop it into our formula.*0984

*Let us go ahead and put in our y value. *0994

*We have our slope and our x value.*0997

*We will clean it up and see what line we have.*1002

*y - 8 = 1/8 - 4/8, my x in there.*1005

*y – 8, 1/8x – ½ and we will go ahead and 8 to both sides, find a common denominator and combine the last of our like terms.*1017

*1/8x + 15/2.*1044

*We can see that its slope is the negative reciprocal of the other line.*1053

*I know it is perpendicular and the way we build that it, for sure it goes through the point 4/8.*1057

*This last one is a little bit more of the word problem but you can see how we can still use these techniques in order to build the equation of the line.*1066

*We want to build a line that represents the following problem.*1074

*It cost a $20 flat fee to rent a drill + $2 every day starting with the first day.*1077

*Let x represent the number of days that we rent this drill and y represent the charge to the users.*1085

*How much will we end up willing to get from them?*1091

*When we are all done, let us see if we can write this line in a slope intercept form.*1095

*Using the techniques that we know about so far, we need to figure out what our slope is and what point this will go through.*1100

*Let us see, one thing that we can think of is the slope being what changes or your variable costs.*1108

*The $20 that you have to pay up front is not a variable.*1120

*It is going to be there no matter what, what does change is a $2 every single day.*1123

*The variable cost, we can think of that as our slope.*1134

*Let us see what we can do with that.*1144

*What happens with that flat cost?*1146

*We will be charged of that no matter what, that is like our y intercept.*1151

*If this one, we can just drop in our variable cost and we can drop in our flat fee.*1158

*What we are left here is the equation of the line but it represents how much we end up charging the person*1177

*and we can see this by substituting some values.*1183

*Suppose they rent this drill for one day, it cost $2 for that day + $20 flat fee, it cost them $22.*1186

*If they rent it for two days, we can put that in for x, $4 for the variable + $20 flat fee, $24.*1196

*You can go on and on figuring out how much it cost for each of the different days.*1206

*But in the end, this formula right here would represent how much we need to charge them.*1211

*Just simply plug in the number of days for x and y would give you the cost.*1215

*You can see that building a line is not so bad and using point slope form is handy in that entire process.*1221

*If you do use point slope form, you need to know the slope of the line and you need to know a point on that line.*1228

*Thanks for watching www.educator.com.*1234

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at an introduction to functions. *0002

*There is quite a bit to cover when it comes to functions, we will go over this bit by bit.*0008

*Some of the first things I will do with functions is look at some of the terms. *0014

*In other words, what exactly makes a function and why is it a special type of a relationship. *0019

*What is it that makes a function a function?*0024

*We will get into a term such as the independent and dependent variable so they can find those parts in a function.*0028

*We will talk about how functions can be like little machines and how they have input and output.*0035

*We will call this out its domain and its range.*0040

*We will get into a little bit on how you can represent function using diagrams that we can better keep track of the inputs and outputs*0044

*and also how you can work with functions by evaluating them.*0051

*Let us take a look.*0056

*A function is a very special type of relation.*0061

*What a relation does is it connects 2 variables and usually we use a pair of ordered pairs to describe that connection.*0065

*Maybe I have an x variable and a y variable and one other connected.*0074

*I can show you that anytime I use a 2 for x, I get a 3 for y.*0078

*There are many other types of connections that you can consider as part of a relation.*0084

*Some of our equations can be considered a type of relation.*0091

*A function is just a special kind of one of these relations.*0097

*It is special because for every x value it has, there is one and only one y value associated with it. *0101

*What that means is that if you use a particular x value, you are never going to get two different values using that one x value. *0111

*When we represent functions, we usually identify some sort of independent variable.*0120

*This is what we get to choose for in terms of the x.*0126

*The dependent variable is usually represented by y and it depends on whatever we choose for x.*0131

*Some of the notation you may see for functions looks like this.*0142

*It is important that we can pick out each of the different parts here.*0147

*The first thing is what is this f out in front? f represents the name of the function.*0152

*This function is called f.*0159

*The x inside the parentheses right next to it is not multiplication, what that is indicating is what the variable is.*0168

*We have our function called f and it takes a variable of x.*0180

*Everything after that on the other side, that represents the relation or how things are connected. *0186

*You can also consider this as the rule of the function or what it does when you use a particular x.*0196

*The way you read this notation is f(x) = 3x + 9.*0204

*Our f part helps us identify what variables are being used inside the function.*0215

*To better understand what type of relations our function is, I like to tell my students that they are a lot like little machines.*0226

*They take some input and you put into and they produce some output.*0233

*A nice little diagram would look something like this down here.*0240

*Maybe my function f here takes an input of 3 and produces an output of 5.*0246

*If we start to gather up together more of its inputs and outputs, we can represent those as a whole bunch of ordered pairs.*0254

*What this list of ordered pairs would say is that my function takes an input of 1 and produces an output of 2.*0262

*If I give it an input of 3 then I get an output of 5.*0272

*Every ordered pair is telling me what the input and output was for that particular value.*0278

*Now, if we collected together all of the inputs, this would be known as the domain of the function.*0288

*If we collected together all of the outputs of the function, this be known as the range.*0294

*Using the example from a whole list of ordered pairs, we can actually identify all the things used as an input and all the things used as an output.*0301

*Remember this first value here, those are our x’s, those are inputs and the second values, those are our outputs.*0311

*Let us simply just list out all of the inputs and call that our domain and list out all the outputs and call that our range.*0329

*We used 1, 3, 4, 7 for input and what we got as output was 3, -2, 5, 10.*0336

*You can also represent a relation in the same way, so it is important to point out that this particular relation here is actually a function.*0351

*The reason why we know it is a function is that every time we are looking at any particular output, it only has one output associated with it. *0360

*Input of 1 output of 3, input of 3 output of -2.*0370

*You do not see anything on this list like 1, 7 because that would be a problem *0377

*because the 1 would go to 3 and 1 would go to 7, you will get 2 outputs.*0384

*I just want to make a little note here that this is a function because it satisfies our definition.*0390

*A great way to visualize what is happening with the inputs and outputs is to use a diagram.*0403

*One way to do that is to list all of your domains and group it together in a circle or a box.*0409

*And you group together all of your range in the same way.*0415

*You can show the relation by using arrows to connect the two. *0419

*That is exactly what I have done down below.*0423

*Here I have two different relations, you can see my inputs and you can see my outputs.*0426

*We will consider these our domain and our range. *0436

*The big question is does it satisfy the definition of a function or is it just simply some relation.*0443

*Let us take a look at it and see what happens to those inputs and outputs.*0449

*Remember, it is a function if every input you give it goes to exactly 1 output.*0453

*Let us see what we got.*0459

*Looking at this input here of 1, it looks like it has an output of 4 and that is the only thing it has for now, I will put just 4.*0461

*Let us check out the 3, if I use an input of 3 it looks like I will get an output of 4 and I will get an output of 0. *0470

*Let me highlight that.*0479

*Notice how this one has an arrow that goes right over here to 4 and has another one that goes out to 0. *0483

*When I use an input of 3, it goes to two different outputs then I can say that this is not a function.*0493

*It is close but it does not work.*0509

*Since we know that one was not a function, let us take a look at the other relation.*0512

*When I use an input of 1 it goes to 8, I will use an input of 3 it goes to 5, input of 6 goes to 1, input of 7 goes to 0. *0519

*Every time I use an input here, it only goes to one output.*0530

*This one was okay, this one is a function.*0534

*You can see these diagrams help out with looking at what is in the domain and range, it is all grouped together nice and simple.*0542

*In some functions, they may not actually specify what domain is being used and do not actually says what the domain is.*0555

*We just assume that is known as the natural domain.*0563

*What this natural domain represents is all of the possible real numbers that you could use as input *0567

*as long as it does not make the function itself undefined.*0574

*You might start with thinking about all possible numbers and eliminating ones that you can not use.*0578

*To help better identify what you can not use in the domain, look for stuff like fractions or even roots. *0586

*The reason for this is, you cannot divide by 0.*0595

*If any value would make the bottom of a fraction 0 then you toss them out, they are not in the domain. *0600

*Also, we do not want to have to deal with imaginary numbers, we do not want negative numbers underneath an even root. *0607

*If that happens, we will toss those out of the domain as well.*0615

*When it comes to writing down a lot of different numbers in the domain, there is a special way that we go about doing that.*0620

*We often use what is known as interval notation. *0628

*What interval notation is, it is a way of packaging up all of the range of numbers on a number line.*0631

*Let me give you a nice quick example.*0640

*I wanted to describe all the numbers between 2 and 3. *0641

*I could do that by shading in those numbers on a number line so you can see that I have shaded in 2, 3, and everything in between. *0647

*To use interval notation to do this, I would list out the starting point and the endpoint of everything that I have shaded.*0658

*Then I would use a bracket to say that the endpoints are included.*0670

*Everything between 2 and 3, I’m using that square bracket because 2 and 3 shall also be included. *0676

*If we can go all the way up to a point but maybe not included, we will still use interval notation for this.*0685

*But usually we use a parenthesis when it is not included.*0698

*I have re adjusted the number line here, now I shaded everything from to 2 to 3 but not including 3.*0704

*I have made the corresponding interval notation to reflect that. *0711

*I started at 2 stop at 3, includes 2, but does not include 3.*0715

*How we work with these functions?*0724

*One of the more common things we do with functions is to simply evaluate them.*0726

*To evaluate a function, what we simply do is substitute in the value given in place of x.*0734

*If I'm dealing with this function f(x) = 3x + 9 and maybe I want to evaluate it at the number 7. *0742

*The way I do this is I just replace all x’s with 7.*0750

*You will see that the process itself is not that difficult.*0759

*It is sometimes just the notation that throws people off because when you look at the left side over here, *0763

*your brain sometimes looks at that and you want to think of multiplication, but it is not multiplication.*0769

*When you have that f(7), what you are saying is that you used an input of 7 and then over here is what you got as your output.*0776

*Keep that in mind when you are evaluating functions.*0789

*Let us get into some examples and see if we can figure out whether certain relations are functions*0795

*and maybe identify a domain, range, and all that fun stuff.*0803

*In this relation, I can see that my inputs would be the first values and my outputs those are the second values.*0808

*Let us represent this using a diagram just to help out. *0820

*We will group together all of our inputs into a giant circle over here.*0824

*We will group together all of our outputs into a giant circle over here.*0828

*First the input, I have -3, 6, I have another -3 so I do not want to list it twice and I have 5.*0834

*Our outputs are 7, -2, 4, and 9.*0845

*When I use an input of -3 it goes over here to 7 and when I use an input of 6, this heads over to -2.*0853

*I have another -3, -3 also goes to 4 and 5 goes to 9.*0866

*That is a bit of a problem isn’t it?*0877

*Notice how we have a single input and it goes to two different spots. *0879

*If we have a single input going to 2 outputs then we can say that this is not a function.*0887

*Here I see another diagram.*0905

*Let us see if we can figure out if this one is a function, 13 goes to 7, 81 goes to 7, 32 goes to 60, 27 goes to 19 and 45 goes to 55.*0906

*This one actually looks pretty good. *0920

*This is a function.*0926

*There is an interesting feature here that I want to point out, *0932

*you noticed in the last one we had one input that went to 2 different outputs and that made it not a function.*0937

*You may be curious if this would also make it not a function? After all, 13 and 81 go to the same spot. *0943

*That is okay, you can have two inputs go to the same spot *0954

*but what you do not want to happen is to have one input split off into two different places. *0958

*This right here, this is okay.*0965

*Let us see if we can identify the domain of a function just from looking at its equation.*0980

*Since no domain is specified, we are going to look for the natural domain.*0986

*That means we will assume that all real numbers can be used unless it makes the function undefined.*0990

*In these examples, I put one with a fraction because we want be concerned about dividing by 0.*0997

*I put another one with a square root because we do not want negatives underneath an even root, like the square root.*1002

*In this first one, the way that it would not work out is if I had a 0 in the bottom.*1010

*Are there any restricted values for x?*1019

*Anything can x can not be would that give us a 0. *1022

*Well, if I had to solve just the bottom part, I can see that if x was equal to -1 I would definitely have a problem. *1026

*I think that is the only issue that we would end up with.*1036

*Let us just write them out, x can not be -1.*1039

*Anything else is perfectly acceptable.*1047

*If I was looking at a number line and trying to shade in all the different things x could or could not be, *1052

*the number line would look something like this.*1059

*It could definitely be any of these negative values over here. *1064

*I’m digging a big open hole at -1 since it can not be that and it could be anything greater than that.*1069

*It can be any value but not just -1.*1076

*I will represent that using some intervals.*1080

*It could be anything from negative infinity all the way up to -1, I would not include that.*1085

*It can be anything from -1 up to infinity, I will use little u’s to connect those.*1089

*It can be anything just not -1.*1096

*Let us try another one, f(x) = √x+ + 5.*1100

*I want to make sure that whatever x + 5 is, we do not want it to be negative.*1108

*It is okay if it is 0 but definitely we do not want it to be negative. *1115

*The only way it would end up being negative is if x was less than -5.*1126

*Let us say x can not be less than -5.*1134

*We will represent this on a number line as well so we can end up making an interval for it.*1147

*Let us see, here I have -5, -4, -3, -2, -1, 0, I have some things that are less than -5.*1153

*x can be anything, it can even be -5 because you can take the square root of 0, it can be anything greater.*1169

*The domain of this one, the natural domain would be from -5 up to infinity and it is okay to include that -5.*1182

*For this one, we want to work on evaluating the function for some given values of x.*1195

*Be careful on this notation over here, we are not multiplying or simply substituting 3 into our function and seeing what the result is.*1201

*I'm going to replace everywhere I see an x with this number 3.*1215

*Once I substituted it in there, we go through and clean it up a bit, I have -12 + 9 or -3.*1224

*Our input was 3 and our output is -3.*1237

*Let us do the same thing for f(-5).*1243

*Same function, we will just put in a different value, -5.*1248

*We will work to simplify it.*1257

*-4 × -5, negative × negative will be positive, 4 × 5 is 20, 20 + 9 and we would get 29.*1260

*In evaluating functions, just simply substitute a value in there for x and end up simplifying it. *1274

*That is all I have for now, thank you for watching www.educator.com.*1281

*Welcome to www.educator.com.*0000

*In this lesson, we are going to work more with graphing functions now that we know a little bit more about them.*0002

*What you will see is that when it comes to graphing functions, we use a lot of the tools that we use with graphing just a normal equation.*0012

*We will see first how you can use a chart to plot a whole bunch of different points and graph out an entire function. *0018

*The good news is if you have a linear function, we can often use our tools for lines to shortcut that process. *0027

*Once we can look at the graph of the function, we will be able to determine whether it is truly a function using something like the vertical line test.*0035

*More importantly, once we have the graph of the function we can test out what its domain and range is just from looking at its graph.*0044

*Let us go ahead and take a look.*0052

*Graphing a function is the same process as graphing an equation.*0058

*We want to look at the relationship between the variables.*0063

*In order to do that, we can simply develop a table of values. *0066

*Remember doing this in our graphing of linear equation section, we may pick a lot of different values for x and see what the corresponding y value is. *0071

*The only difference that we might make with function is simply use different notation.*0081

*I would still pick a lot of different values for x but then I would simply see what the corresponding output is for my y values. *0086

*Always keep in mind that when we are dealing with our inputs, those are going to be our x values.*0096

*When we are dealing with our outputs, those are going to be our y values.*0102

*We will still be able to plot them on a coordinate axis.*0107

*Our x and y will correspond to the inputs and outputs of that function. *0109

*Now if a function represents a line then I have some good news for you.*0119

*You can use a lot of your techniques, especially about slope intercept form. *0122

*Here is a function written in slope intercept form, you can see that it still has the n being slope and it still has b being the y intercept.*0127

*The only change that I have done here is instead of writing out the y, I'm using my function notation. *0141

*This just gives us the name of the function and tells us that our independent variable here is x.*0153

*Since it is in slope intercept form, I could simply graph something like this by first using the y intercept as a starting point*0161

*and then using the slope to identify another point on that graph.*0167

*I have 2 points that I could connect them and then I'm good to go.*0171

*To determine if the graph is a function or not we can use what is known as the vertical line test.*0182

*The way the vertical line test works is you imagine a vertical line on the graph.*0189

*As long as it only crosses the graph in one spot, then you can consider it a function.*0195

*If the vertical line crosses the graph in only one spot and then we can consider it a function.*0201

*If it crosses in more than one spot that is where we will get into trouble.*0208

*Here I have two little diagrams, this one crosses here and here, we would say that this is not a function.*0212

*This one, it only crosses in one spot and if I was to move that dotted line into a different spot over here, it still crosses only one spot.*0226

*In that situation, I would say that this is a function.*0247

*To determine the domain and range of a function when you are looking at its graph,*0262

*think of tracing back all of the values that we used back to the x values on the x axis and back to the y values on the y axis. *0266

*This looks a little difficult to do at first but it is not that bad.*0276

*I imagine picking out some point out on the graph but if I am looking for the domain, I will trace it back to figure out what value on the x axis it came from.*0281

*If I pick another point, trace that back where did it come from on the x axis.*0293

*I do this for many, many different points I’m always tracing it back *0299

*What I'm looking to do is trace back essentially every single point on that graph.*0303

*Now what this would end up doing is I will end up plugging back many different points *0310

*and they would end up shading in the domain of all the x values that we have used in the function.*0315

*This one, if I trace this back trace it back, you can see that it creates that entire line. *0327

*In this part of the line out here comes from tracing back values on this side.*0332

*Even the ones that it can not see it, sure enough they traced all the way back.*0338

*In a similar fashion, you can figure out the range by taking these points and going to the y axis.*0343

*It is because the y’s represents the outputs, shading the axis, so you knew what was in your range.*0349

*That way we get a little bit better sense of how to graph functions and things about them.*0365

*Let us practice.*0371

*Let us use the vertical line test to see whether these following relations are functions or not.*0373

*The way a vertical line test works is we imagine a vertical line or test it to see if it crosses in only one spot.*0379

*On this first one over here, let us go ahead and put down a vertical line.*0387

*No matter where we move that vertical line, it looks like it will only end up crossing in one spot.*0394

*Since it only crosses once, we will say that it is a function.*0402

*With this one over here, when I put down a vertical line it is easy to see that it crosses in 2 spots.*0412

*It crosses in two spots, not a function.*0422

*The vertical line test says it must only cross in one spot at the most.*0430

*Let us go ahead and see if we can graph one of our linear functions.*0440

*In this one, it is a special type since it is linear.*0443

*We want to look at the form to see if that will help us out.*0448

*This one is written in slope intercept, I know that the y intercept is the 3 and I have a slope of -1/4.*0451

*Let us start with our first being right at 3 and from that point I will go down 1 to the right, 1, 2, 3, 4.*0464

*I have a second point, so I will connect the two.*0473

*There is my line.*0483

*I can also graph out this line by simply choosing a whole bunch of different values for x and evaluating them one at a time. *0485

*I simply use the slope intercept form because it will be a lot quicker.*0495

*I do want to point out that either way would be fine, just pick out some different things for x, plug them in and see what you will get for y.*0500

* You use the graph to determine the domain and range of the function.*0513

*This is unusual in that with last time we actually looked at the equation and tried to pick out what the natural domain was.*0518

*In here, I just have the graph and I have no idea what is being used in here but I can see the inputs and outputs.*0525

*Remember that is every single point on this graph here.*0531

*To first figure out the domain, I will imagine all the points and trace them back to this x axis.*0536

*What I'm doing is I’m figuring out what points we get shaded in on that x axis when I start tracing them all back.*0545

*What it looks like it is doing is it is tracing out a lot of different values here.*0555

*In fact, even my little point way out here we get trace back.*0561

*I will end up shading quite a bit of the x axis. *0567

*One thing to note is nothing is over on the side.*0571

*The reason why I have nothing over there, is there is no graph to trace back to the x axis.*0576

*Our domain looks like it would start here at -3 and it will go on and on forever from there.*0581

*We can use the same process to figure out what the range is.*0596

*We will simply take all our points now and trace them back to the y axis.*0600

*We will see what this shades in as we do this, bring that one back and you can see I’m shading a whole bunch of different values here.*0605

*In some places you might have more than one spot it traces back, but that is okay. *0614

*Let us see, keep shading going to the y axis, this guy will go back to -2.*0620

*Notice how below that I'm not going to shade anything on that part because there is no graph to trace back to the y-axis.*0630

*What do we have for our range? Well, the lowest value I have here is that -2.*0642

*We will start there and then it keeps going on from there since the rest of it is shaded.*0648

*Let us do one more domain and range.*0660

*Graph the relation and determine if it is a function then state its domain and range.*0665

*We got a little bit to do with this one, let us first just develop a graph in it.*0670

*This one is not a linear equation so I do not have too many shortcuts at my disposal.*0675

*I’m just going to end up creating a table of values to help me out.*0680

*Let us choose some different values like 4, 5, 8, and 13.*0687

*These values will make it a little bit easier to evaluate this.*0698

*If I was to use 4, I would end up with 2 × √4 - 4 or 2 × √0 which is 0.*0702

*That is one point I know is on my graph, at 4, 0.*0721

*Let us go ahead and put in our next value and that would be 5.*0731

*2 × √5-4 that would be 1, √1 = 1 so I have 2 as this value.*0736

*I will plug in 5 and I have 2 as my output.*0748

*Looking good, let us try some more.*0752

*Let us go ahead and put in the 8.*0756

*8 - 4 would be 4 and the √4 is 2, 2 × 2 =4.*0762

*This shows that when I put in 8 my output is 4.*0773

*Let us put in 1, 2, 3, 4, 5, 6, 7, 8, 4, there you go.*0792

*It looks like 13 is going to be off my-*0803

*13 - 4 would be 9, 2 × √9, 2 × 3 or 6.*0815

*That is another point on our graph.*0823

*Looking at our points, I might as well start putting them together so we can have a nice little curve right here.*0827

*Onto our first question, I was able to graph the relation, but is it a function or not?*0835

*Does it pass the vertical line test? *0844

*That is our question that we should be asking. *0846

*If we imagine a vertical line on here, does it cross once, more than once? What is happening?*0849

*What was this vertical line? I can see that no matter where I put it, it is only going to cross this graph in one spot.*0856

* I will say yes it is a function since it passes the vertical line test.*0862

*Now we have to figure out what is its domain and range.*0877

*What are all the inputs we could use and what are all the outputs?*0880

*In making this chart, I can already see some of the inputs that I used that you could potentially use even more than that.*0884

*If we trace back all these values, it also includes all the numbers between the ones we used.*0890

*I’m tracing things back to the x axis and it looks like I would shade in all of this.*0896

*This starts at 4 and continues on from there.*0903

*The domain would start at 4 and just go on from there, 4 to infinity.*0907

*If I take all of the same values then I start to trace them back to the y axis it will shade in a lot of other values but it looks like nothing less than 0.*0917

*We would shade in all that and now we have our range from 0 up to infinity.*0933

*You can see that graphing functions are the same process as just graphing any type of relation. *0944

*Keep track of your inputs and outputs. *0949

*If it is a special type of function like a linear function, then use your tools for graphing lines.*0953

*When it comes to the domain and range, look at your inputs and outputs by tracing all of the values back*0959

*and then show the intervals of all the numbers that should be included. *0967

*Thanks for watching www.educator.com.*0970

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at systems of linear equations, pairs of linear equations *0002

*and get into their solutions and how we can graph them.*0008

*More specific things that we want to know is, what exactly is a system of equation and how can we find their solutions?*0015

*One of the greatest ways that we can do to find their solutions is just looking at their graphs.*0023

*Some special things that we want to know in terms of their types of solutions is how do we know when there is going to be no solution*0030

*and how we will know when they will actually be an infinite amount of solutions.*0036

*Watch out for those two as we get into the nuts and bolts of all this.*0041

*What is a system of equations?*0049

*A system is a pair or more of an equation coupled together.*0052

*Sometimes you will see a { } put onto equations just to show that they are all being combined and coupled together.*0059

*Here I have a pair, but I could have 3, 4, 5 and many more than just that. *0070

*Now if all of the equations that are being coupled together are lines, then we will call these linear equations.*0076

*You will notice in the examples that I have up here, both of these are lines in standard form.*0083

*I have a system of linear equations.*0089

*In order to be a solution of a system, it must satisfy all of the equations in that system all simultaneously, all at once.*0094

*If it only satisfies the first equation then that is not good enough to be a solution of the system, it must satisfy all of them.*0101

*Let us quickly look at how we can determine if a point is a part of the solution or not.*0116

*We are going to do this by substituting it into an actual system.*0123

*My system over here is 2x + 3y = 17 and x + 4y = 21.*0126

*Is the point 7, 1 a solution or not?*0133

*Let us go ahead and write down our system.*0138

*Let us first put it into equation 1, we have (2 × 7) + (3 × 1) is it equal to 17, I do not know.*0144

*Let us find out.*0157

*2 × 7 would be 14 + 3 and sure enough I get 17.*0158

*I know it satisfies the first equation just fine.*0168

*Let us take it and substitute it into this second equation.*0174

*I have 7x + (4 × y value × 1).*0182

*4 × 1 = 4, 7 + 4, that one does not work out because 11 is not equal to 21.*0193

*Even though it only satisfies one of the equations, I can say that it is not a solution.*0206

*It only satisfies one equation.*0223

*Let us try this with the point 1, 5 and let us see if that is a solution of the system.*0234

*We will put in 2 or 3 if we want to know if this is equal to 17.*0241

*I will put in 1 for x and we will put in 5 for y and let us simplify it.*0248

* 2 × 1 is 2 ,3 × 5 is 15 and 2 + 15 =17 and this one checks out.*0258

*Now let us substitute it into the second equation.*0273

*I will take 1 and put in for x, we will take 5 and put it in for y.*0278

*I have 1 + 20 this that equal 21? It does 21 equals 21.*0287

*I can say it is a solution and it satisfies both of the equations.*0296

*There are few ways that you can go about solving a system of equations and we will see them in later lessons*0316

*or the first ways that you can sense what the solution should be is to the graph the equations that are present.*0321

*If these are linear equations that you are dealing with then you can use the techniques of graphing your lines to make this a much easier process.*0328

*Let us work on finding our solutions graphically.*0335

*To do this, all you have to do is graph the equations and then find out where those equations cross, when they do that is going to be our solution.*0339

*There is one downside to this so be very careful.*0349

*In order for this process to work out, you must make accurate graphs. *0353

*If your lines are a little wavy or you do not make them very accurate then you might think you know where it cross, but actually identify a different point.*0358

*Be careful and make accurate graphs. *0369

*Since finding solutions using a graph is as nice and visual, we definitely will start with it*0381

*but we will use some of those later techniques for solving because the accuracy does tend to be an issue with the graphs.*0388

*Let us get into the graphing and finding a solution that way, I will try.*0400

*Here I have a system of linear equations and it looks like both of them are almost written in standard form. *0405

*The second one is not quite as in that form because it has a negative sign out front.*0412

*I think we will be able to graph it just fine.*0417

*I'm going to graph it by using its intercepts.*0419

*We will make a little chart for our first equation and I'm going to identify where it crosses the y axis and the x axis by plugging in an appropriate value of 0.*0424

*If I use 0 in my equation for the first one, I just need to solve and figure out what y is.*0436

*Putting in a 0, we will limit that term and solving the rest out, divide it by 3 and I will get that y =2.*0445

*I know that is going to be one point on my graph, 0, 2.*0457

*We will go ahead and plug in 0 for y and we can see that it will eliminate our y terms and now we just have to solve 2x = 6.*0467

*If we divide both sides of that by 2, we will get x =3.*0484

*We have a second point that we can go ahead and put on our graph, 3, 0.*0492

*Now that we have two points, let us be accurate on graphing this out.*0501

*If you want to make it even more accurate, one thing you can do is actually use more points to help you graph this out.*0517

*We have one line on here, it looks pretty good.*0526

*Let use another chart here and see if we can graph out the second one. *0528

*Once we have both lines on here, we will go ahead and take a look at where they cross.*0540

*This will be for line number two, we will get x and y. What happens when we put in some 0?*0545

*-0 + y = 7 that is a nice one to solve.*0554

*-0 is the same as 0, the only thing left is y = 7.*0560

*Let us put on the point 0, 7 right up there.*0567

*If we put in 0 for y, -x + 0 = 7, I will get that –x =7 or x = -7.*0578

*There is another point I can put on there.*0591

*Now that I have two points here, let us go ahead and connect the dots.*0599

*We can see where our two lines cross.*0614

*This is from our first line and the blue line is from our second equation.*0618

*Right here, it looks like they definitely cross.*0626

*That is the point -3, 1, 2, 3, 4. *0630

*The reason why we went over a little bit of work on testing solutions is, if our graph was inaccurate and this was incorrect, *0638

*we can take it and put it into our system just to double check that it actually works out. *0645

*If you want you can take -3 and 4, plug it back in and let us see what happens. *0650

*Starting with the first equation 2, 3y = 6 and let us plug in our x and y.*0664

*Checking to see if this is the solution, to see if it satisfies our first equation.*0678

* -6 + 12 does that equal 6? -6 + 12 does equal 6.*0683

*It checks out for our first equation.*0691

*Let us put it into the second one, our x value is -3 our y value is 4.*0696

*Negative times negative would be 3.*0710

*I’m sure enough 3 + 4 does equal 7 so it satisfies the second equation as well.*0716

*I know that -3, 4 is definitely my solution.*0721

*Let us try and find a solution to another system. *0730

*Here I have -2x + y = -8 and y = -3x + 2.*0735

*The first one I can go and ahead and make a chart for and maybe figure out to where its intercepts are.*0744

*That seems like a good way to find out that one.*0752

*What happens when x = 2, what value do we get for y?*0755

*This term would drop away because of the 0 and I have only be left with y = -8.*0765

*I know that is one point I will need.*0772

*Let us go ahead and put a dot on there.*0775

*x =0, y = -1, 2, 3, 4, 5, 6, 7, 8 way down here.*0777

*Now we put in a 0 for y, -2x + 0 = -8, let us see how this one turns out.*0786

*My 0 term will drop away and we continue solving for x by dividing both sides by -2, -8 ÷ -2 is 4.*0799

*There is another point, 4, 0, 1, 2, 3, 4, 0.*0818

*Now that we have two points, let us go ahead and connect them together and we will see our entire line.*0824

*We got one of them down and now let us graph our second one.*0843

*Notice how the second equation in our system is written in slope intercept form*0847

*which means we will take a bit of a short cut by using just the y intercept and also using its slope.*0852

*I will make things much easier. *0862

*Our y intercept is 2, I know it goes through this point right here.*0864

*Our slope is -3 which I can view as -3/1.*0868

*Starting at our y intercept we will go down 3 and to the right 1.*0873

*1, 2, 3 into the right 1, there is a point.*0878

*I’m bringing out my ruler and we will connect these.*0885

*Now we can check if each of them cross.*0897

*Let us highlight which equations go with which line, there we go. *0902

*Also, it looks like they do cross and I would say it crossed down here at 2, -4. *0906

*Even though I made some pretty accurate graphs, it does look like it is a hair larger than 2, and sometimes this happens.*0916

*Even when we do make some pretty accurate graphs, sometimes where they cross it looks like it is just a hair off.*0923

*The frustrating part is the actual solution may be different from 2, -4.*0932

*About the only way I'm sure when using this graphing method is to check it by putting it in to the system.*0937

*Later on we will look at more accurate systems where we would not have to check quite as much.*0944

*Let us go ahead and put in our values for x and y to see if it satisfies both equations.*0950

*x is 2 and y is -4, -2 × 2 = -4 + -4 is a -8.*0958

*It looks like the first one checks out, everything is nice and balanced so it satisfies the first equation.*0978

*Let us go ahead and put it into the second one and see if that one also works.*0986

*X is 2 and y is -4, let us see if this balances out.*0994

*-4 is not equal to -6 + 2 I think it is, because when you add -6 and 2 you will get -4.*1003

*It satisfies both equations.*1014

*2, -4 is a solution. *1017

*When you are going through finding different solutions for a system, several different things could happen. *1023

*The first situation that could happen is a lot like the examples we just covered.*1030

*You are going to go through the graphing process and you are going to find one spot where the two actually cross.*1035

*You will notice this kind of case that you like to be in. *1040

*However, when graphing these lines, you might also find that the lines are completely parallel.*1043

*It does pose a little bit of a problem because it means that the lines would not cross whatsoever.*1049

*With parallel lines, we will not have a solution to our system. *1065

*Another thing that could possibly happen is you go to graph the lines and they turn out to be exactly the same line.*1071

*In that situation, we will not only get solutions but we will get lots of solutions because if they are the same line, they will cross an infinite amount of spots.*1078

*Basically everywhere on a line will be a solution.*1087

*Watch for these to show up when you are making these graphs, you will either find one spot where they cross then you have a solution. *1090

*You will see that the lines are parallel and they do not cross so you have no solution.*1098

*You will find that they are exactly the same line and then you will have an infinite number of points as your solution.*1103

*In this next example, I want to highlight that the two special cases that could happen. *1114

*One thing I want to point out is that when you are looking at the system, *1119

*it is sometimes not obvious whether it has no solution or an infinite amount of solutions. *1122

*You got to get down into the graphing process or the solving process before you realize that.*1127

*My first equation is x +2, y= 4 and the other one is 3x + 6, y =18.*1133

*I’m going to graph these out using my x and y intercepts.*1140

*What happens when x =0, what happens when y= 0?*1146

*For the first one, I put in 0 for x and we will go ahead and we start solving for y.*1149

*You can see my 0 term is going to drop away, which is good.*1157

*I will divide both sides by 2 and I will get that y = 2.*1161

*There is one point that I will mark on my graph 0, 2.*1169

*I will put in 0 for y and let us see what happens with that one. *1179

*The 0 term is going to drop out and the only thing I'm left with is x = 4.*1189

*I will put that point in there, 4, 1, 2, 3, 4, 0.*1199

*I will go ahead and graph it out.*1204

*Now that we have one line, let us go ahead and graph the other.*1219

*We will get its intercepts by putting in these 0 for x and 0 for y.*1228

*(3 × 0) + 6 time to solve for y.*1243

*Our 0 term will go away and I will just be left with 6y =18.*1251

*6 goes into 18 three times so I have 0, 3 as one of my end points.*1259

*Putting in 0 for y, let us see what happens there.*1273

*3x + 6 × 0 = 18.*1275

*It looks like that term will drop away and I will just have 3x =18.*1282

*3 goes into 18 six times, x = 6.*1290

*1, 2, 3, 4, 5, 6 I have that one.*1300

*Let us go ahead and graph it out now.*1306

*What we can see from this one is that it looks like the two lines are parallel and they are not crossing whatsoever.*1320

*If it is a little off and it looks like one of your lines could potentially cross but maybe off the graph, *1330

*one thing you can do is you can rewrite the system into slope intercept form so you can better check the slopes.*1335

*I'm pretty sure that these are parallel, I’m going to say that there is no solution to the system. *1344

*They do not cross whatsoever.*1353

*Let us try one more and see if this one has a solution. *1359

*This is y = 4x - 4 and 8x -2y =8.*1362

*The first one is written in slope intercept form so I will graph it by identifying the y intercept and its slope.*1370

*This starts at -4 and starting at the - 4 I will go up 4/1and just like that we can go ahead and graph it out.*1382

*There is our first line.*1406

*The second one is written more in standard form so I will go ahead and use its intercepts to track that one down.*1408

*What happens when x =0 and what happens when y = 0.*1415

*When x =0, that will drop away that term right there and I'm looking at -2y = 8.*1422

*We will be dividing both sides by -2 and I will end up with -4.*1434

*There is one point I can put on there, notice it is on the same spot.*1443

*Putting in 0 for y, I can see the term that will go away.*1451

*I am left with 8x = 8.*1468

*Dividing both sides by 8, I am simply be left with x = 1.*1471

*We will put that point on there and both points ended up on the other line.*1482

*When I go to draw this out, you will see that one line actually ends up right on top of the other one. *1487

*They are essentially the same line. *1492

*In this case, we end up with an infinite amount of solutions because they still cross, but they cross in lots of different spots. *1501

*I could say they cross everywhere, there are infinite number of solutions.*1508

*One of the best ways that you can figure out a solution to a system of equations *1530

*is simply graph both of the equations that are present in the system and see where they cross.*1533

*We will look at some more accurate methods in the next lesson. *1539

*Thank you for watching www.educator.com.*1542

*Welcome to www.educator.com.*0000

*In this lesson, we are going to work on solving a system of equations using the substitution method.*0002

*What we want to focus on is how we use the substitution method in order to find solutions and I will walk into that process.*0013

*We will also keep in mind that since sometimes there is no solution or there is an infinite amount of solutions, *0021

*how we can recognize those cases when using the substitution method.*0026

*We are called the reason why we need more methods for solving a system is that when we use the graphing method, *0035

*you have to be very accurate in order for that method to work and if your lines are a little bit wavy *0042

*or you do not make them just right then you may not find a solution of that system.*0048

*Since, accuracy is such a problem that is why we are going to focus more on those algebraic methods.*0054

*How can we manipulate the system in such a way so that it gives us a solution.*0060

*Two of these methods of our algebraic methods are substitution and elimination.*0067

*In this lesson, we focus on substitution and then look at elimination in the next one. *0073

*Let us see how the substitution method works.*0080

* In the substitution method, what we first end up doing is taking one of our equations and solving it for one of the variables.*0086

*Once we have taken one equation and we have solved it for variable, we will take that and substitute it into the other equation.*0095

*What this will do is it will give us an entirely new equation with only one type of variable in it *0105

*and will end up solving for that one remaining variable that is in there.*0110

*We will have half of our solution but since we are looking for a system and we need an x and y coordinate, *0117

*then we will end up using back substitution to find the rest of our solution.*0123

*We will take what we have found and end up substituting back into one of the original equations. *0129

*This will help us find the other variable. *0135

*As long as everything works out good, that it should be our solution, but it is not a bad idea to just take that and check it in the original system. *0138

*If you do make a mistake, sometimes you would not catch it, so playing it back into the original system is always a good idea.*0147

*Let us walk through this in substitution method using the following example.*0157

*I have - x + 3y = 10 and 2x + 8y = -6.*0162

*I want to start solving for at least one of the variables in one of the equations.*0170

*I have lots of different choices that I can do. *0175

*I can solve for x over here. *0178

*I can solve for y in the first one or you can even pick on the second equation a little bit and solve for its x or y. *0180

*It does not matter which one you solve for, just pick a variable and go ahead and solve for it.*0187

*I only choose x in the first equation, it looks like it will be one of the easier ones to get all by itself. *0194

*What I will do is solve for x and I'm moving everything to the other side.*0200

*I'm subtracting 3y on both sides.*0209

*It is almost all by itself but it looks like I have to multiply it by -1 to continue on from there. *0213

*I’m multiplying through by -1 and I will end up with x = 3 and y -10.*0220

*This entire thing right here is equal to x.*0228

*For the substitution method happens is I will take in the x in the other equation and replace it with what it is equal to, 3y – 10.*0235

*It will leave lots of room for this so you can see what it is going to turn that second equation into.*0250

*Instead of writing x, I’m going to write 3y -10.*0258

*I have taken that second equation, substituted in those values and notice how this new equation we created it only has y.*0269

*Since this new equation only has y’s in it, we will be able to solve for y and be able to figure out what that is.*0279

*We still have a little bit of work to do, of course we will have to simplify and combine terms but we will definitely be able to solve for that y.*0287

*Let us give it a shot.*0293

*I’m going to distribute through by my 2 and put 6y – 20 + 8y = -6.*0295

*Let us go ahead and combine our y terms, so 6 + 8 is a 14y and then let us go ahead and add 20 to both sides.*0307

*14y – 6 + 20 is 14 divide both sides by 14 and get y = 1.*0319

*It looks like I have a solution but remember that this is only half of our solution.*0330

*We still need to figure out what x is.*0333

*What this point, now that we know what y is I can substitute it back into one of my equations and figure that out. *0336

*Let us do that, -x + 3 and I will put in what y is equal to 1.*0346

*I will end up solving this equation for x, -x + 3 = 10, so I'm going to subtract 3 from both sides.*0355

*I will go ahead and multiply both sides by -1 and get x = -7.*0372

*I have both halves of my solution.*0383

*The x is written first, so -7 and y is written second.*0387

*My solution is -7, 1.*0393

*If I'm looking at that and I'm curious if it is the solution or not, it is not a bad idea to just take it and plug it back into the original.*0396

*Let us see how that works out.*0404

*My value for x is -7 + 3 and my value for y is 1.*0408

*Does that really equal 10? Let us find out.*0413

*Also a- -7 would be 7 + 3 = 10, sure enough, 7 + 3 is 10.*0416

*That works out in the first equation.*0427

*Right now, it is still the same way with the second one.*0430

*(2 × -7) + (8 × 1) is that =equal to -6, let us find out.*0432

*-14 + 8 sure enough -6 is equal to -6, it checks out in both the equations. *0440

*I know that -7, 1 is my solution.*0449

*There is a lot of substitution, of course, in the substitution method, but is not as bad as you think.*0454

*You do have some choices on what you will solve for and where it goes.*0460

*One thing that we saw when we are finding solutions graphically is that several things could happen.*0472

*We might have one solution, we might have no solution or we actually might have an infinite amount of solutions. *0477

*Now when we are looking at all of these graphically, it was pretty simple on figuring out which case we were at. *0483

*We either looked at both our lines, actually cross and then we said they have a solution. *0489

*We could see if there was no solution if the lines were parallel and we can see that it was an infinite amount solution if they were on the same line.*0495

*Now that we are doing things more algebraically, we would not be able to visually see what case we are in.*0503

*How is it we figure out if it has no solution or an infinite amount of solutions?*0509

*It all depends on what values you will get while going through that solving process.*0514

*If you go through the substitution method and you can find an x and y that work, then you will know that it has a solution.*0520

*It has one solution, nothing strange happens.*0526

* If however you go through that substitution method and end up creating a false statement, it means that there is actually no solution to the system.*0530

*Now if you do come across a false statement, what I always tell everyone is check your work first to make sure that you have made no mistakes.*0539

*If all your work looks great and you still create a false statement then you can be sure that it has no solution.*0548

*If you go through and you create a true statement then you can not see what your solution is.*0555

*That is a good indication that the two lines are exactly the same and you have an infinite amount of solutions.*0560

*Again, check your work with this one, but if all your work looks good, then you know that you have an infinite amount. *0566

*Let us look at some examples like these ones and see how we can recognize what I mean by a false statement and what I mean by a true statement. *0576

*We want to solve the following system 5x - 4y = 9 and 3 - 2y = -x.*0589

*The first thing I need to do is solve for either x or y.*0597

*It does not matter which one we choose, it looks like it might be a good idea to solve for x down here since it is almost all by itself anyway.*0601

*We will just go ahead and multiply everything through by -1, -3 + 2y = x.*0609

*Now that we have what x is equal to, we will substitute it into the first equation for that x.*0619

*It will leave lots of space and we will just go ahead and put it in there, -3 + 2y.*0627

*We can see in this new equation it only has y, so I need to work on getting them together and isolating them.*0636

*-15 it will be from distributing through +10y -4y = 9.*0645

*It have some like terms I can put together 10y and -4 that will be 6y =9 and adding 15 to both sides, + 15 = 24.*0656

*Now dividing both sides by 6, I will get that y = 4.*0680

*We have half of our solution and we always figure out what the other half is by putting it back in one of the original equations. *0685

*I often have students ask me should you put it back in the original, after all could not I just stick that y back and up here.*0692

*What I tell them is yes you could put it back into this equation for y, but be very careful because that equation, we already have manipulated in some sort of way.*0700

*If you want to be sure that you are not making mistakes on top of mistakes, take that and plug it into of the original ones, *0711

*something that you have not mess around with too much.*0718

*Let us see 3 – 2 and I will put the 4 in there.*0721

*We take this and solve for x, 3 -8 = -x.*0729

*3 - 8 would be -5 and dividing both sides by negative, I have that x = 5.*0737

*I have both halves of my solution x = 5, y = 4.*0747

*My solution is 5, 4.*0755

*Again, if you want to make sure that is the solution, feel free to plug it back into both equations and check it to make sure it works out.*0759

*In this next example, we will see if we can solve this one using the substitution method.*0769

*It says 4x - 5y = -11 and x + 2y = 7.*0773

*Let us see what variable shall we solve for.*0780

*This x looks like it would be nice and easy to get all alone, let us do that one.*0784

*We will go ahead and say x = -2y + 7.*0790

*I took the 2y and subtracted it from both sides.*0797

*I will take all of this and substitute it into the first one.*0801

*It will give me 4 - 5y = -11.*0805

*I left a big open spot so I can drop what x is equal to in there, 2y + 7, it looks good.*0812

*Now I’m going to distribute through with my 4, -8y + 28.*0820

*And I'm going to continue trying to solve for y.*0832

*I will also see what I got here, -8 - 5 putting those together I will end up with -13, subtracting 28 from both sides, I will end up with -39.*0835

*It looks like I can finally do dividing both by it sides by -13, in doing that y = 3.*0853

*Let us go ahead and take that and substitute it back into one of our original equations.*0863

*Let us go ahead and put in here, x + (2 × y).*0867

*Multiplying together the 2 and the 3 would be 6 and I'm left with x + 6 = 7.*0878

*Subtracting 6 from both sides x = 1.*0886

*My solution for this one is x = 1 and y = 3.*0892

*In some systems, remember we could have that situation of having no solution or an infinite amount of solutions.*0903

*Sometimes when you are looking at an equation, it is tough to tell whether you have that situation or not.*0909

*Watch very carefully what happens with this one.*0914

*I’m going to start off like I what did with the other examples.*0917

*We just want to take these and solve it for one of our variables.*0920

*I’m going to solve the first one for x since that one looks like it can be nice and easy to get isolated.*0924

*That would give me x - 2y from both sides, + 4.*0932

*Now that I have x isolated, let us substitute it into the second equation.*0940

*3 + 6y = 13 and we will drop in -2y + 4.*0947

*Now I only have y in this new equation so we will continue solving it by getting y all by itself. *0959

*Let us go ahead and distribute -6y + 12, + 6y = 13.*0965

*There is something interesting happening here.*0975

*Over here I have -6y and over here I have 6y.*0977

*Since they are like terms, I can put them together by -6 and 6 would give me 0y, it cancel each other out.*0982

*That means the only thing left is 12 on the left and 13 on the right but that does not make any sense.*0993

*12 is not equal to 13 so I’m going to write not true.*0999

*If you check through back all of my steps, I did not make any errors.*1008

*I moved the -2y to the other side that turnout okay and I did my distribution fine.*1012

*I know my steps worked out perfectly good.*1019

*What this statement is telling me at the very end, since it is not true is that these two lines are actually parallel. *1023

*There is no solution and they do not cross.*1030

*Watch out for these false statements that might come along the way and they will help you recognize when a system does not have a solution.*1039

*One more system, this one is 2y = 4x and the other equation is 4x - 2y = 0.*1051

*Let us take the first one and solve it for y.*1061

*We can easily do it just by dividing both sides by 2, y = 2x*1065

*Now that it is solved for y, let us go ahead and substitute that into the second equation.*1075

*You will see that this is turning out a lot like the previous example.*1091

*If I combine my x’s, the other since they are like terms, they will cancel each other out.*1097

*This is slightly different though because after they do cancel out, all that I get is 0 = 0, which happens to be a true statement 0 does equals 0.*1103

*Let us check our work and make sure you did everything okay.*1118

*I divide both sides by 2 and that worked out good and I combined my 2’s together and cancelled.*1122

*All my work looks great.*1127

*The fact that I have a true statement and all my steps were good means that what is going on here is that the lines themselves are exactly the same line.*1129

*We might even see that at the very beginning if we end up rewriting them.*1139

*When I moved the 2y to the other side on the second equation and then end up rearranging them, they are exactly the same equation.*1146

*Visually these would be right on top of one another.*1160

*This tells us we have an infinite number of solutions and everywhere on that line is a solution.*1164

*These more algebraic methods are definitely more accurate when it comes to finding solutions.*1179

*You want to be very familiar and being able to go through them.*1184

*Be on the watch out for these two special cases, if you get a false statement then you know you have no solution.*1188

*If you get a true statement then know you have an infinite amount of solutions.*1195

*Thank you for watching www.educator.com.*1199

*Welcome back to www.educator.com.*0000

*In this lesson we are going to work on solving a system of linear equations using the method of elimination.*0003

*We focus on how to use this elimination method or how it differs from substitution *0011

*and how we can recognize the many different types of solutions that could happen.*0017

*Do we get one solution, and infinite amount of solutions, or possibly no solution or whatsoever. *0021

*The goal with the elimination method is to combine the equations in such a way that we eliminate one of our variables.*0030

*In that way, we will only have one that we will need to solve for.*0037

*In order to do this, sometimes we will have to multiply one of the equations or even both of the equations by some sort of a constant. *0042

*Now what we are looking to do when we multiply is we want a pair of coefficients that will end up canceling each other out.*0050

*Do I have - 3x and 3x?*0059

*When we combine those they would end up canceling out and that is what I want.*0062

*Once I get a pair of coefficients that I know will cancel, we will go ahead and actually add the 2 equations together.*0067

*In fact, some people call the elimination method the addition method because we end up adding the equations.*0073

*We will get a new equation after doing that and we will only have one type of variable in it.*0081

*We will end up solving that new equation.*0086

*We will have half of our solution at this point so we will be able to take that and substitute it back into one of the originals.*0090

*In fact, if you look at all the rest of stuffs from here on out, these are the same as the substitution method.*0096

*We simply have half of our solution and we are back substituting so we can find the other value. *0104

*Once we are all the way done through this entire process, it is not a bad idea to check the solution to make sure that does satisfy both of the equations.*0110

*Let us walk through this method, so you get a better sense of how it works.*0123

*Here I have 2x – 7y = 2 and 3x + y = -20.*0129

*What my goal is to either eliminate these x values here or I need to eliminate my y values. *0136

*If I was just going through and I decide that I was going to add them together as they are, *0145

*you will see that this would not be a good idea because nothing happens.*0150

* 2x + 3x would give me 5x and – 7y + y would give me -6y.*0154

*2 and -20 is -18, so I definitely created a new equation, but it still has x’s and it still has y’s.*0163

*There is not much that I can do with it.*0171

*What I want to happen is for some things to cancel out, so I only have one type of variable to solve for. *0174

*Let us go ahead and a do a little work on this so that we can get something to cancel out.*0181

*The first thing is we get to choose what we want to cancel out.*0188

*Is it going to be the x's or y’s?*0191

*I want to choose my y since one of them is already negative and the other one is positive.*0194

*In order to make these guys came flat, let us take everything in the second equation and multiply it by 7.*0200

*The first equation exactly the way it is, no changes, and everything in the second equation will get multiplied by 7.*0210

*3 × 7 = 21x, I have 7y, -20 × 7 would be -140.*0218

*I still have a system and I still have a pair but now I noticed what is going on here with the y.*0231

*Since 1 is -7 and one is 7, when I add those together, we will end up canceling each other out.*0236

*Let us do that, let us go ahead and get to the addition part of the elimination method.*0244

*Adding up our x's, we will get 23x.*0248

*When we add up the y, we will get 0y so that term is gone.*0254

*Over on the other side - 138 and in this new equation the only thing I have in here is simply an x.*0259

*I can solve this new equation for x and figure out what it needs to be which I can do by dividing both sides by 23.*0270

*How many times 23 is going to 138?*0292

*6 × 3 = 18, looks like 6 times so x = -6.*0303

*That is half of our solution and now we need to work on back substitution.*0313

*We will take this value here for our x and plug it back into one of the original equations.*0318

*It does not matter which one you plug this into, just as long as you put it into one of them it should turn out okay.*0323

*I end up solving this one for y, 2 × -6 = -12 and we are adding 12 to both sides would give me 14 and y divided by -7, I have -2.*0335

*I have both halves of the solution.*0360

*I have x = -6 and that y = -2.*0362

*In the elimination method, we are working to eliminate one of our variables.*0370

*You might be curious what would have happened if we would have chosen to get rid of those x's instead.*0377

*We could have done it, but you would have to multiply your equations by something different.*0383

*I’m not going all the way through this, but just to show you how you could have started.*0390

*For example, if you multiply the first one by 3, it would have given you 6x - 21y = 6.*0394

*If you multiply the second equation by -2, - 6x - 2y = 40.*0405

*You would see that using those multiplications, the x's would have cancel out when you add them.*0414

*In the elimination method, eliminate one of your variables.*0421

*Onto the next important thing with the elimination method.*0428

*We have many different things that could happen. *0432

*We could have one solution, no solution or an infinite amount of solutions.*0433

*Visually, we can see that the lines either cross, do not cross, or perhaps they are the same line.*0439

*The way you recognize this when using the elimination method is that you might go through that method and everything will work out just fine.*0457

*You actually be able to find your solution.*0464

*That is when you know it has one solution, everything is good.*0466

*If you go through the system and all your work looks good, but you create a false statement *0471

*then you will know that there is actually no solution to the system.*0476

*In fact, they are 2 parallel lines and they never cross. *0479

*I think I mentioned this earlier, but go ahead and check your work if it looks like you are getting a false statement*0483

*just to make sure that the false statement is created from them being parallel and not from you making a mistake.*0488

*If you go through and you create a true statement, then this is an indication they have an infinite amount of solution.*0496

*Again, check your work, but make sure that you know whether it has an infinite amount or not.*0503

*These are exactly the same criteria that using the substitution method, so they are nice and easy to keep track of.*0512

*False statement, no solution, true statement, infinite amount of solutions and everything works out normally, then you just have one solution.*0518

*Let us get into some examples and see this in action.*0526

*In this first example, we will look at solving 2x - 5y = 11 and 3x + y = 8.*0534

*Think about our goal with the elimination method.*0543

*We want to get rid of these x's or we want to get rid of the y.*0546

*You get to choose which one you want rid of just a matter how you will end up manipulating these to ensure that they do cancel each other out.*0550

*I think the better ones to go after are probably these y, since one is already positive and one is already negative.*0561

*The way we are going to do this is when I take the second equation and we are going to multiply everything there by 5.*0570

*Let us see the result this will have.*0579

*We have not touched the first equation and everything in the second equation by 5 would be 15x + 5y = 40.*0582

*I can see that when we add together our y values, they will cancel each other out.*0597

*2x + 15x that is 17x, -5y + 5y they would cancel each other and get 0y.*0605

*11 + 40 =51, so in this new equation I can see that we only have x to worry about and we can continue solving for x.*0617

*We will do that by dividing both sides by 17, x =3.*0625

*Now that I have half of my solution and I know what x is, let us take it and end up substituting it back into one of our original equations.*0636

*(2 × 3) - 5y = 11.*0647

*Our goal here is to get that y all by itself.*0656

*We will multiply the 2 and 3 together and get 6, then we will subtract 6 from both sides, - 5y is equal to 5.*0660

*Now, dividing both sides by -5, we will have y = 8, -1.*0680

*We have both halves of our solutions and I can say that our solution is 3, -1.*0693

*Let us try another one.*0703

*For this next one we will try solving 3x + 3y = 0 and 4x + 2y = 3.*0709

*In this one, it looks like all of my terms on x and y are all positive.*0716

*It is not clear which one will be the easier one to get rid off, we just have to pick one and go with it.*0721

*Let us go ahead and give these x's a try over here and see if we can eliminate them.*0726

*Some things that you can do to help you eliminate some of these variables, is first just try and get them to be the exact same number.*0732

*What some of my students realizes is that if you just take this coefficient and multiply it up here, *0740

*then take the other coefficient and multiply it into your second equation.*0746

*That is usually enough to make them exactly the same .*0751

*I’m going to do that here. *0753

*I’m going to take the entire second equation and we will multiply it by 3.*0755

*We will take everything in the first equation and we will multiply that by 4.*0762

*Let us see, 4 × 3 = 12x + 12y, 0 × 4 = 0, so that will be my new first equation.*0768

*Everything in the second one by 3, 12x + 6y = 9 will be the second one.*0783

*We are off to a good start and I can see at least that the x's are now exactly the same.*0792

*We want them to cancel out, let us go ahead and take one of our equations and multiply it entirely through by a negative number.*0798

*Let us do it to the second one, everything through in the second one by -1.*0806

*It will give us -12x - 6y = -9, the first equation is exactly the same so we would not touch that one.*0810

*By multiplying it just that way, we are doing a little bit of prep work, now I can be assured that my x's will definitely cancel out.*0823

*Let us add these 2 equations together and end up solving for y.*0831

*12x - 12y that will give us 0x, that is gone.*0836

*12y - 6y will be 6y and 0 + -9 = -9.*0841

*y = -9/6 which we might as well just reduce that call it to say – 3/2 and there is half of our solution.*0852

*Let us go ahead and take this, plug it back into the original and see if we can find our other half.*0866

*I’m just going to choose this one right here, 4x + 2 =- 3/2 equals 3.*0873

*In this equation, the only thing I need to solve for is that x, let us go ahead and do that.*0886

*2 × -3/2 = - 3, adding 3 to both sides here I will be left with 4x = 6 and dividing both sides by 4, I will have that x = 3/2.*0895

*My final solution here, x = 3/2 and y = -3/2.*0917

*As long as you do proper prep work, you should be able to get your final solution just fine.*0926

*Remember that it has 2 parts, the x value and y value.*0931

*Let us see if we can solve this one using elimination.*0939

*I have my x's that are both positive and my y’s are both negative, but they are almost the same.*0947

*Let us start off by multiplying the second one by 5 and see if it gets a little bit closer to canceling.*0955

*5x - 5y, 5 × 12 = 60 we multiply the entire second equation by 5.*0965

*It still looks like nothing will cancel out, so I will also multiply the second equation by -1.*0977

*-5x + 5y = - 60 looking pretty good here.*0983

*I will add these together and I can eliminate my x's but you will notice that it looks like the y will also eliminate.*0998

*If the x's and y’s are gone, the only thing on the left side is just 0.*1007

*What is on the other side? 3 - 60 would be a -57 and that is a bit of a problem because the 0 does not equal -57.*1013

*Let us say that this is a false statement.*1026

*This is one of these situations and it is a good idea to go back through your work, check it and make sure it is all correct.*1031

*All of our steps do look good here, we multiplied it by 5, we multiplied by - 1.*1038

*We do not actually have a whole lot of spots where we could make mistakes.*1042

*What this false statement is telling us here is that there is no solution to the system.*1046

*These 2 lines are actually parallel and they are not crossing whatsoever. *1057

*Watch out for those special cases.*1061

*Let us try one last example, x – 4y =2 and 4x – 16y = 8.*1068

*I want to try and maybe eliminate, let us do the x's.*1077

*We will do this by multiplying the first equation by -4.*1081

* -4x, -4 × -4 =16y equals -8.*1094

*In the second equation I do not need to manipulate that one, I will leave it exactly the same.*1105

*This is a lot like our previous example.*1113

*The x's are going to cancel out when they are added together, but then again, so are the y values. *1117

*Both of them are going to cancel out.*1122

*I have 0 on one side of my equal sign.*1124

*This one actually is not quite so bad because if I look at my -8 and 8, they will cancel each other out as well.*1128

*I will get 0 = 0, which happens to be a true statement. *1137

*Now, one statement is true but what is your x and y?*1141

*In indicates that the lines are exactly the same and there are on top of one another.*1145

*We have an infinite number of solutions.*1151

*Using one of these algebraic methods is a great way to figure out what the solution is.*1164

*Keep in the back of your mind what the graphs of these look like*1169

*so we can interpret what it means to have one solution, no solution, or an infinite amount of solutions.*1172

*Thank you for watching www.educator.com.*1178

*Welcome back to www.educator.com.*0000

*In this lesson we are going to look at applications of systems of equations. *0003

*Think about word problems involving the systems of equations that we saw earlier. *0007

*Our main goal will be looking at these word problems and figuring out how we can interpret them and build the system from there.*0015

*Here is a part of this that we will be actually looking at our solutions and be able to figure out how it fits in the context of that word problem.*0022

*A good way to think of how to start picking these apart is to take every little bit of information in there.*0033

*Say 2 different situations and create an equation for each of those situations.*0039

*You will notice some of these problems look similar to ones we did in the earlier applications of linear equations.*0045

*In that lesson we tried to write everything in terms of one variable. *0051

*Now that we know more about systems, we can use 2 variables to help us out and create an equation for each of the situations present.*0055

*It will actually make things a little bit neater.*0062

*Once we have built our system of equations, we are getting down to being able to solve that system.*0067

*We will use our techniques for solving a system such as elimination method or the substitution method.*0072

*We can also use tables like we did earlier to help you organize this information *0079

*but I will mainly focus on just the methods of substitution and elimination to tackle the systems of equations.*0083

*One very important part to remember is that all of these are going to be word problems.*0094

*Even though we get a solution like x = 3 and y = 5, we have to take those and interpret them and say what is x and y.*0099

*Always interpret these in the context of the problem. *0108

*One thing that will make this very easy and not so bad, is to go ahead and write down what your variables are at the very beginning.*0112

*In that way, once you are all done with the problem you can reference that and say okay my x represents this and my y represents this.*0122

*That way you can easily say here is how it fits in the context of the problem.*0130

*Let us jump into our examples and see how we can actually start creating a system of equations.*0140

*In this first one, we have a restaurant that needs to have 2 seat tables and 3 seat tables.*0146

*According to local fire code, it looks like the restaurants maximum occupancy is 46 costumers.*0151

*If the owners have hardly enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?*0158

*Notice that we are under a few different constraints.*0165

*One has to do with simply the number of people that you can fit in this restaurant.*0168

*We want to make sure that we have no more than 46 customers.*0173

*Another restriction that we are under is the number of tables that our servers can handle and provide good service and we can only do 17 tables.*0177

*From the owner's perspective, we do not want to go less than these numbers. *0187

*After all, if we have less than 46 customers, we are not making as much money.*0190

*If you have less than 17 tables then we have servers that are not doing a lot of work.*0195

*We are going to try and aim to get these exact when we set up our equations.*0200

*Let us hunt down some unknowns and first write those down.*0205

*I'm going to say let x that will be the number of 2 seat tables.*0210

*Let y be the number of 3 seat tables.*0228

*We can use these unknowns to start connecting our information gathered.*0243

*In the first situation, we are going to focus on the number of people that we want to fit in this restaurant.*0248

*We are looking to put 46 people in this restaurant and it will come from seating them either at these 2 seat table or at the 3 seat table.*0254

*There will be 2 people for every 2 seat table so we can represent that by saying 2 × X.*0264

*There will be 3 people at every 3 seat table or 3y.*0272

*That expression right there just represents a number of people and we want it equal to 46 customers.*0277

*That is dealing with our people, maybe I will even label that people.*0285

*The other restriction is what we can do with our servers in handling all of these tables.*0291

*We want the total number of tables and that is our 2 seat tables and our 3 seat tables to equals 17.*0298

*You can say we have encapsulated all the information to one equation or the other.*0311

*We have actually set up the system and it is just a matter of going through and solving it.*0316

*You could use the elimination method or you could use the substitution method, both of them should get you to the same answer.*0321

*I’m going to go through the elimination method.*0327

*I will do this by taking the second equation here and we will multiply that one by -2.*0331

*Let us see what the result will be.*0340

*The first equation will remain unchanged and everything in the second one will be multiplied by -2.*0341

*We have done that, we can go ahead and combine both of those equations together.*0365

*You will notice that the x’s are canceling out since one is 2 and one is -2, that is exactly what we want with the elimination method.*0371

*3y + - 2y will give us 1y, and now we have 46 - 834 and that will give us 12.*0379

*One of the great thing is about writing down our variables at the very beginning is not only do I know that y = 12, *0397

*but I can interpret this as the number of 3 seat tables since I have it right here that y is the number of 3 seat tables.*0402

*We are going to continue on, and I will figure out what x is so we can figure out the number for both of the tables.*0410

*It is not so bad, just always remember that the total number of tables must be equal to 17.*0415

*I will borrow one of our original equations down here and just substitute in the 12 for y.*0421

*To solve that one we will subtract 12 from both sides and get that x = 5.*0433

*Now we can fully answer this problem.*0441

*The number of 3 seat tables is 12 and the number of 2 seat tables that will be 5.*0446

*We will keep that as our final answer.*0462

*Set down your unknowns, set up your system, solve it and interpret it in the context of the problem.*0468

*Let us try another problem, it has a lot of the same feel to it.*0478

*This one says that at a concession stand, if you buy 5 hotdogs and 2 hamburgers the cost is $9.50, 2 hotdogs and 5 hamburgers cost $13.25.*0483

*We are interested in finding the total cost of just one hamburger.*0495

*First, we need to go ahead and label our unknown. *0499

*We have the number of hotdogs and we have our hamburgers.*0504

*Let us work on that. *0509

*I will say that we will use and see d for hotdogs and let us use h for our hamburgers.*0512

*Both of these situations here are involving costs.*0529

*In the first one, we are looking at so many hotdogs, 5 of them and 2 hamburgers all equal to $9.50.*0534

*In the next situation, we have a different combination of hotdogs and hamburgers to equal the $13.25.*0543

*We will take each of these and package them into their own equation. *0549

*5 hotdogs 2 hamburgers equals to $9.50.*0554

*2 hotdogs 5 hamburgers equals to $13.25.*0569

*There is our system of equations.*0581

*When we are done solving we should be able to get the price for just one hotdog and a price for just one hamburger. *0583

*In this system, you could solve it using elimination or substitution. *0591

*I think I'm going to attack this using our elimination method.*0595

*I will have to multiply both equations by something to get something to cancel out.*0599

*Let us multiply the first one here or multiply everything in there by -2.*0604

*We will take our second equation and I will multiply everything there by 5.*0614

*Let us see what the result is, -10 - 4h -19.*0621

*Then we will take our second equation 10 d + 25h (5 × 13.25).*0639

*Adding these 2 equations together, let us see what our result is.*0661

*The 10 d and -10 d both will cancel each other out and we will have 25h - 4h = 21h.*0665

*We can add together the 66 + 25 -19.*0679

*In this new equation, we only have h to worry about and we can get that all by itself just by dividing both sides by 21.*0689

*This will give me that h = $2.25.*0702

*By looking at what we have identified earlier, I know that this is the cost of one hamburger $2.25.*0708

*Even though the problem is not asking for it, let us go ahead and figure out how much a hotdog is*0715

*by taking this amount and substituting it back into one of our original equations.*0719

*You can see that I have put this into the first equation, now I need to multiply 2 by $2.25 that will $4.50.*0737

*Subtract $4.50 from both sides and divide both sides by 5.*0750

*Now we have our entire solution. *0768

*I know for sure that the hamburgers will cost $2.25.*0769

*I also know that one hotdog will cost exactly $1. *0774

*Try and pick out each situation there and interpret it into its own equation.*0780

*Let us try this one out.*0789

*In the U.S. Senate, there are 100 representatives, if there are exactly 10 more democrats than republicans, how many of each are in the senate?*0791

*One assumption that we are going to use here is that all representatives are either democrats or republicans.*0799

*We are not even going to consider any third parties here.*0804

*Let us see if I can break this down but first let us identify some variables.*0808

*Let us say r is the number of republicans and we will let d be the number of democrats.*0813

*We need to set up an equation for each of these and one of the first big bits of information I get is that the total number of representatives should be 100.*0846

*We can interpret that as the number of republicans + the number of democrats that should be equal to 100.*0856

*We need to interpret that there are exactly 10 more democrats than republicans.*0870

*To look at that, let us compare the 2.*0879

*If I was to take republicans and democrats and put them on each side of the equal sign,*0881

*they would not exactly be equal because I have 10 more democrats than republicans.*0887

*The way you want to interpret is that right now this side got a lot more, it has 10 more than the republican side.*0893

*If I want to balance out this equation, I need to take something away from the heavier side.*0901

*Since it is exactly 10 more, I will take away 10 and that should balance it just fine.*0906

*My second equation is r = d -10.*0914

*I will get that relationship that there are exactly 10 more democrats than republicans.*0917

*When we are done setting up the system, notice how it set up a good for substitution.*0924

*The reason is if you notice here our r is already solved.*0929

*Let us use that substitution method to help us out.*0940

*We will take the d – 10 and substitute it into the first equation.*0943

*In this new equation, we only have these to worry about.*0958

*Let us combine the like terms, 1d and another d will give us a total of 2d -10 =100 and we can add 10 to both sides.*0961

*One last step, let us go ahead and divide both sides by 2.*0979

*This will give me that the number of democrats should be 55.*0987

*Since I also know that there are exactly 10 more democrats than republicans, *0992

*I can take this number then subtract 10 and that will give me my republicans.*0997

*Let us say we have 55 democrats, 45 republicans.*1011

*I have one more example with these systems of equations and this involves a little bit more difficult numbers.*1038

*We are going to end up rounding these nice whole numbers, so we do not have to deal with too much decimals.*1045

*Be careful along the way because some of the numbers do get a little messy.*1050

*In this problem, I have a Janet that blends coffee for a coffee house*1054

*and what she wants to do is she wants to prepare 280 pounds of blended coffee beans and sell it for $5.32 per pound.*1058

*The way she plans on making this mixture is she is going to blend together 2 different types of coffees.*1066

*She is going to blend together a high-quality coffee that cost $6.25/pound.*1072

*And she is also going to blend together with that a cheaper coffee that only costs $3/ pound.*1080

*The question is to the nearest pound, how much high quality coffee bean and how much cheaper quality coffee bean should she mix in order to get this plan.*1088

*Let us set down some unknowns and see if we can figure this out.*1098

*I will call (q) the number of pounds and this will be for our high quality coffee.*1105

*Let (c), even number of pounds for our cheap quality coffee.*1126

*We are going to try and set up a how many pounds of each of these we need.*1149

*One of the first bits of information that will help out with that is we know that we have a total of 280 pounds.*1153

*The number of pounds for my high quality coffee + the number of pounds for my cheaper coffee better equal to 280 pounds. *1160

*This entire equation just deals with pounds.*1170

*The second part will have to deal with the cost.*1176

*We want the final mixture to be $5.32, so we will use the cost of the high quality coffee bean and the cost of the low quality coffee bean*1178

*and blend the 2 together and get that final cost.*1186

*Let us see what we have here.*1191

*Normally, high quality coffee costs $6.25 every pound of high quality coffee.*1191

*The cheaper coffee usually costs $3/pound.*1200

*We are hoping to make is $5.32 for that final 280 pounds of coffee.*1206

*Let us mark this other one as just costs.*1215

*In our system here, you can see the numbers are not quite as nice but we can definitely still work with them.*1221

*Let us do a little bit of cleaning up then we will go ahead and try the substitution method on the system.*1227

*I’m going to multiply these together first, so $5.32 × 280,*1234

*I'm doing the new system 625q + $3, c =1489.6.*1248

*To use the substitution method, I’m going to take the first equation and let us go ahead and solve it for q.*1265

*Q = 280 – c.*1274

*Once we have that, we can go ahead and substitute it into our second equation.*1280

*$6.25 q is 280 – (c + $3.04c) = 1489.6.*1289

*Our new equation only has c and we can go ahead and solve for that.*1307

*Let us distribute through by the $6.25 and see what we will get.*1312

*17.50 – 6.25c + 3c = 1489.6.*1323

*Let us combine our c terms, -3.25c = 1489.6.*1338

*We almost have c all by itself, let us get there by subtracting the 17.50 from both sides, -260.4.*1354

*One last step, let us go ahead and divide both sides by the -3.25.*1378

*I’m getting 80 and then something after the decimal 12307692, it just keeps going on and on.*1396

*I can see that I’m going to have just a little bit more than 80 pounds, but we want to round this to the nearest whole pound.*1407

*What I will say is that c is about equal to 80 pounds.*1415

*Now, once I know how many pound the cheaper coffee I will need, we can quickly go back to one of our originals*1421

*and figure out how much of the high quality coffee we need.*1427

*I need the total poundage to be 280 and 80 of that will be in the cheaper coffee, then I know that the high quality coffee I will need 200 pound.*1433

*Our rounded solution, I have that 80 pounds of cheap coffee and 200 pounds of high quality coffee.*1446

*If you pick apart your word problems and try and set down an equation for some of the situations in there,*1454

*you should be able to develop your system of equations just fine.*1459

*Then use one of the methods that we have learned especially those algebraic methods to solve the system *1463

*and do not forget to interpret your solution in the context of the problem.*1467

*Thanks for watching www.educator.com.*1472

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at solving linear inequalities in one variable.*0003

*To solve linear qualities, you will see that the process is not so bad. *0010

*But we are willing to take a little bit of time just to highlight the difference between inequalities and equations, *0013

*after looking at the difference between the two and that will cause a little bit of a problem.*0020

*You have to be careful on how we describe our solutions for inequalities.*0025

*Then we will get into a bunch of different examples on how we can finally solve these linear inequalities. *0030

*Let us get started.*0035

*When you are looking at an inequality versus an equation and you try to figure out what is the difference.*0040

*One of the main differences, of course you will see a different symbol in there, one of our inequality symbols.*0045

*It is these guys that we saw earlier.*0051

*We have our greater than symbol, greater than or equal to symbol, our less than symbol or less than or equal to symbol.*0054

*It is a little bit more than just having an additional symbol in there that makes this different. *0063

*What makes an inequality different from an equation has to do with the solutions that you get out of them.*0068

*To highlight this I will go over some quick examples of equations and get to one of those inequalities.*0075

*Consider this first one here, 3x = 15.*0081

*If you are going to solve that, it would not take probably that long.*0087

*Just simply divide both sides by 3 and you get that x = 5. *0091

*Now what that says the only number, the only one out of all possible numbers that will make that equation true is just 5.*0099

*It is like this little isolated point out on the number line.*0106

*Some equations might actually have more than one solution, for example, this next one x*^{2} = 9. 0110

*If you are going to look for numbers that will make that equation true, you will actually find two of them.*0117

*When x = 3 and you square it you will get 9, but also when x = -3 and you square it you still get 9.*0123

*You know it is a little different that you have two solutions, but both of the solutions those are the only two things that will make it work.*0132

*It is like two little isolated points on your number line.*0138

*To contrast that to this first inequality over here, 4x is greater than or equal to 8.*0142

*If you try to figure out what numbers make that true, you would end up with a huge, huge list of numbers that work.*0149

*Let us start picking out some examples. *0157

*Also, what if I said that x was equal to 2.*0159

*If you substitute that in, you would have 4 × 2 is greater than or equal to 8, *0163

*which would simplify to 8 is greater than or equal to 8, which is, of course true.*0167

*But it is not the only thing that works out. *0174

*You could also put in 3, 4, you can get a little creative and put in some other ones like 4 1/2, and all of these would be solutions.*0176

*Trying to list out solutions for an inequality is simply not the way to go.*0191

*The reason for that is if you look at the numbers that would make an inequality true, you will get an entire range of values.*0196

*So, not just in a little isolated point, they are in whole ranges of values on our number line.*0204

*Since we are after a whole range of values other than these isolated points, we are going to have to be a little bit careful on how we describe those.*0211

*In fact, there are many ways that you can describe the solution to an inequality.*0223

*Some you seen before and some might be a little new, let us go over them.*0227

*The first way that you can describe the solution is to simply use our inequality symbols that will be like this guy down here.*0231

*Maybe I’m trying to describe all the numbers that are less than 4, I can simply write x is less than 4.*0239

*We can also use our interval notation.*0246

*This is a good way to mimic the number line and that it gives us the lowest number or the starting point and the endpoint. *0250

*It also gives us information on whether we should include the number or not.*0262

*Since this one has parentheses, I know that the 4 is not included.*0266

*It is simply a different way of writing the same information.*0273

*This interval says I'm looking at all the values that are less than 4. *0276

*We could also describe our solutions a little bit more visually using a number line.*0283

*The way we do this is we have a number line and we shade in the solution.*0288

*To indicate whether the endpoint is included or not, you can use an open circle or close circle.*0300

*The open circle here means it is not included, so I would have used a close circle if I had x is less than or equal to 4.*0310

*And one last way that you might see, I do see this in a few Math books is known as set builder notation. *0323

*It looks a little clunky and that we have a couple of curly brackets thrown in there *0329

*but let me show you how you can interpret the set builder notation.*0335

*First of all, we identify some sort of variable at the very beginning. *0341

*We are looking for all x values.*0346

*And after that you will notice that we borrow our inequality. *0352

*This is the rule of all of the numbers that we will include.*0356

*That little line you see in between, we say that is L and it is just the divider or it separates both those sections. *0370

*The way you would read the set builder notation is the set of all x’s such that x is less than 4.*0384

*We know that we are building for x values and we know which x values would get in there.*0393

*I’m not going to focus a whole lot on that set builder notation but you will definitely see me use some inequalities, some number lines *0397

*and some interval notation just to describe our solutions.*0404

*Now that we know little bit about what makes an inequality different, we know little bit more on how to describe the solutions, *0410

*let us get into how we can actually start solving these things.*0416

*The good news is when it comes to solving linear inequality, we use the same techniques as we do with solving a normal linear equation.*0421

*There is one additional rule that you want to be aware of.*0432

*When you multiply or divide by a negative number then you want to flip the inequality sign.*0435

*Now that rule applies to multiplying and dividing, so be careful not to try and use it if you are only adding and subtracting.*0444

*Use it only for multiplying and dividing.*0453

*I often get lots of questions like why is it that you have to flip the sign when your doing the multiplication and division by negative numbers.*0458

*I think a quick example will help you understand why that is.*0465

*Let us take a nice inequality like 2 < 3, so seems pretty simple. *0469

*We know that 2 < 3 but watch what happens if I multiply both sides by a negative number.*0476

*-2 and -3, now which one is less than the other one?*0483

*If we take a peek at our number line, you will see that it is the -3 that is now the smaller number since it is on the left side of -2.*0491

*This means that if we want to preserve the trueness of our statement, we also must flip that sign to compensate for that negative that we threw in there.*0510

*It is easy to see with numbers like this but in a moment when we start dealing with x’s and unknowns, *0520

*we still have to remember to flip the sign when we divide or multiply by that negative number.*0525

*I think we have all the information we need.*0534

*Let us go ahead and get into the solving process and see if we can tackle some of these inequalities.*0537

*We want to solve the following inequality then write our answer using a number line and using interval notation.*0543

*Okay, when I solve an equation usually I try and isolate the x's and I'm going to do the same thing here. *0549

*I will do that by first moving the 5 to the other side, 3x < 11 - 2x and then we will go ahead and we will add 2x to both sides. *0557

*5x < 11, let us go ahead and divide both sides by 5.*0578

*I will get that x < 11/5.*0593

*This could represent my solution using just the inequality symbol. *0597

*Now I want to go and have a look at the number line and its interval notation.*0601

*Here is a nice quick sketch of a number line. *0615

*I'm looking to shade in all values that are less than 11/5.*0620

*If I want to rewrite that, 5 goes into 11 twice with the remainder of one, so it is the same as 2 1/5.*0625

*It is a little bit greater than 2, but not much.*0638

*Notice how I’m using an open circle there because our inequality is strict. *0643

*We do not want to include that value.*0650

*I will shade in everything that is less than my 2 1/5.*0653

*This number line represents all of the solutions to my inequality, I could use anything on that side and it worked. *0662

*As long as we have our number line, let us go ahead and also represent our solution using interval notation.*0670

*We want to think of where we start on the left side, it looks like we are starting way down at negative infinity.*0680

*We go all the way up to 11/5 but I will use a parenthesis since we do not include that value.*0686

*I have represented my solution now in 3 different ways. Not bad.*0694

*Let us look at another, for this one we are looking to solve 13 - 7x = – 4.*0697

*Let us start off by trying to get those x's isolated and we will do that by first getting them together. *0710

*I'm going to move this 10x to the other side, so I'm subtracting here. *0716

*I do not have to worry about flipping any signs nothing like that yet, 13 - 17x = -4.*0722

*Let us go ahead and subtract 13 from both sides, we have -17x = -17.*0735

*Be very careful in this next step.*0750

*I need to get the x all by itself, in order to do that I will divide by -17.*0751

*I will do all of the algebra normally -17 ÷ -17 = 1. *0763

*I'm also going to remember what to do with that sign, we are going to flip it around.*0770

*I have it x = 1.*0777

*Let us get to our number line and go ahead and represent the solution.*0784

*I have the number 1,we will make a nice solid circle since this is or equal to *0793

*and we will go ahead and shade in everything less than that since it says x < 1.*0804

*Now that I have my number line, let us represent this using our interval notation.*0811

*On the left side we are starting at way down to negative infinity and going all the way up to 1.*0815

*We want to include the one so we will use a bracket.*0821

*Let us try one more example.*0838

*In this last one I picked a little bit more of a word problem so you can see that inequalities are important even in applications.*0843

*This one says that Joanna is hiring a painting company in order to get her house painted*0850

*and the company has two plans to choose from, you can choose plan A.*0855

*In plan A they charge you a flat fee of $250 and $10 for every hour that it takes to paint the house.*0861

*For plan B, they do not charge a flat fee but they will charge $20 per hour.*0869

*The question is when will plan B be more expensive than plan A?*0874

*Pause for a moment and think why would we be using an inequality here?*0878

*Why do not we just set up an equation?*0882

*What I'm looking for is not one specific time, but all times when plan B will be more expensive.*0885

*It makes a little bit more sense to go ahead and set up an inequality for this word problem.*0891

*Let us go ahead and hunt down both of our situation.*0896

*Let us have plan A and let us go ahead and have plan B.*0899

*We need an unknown here, let us say x is the time to paint the house.*0907

*Plan A cost a flat fee of $250 + $10/hour.*0924

*$10/ hour is like our variable cost, that would be 10x, flat fee does not vary so +250.*0929

*That expression will just keep track of the cost for plan A depending on how many hours it takes.*0938

*Plan B has only a variable cost of $20/hour, I will just say 20x.*0944

*Now comes the big question, when is plan B more expensive than plan A.*0951

*B would be more expensive, I would use an inequality symbol of plan A < plan B *0958

*and that is exactly how we will connect our expressions as well. *0964

*10x + 250 < 20x and to completely answer this problem we just have to solve the inequality.*0967

*Let us work on getting our x's together by subtracting 10x from both sides.*0976

*250 < 10x, now we will divide both sides by 10.*0986

*25 < x, now it is time to interpret exactly what we have here.*0999

*x represents the time that it will take to paint the house.*1006

*What I can see here is that anytime my time is more than 25 hours then I can be sure that plan B will be more expensive, there you have it. *1011

*Also, be careful when setting up and solving these inequalities.*1021

*Remember to go ahead and flip your inequality anytime you multiply or divide by negative number.*1024

*Thank you for watching www.educator.com.*1031

*Welcome back to www.educator.com.*0000

*In this lesson, we are going to take care of compound inequalities.*0002

*In compound inequalities we will look at connecting inequalities using some very special words and, and or.*0010

*We will have to first recognize some subtleties between using both of these words.*0017

*Blabbering about some nice new vocabulary, we will learn about union and intersection.*0023

*Once we know more about how to connect them then we can finally get to our compound inequalities and how we can solve these.*0029

*In some situations, you have more than one condition and they could be connected using the word and or they could be connected using or.*0040

*To see the subtle difference between using one of these things, let us just try them out on a list of numbers.*0050

*I have the numbers 1 all the way up to 10 and we will see what number should be included in the various different situations below.*0056

*Let us see if we can first take care of all the numbers greater than 3 and less than 7.*0066

*Notice in this one I’m using that word, and.*0071

*Numbers that are greater than 3, we will have 4 or 5 and 6 and all of these are also less than 7.*0075

*The things I will include in my list 4, 5, and 6.*0083

*Let us think on how we will approach this problem a little bit different for the next one.*0090

*Maybe list out all numbers that are greater than 4 or they are less than 2.*0095

*The numbers greater than 4, that would we 5, 6, 7, 8, 9, and 10 and the numbers that are less than 2, 1.*0101

*Notice how in that situation I was looking for 2 things.*0112

*I was looking for all numbers that were greater than 4 and I have highlighted those first.*0115

*And then I went back and I looked for ones that were less than 2.*0119

*Let us give it a few little marks here and try the third situation. *0127

*All numbers greater than 6 and they are less than 3.*0132

*That is a tough one to do because I can go ahead and highlight the numbers greater than 6, that will be 7, 8, 9, 10 but they are not less than 3.*0137

*I can not include them.*0148

*Notice how in that situation since I'm dealing with and, it is like I have to satisfy both of these conditions must be greater than 6 must be less than 3.*0150

*If it does not satisfy both of them then I cannot include them.*0164

*Let us try another one and see how it works out.*0167

*All numbers greater than 4 so I have 5, 6, 7, 8, 9, 10 or they are less than 7. *0170

*What numbers less than 7 would be 6, 5, 4, 3, 2,1.*0178

*All the numbers fall into one category or they have fallen to the other one. *0183

*In fact, using or seems like it is a little bit more relaxed as long as it satisfies one of my conditions, then I will go ahead and include it in my list. *0188

*In fact, that highlights the difference between using and using the word or to connect these 2 conditions.*0197

*Let us make it even more clear.*0204

*When you use and to connect two conditions, then you must have both conditions met in order to include it.*0207

*However, when you are using or, then as long as you have one of the conditions met then you can go ahead and include that.*0214

*The way you will see and and or used is when we combine our intervals in our solutions.*0222

*If an object meets both conditions using our and connection, then it is said to be in the intersection of the conditions *0230

*and we can use the symbol to connect those.*0239

*Some people might think of this as and but it stands for the intersection.*0242

*On the flip side, if an object meets at least one of the conditions using the word or, *0253

*then it is said to be in the union of the conditions.*0257

*We will use the symbols, think of or and union.*0261

*You will see these symbols for sure and watch how we connect our solutions using either and or or.*0267

*We know a little bit more about the connections, let us get into how we can actually solve these things.*0277

*We will connect these things using and or or and the way you go about solving is you can actually solved each of the inequalities separately.*0285

*Just take care of one at a time and the part where these and or or come into play is when we want to connect together our solutions. *0294

*There are some situations where we can actually connect the inequalities at the very beginning.*0305

*One of those situations is when you are dealing with and and 2 of the parts are exactly the same.*0310

*Just like this example that I have highlighted below.*0316

*I have 5x < or = 3 + 11x and I also have a -3 + 11x on the other side over here.*0321

*We are connected using and. *0329

*I'm going to put these together into what is known as a compound inequality.*0332

*You will notice that everything on this side of the inequality comes from the left side.*0336

*Everything on the other side of inequality over here comes from this inequality.*0344

*It encapsulates both of them at the same time.*0351

*If you do connect one like this, the way you end up solving it is just remember that whatever you do to one part, do it to all 3 parts.*0354

*If you subtract 3, subtract 3 from all parts.*0362

*This also includes if you have to flip a sign.*0366

*If you multiply by a negative number then flip both of those inequality signs that are present then you should be okay.*0368

*Just remember if you want to solve each one separately, that works to.*0377

*Let us see some of our examples, see the solving process in action.*0383

*We want to solve the following inequality and of course write our answer using a number line and using interval notation.*0388

*I can see I have 2x -5 < or = - 7 or 2x – 5 > 1.*0395

*I’m going to solve these just separately, just take care of one at a time.*0403

*Looking at the left, I will add 5 to both sides, 2x > or = -2.*0406

*Now I can divide both sides by 2 and get that x < or = -1.*0418

*There is one of my solutions, let us focus on the other one.*0429

*You will add 5 to both sides of this inequality, 2x > 6, now divide by 2 and get that x > 3.*0434

*I have 2 intervals and I will be connecting things, let us drop this down using or.*0450

*I want to think of all the numbers that satisfy one of these, or satisfy the other one.*0458

*As long as they satisfy one of these conditions, I will go ahead and include it in my overall solution.*0463

*Let us take this a bit at a time.*0469

*I will do a little number line here, a number line here and I work on combining them into one number line.*0472

*First I can look at all the numbers that are less than or equal to 1.*0481

*Here is -1, I'm using a solid circle because it is or equals to and we are less than that so we will shade in that direction.*0489

*For the other inequality, I'm looking at numbers like 3 but not included or greater so I’m using an open circle there.*0501

*What will I put on my final number line is all places where I shaded at least once. *0509

*Let us see how that looks, I have my -1 and everything less and I have 3 or everything greater.*0523

*This number line which has both of them shows one condition or the other, and I can include both.*0535

*Let us go ahead and represent this using our interval notation.*0543

*We have everything less than - 1 and we include that -1 and we have everything from 3 up to infinity. *0546

*Since we are dealing with or, let us go ahead and use our union symbol to connect the two.*0558

*I have the many different ways that you can represent the solution for this inequality.*0563

*Let us take a look at another one, solve the following inequality and write your answer using a number line and interval notation.*0571

*This is a special inequality, this is one of our compound inequalities because it have 1, 2, 3 different parts to it.*0578

*As long as you remember that whatever we do to one part, we should do it to all three,I think it will turn out okay.*0585

*Let us work on getting that x all by itself in its particular part.*0591

*We will go ahead and subtract 5 from all three parts and get – 8 < 2x is < or = 2.*0597

*Let us divide everything by 2.*0612

*-8 ÷ 2 = -4, x < or = 1.*0620

*In this one I am looking for all values between -4 and 1.*0628

*I think we can make a number line for that.*0633

*Okay, I need to shade in everything in between -4 and all the way up to 1. *0649

*It looks like the -4 is not included, I will use a nice open circle, but the 1 is included, so we will shade that in.*0657

*Now that we have our number line, we can describe this using our interval notation.*0667

*We are starting way down at -4 not included and going all the way up to , that is included. *0671

*In some of these inequalities, you have to be careful on which conditions it satisfies.*0689

*If you end up with no numbers that simply do not work, watch how that turns out.*0695

*We want to solve the following inequality and write our answer using a number line and interval notation.*0701

*I have 4x + 1 > or = -7 or -2x + 3 > or = 5.*0706

*Let us begin by solving each of these separately.*0716

*With 1 on the left, I will start by adding 1 to both sides, 4x < or = -6 and now divide both sides by 4.*0721

*We are reducing that, I get that x > or = -3/2.*0737

*We will set that off to this side and solve the other one.*0744

*Let us subtract 3 from both sides and we will go ahead and divide both sides by -2.*0749

*Since we are dividing by negative, I’m going to flip my inequalities symbol.*0758

*This one is very interesting.*0766

*Notice how I'm looking for numbers that are less than or equal to 1, but I’m also looking for numbers that are greater than or equal to -3/2.*0768

*If I look at where those two are located, here is -3/2 and here is 1.*0779

*You will see that we get actually all numbers because I'm looking for things that are greater than or equal to -3/2, *0785

*that will shade everything on the right side of that - 3/2.*0792

*Everything less than or equal to 1 would shade everything in the other direction.*0796

*It will end up shading the entire number line.*0801

*We will say we would not include all numbers from negative infinity up to infinity. *0805

*I might make a note, all real numbers.*0813

*Watch for a very similar situation to happen when we deal with and.*0824

*Solve the following inequality and write our answer using a number line and interval notation.*0830

*I have two inequalities and a bunch of different x's, let us work on getting them together first. *0837

*I'm going to subtract an x from both sides on the left side here, 1x < -5.*0842

*I can add up 8 on both sides and we will get that x < 3.*0851

*Let us go over to the right and see what we can do there.*0861

*I will subtract 15 from both sides, that will give me –x < -10.*0864

*Now divide both sides by -1, since we are dividing by a negative number, let us flip out the inequality sign.*0871

*I have x < 3 and x > 10.*0882

*This one actually proves that when we are dealing with and, it must satisfy both of these conditions in order to be included.*0889

*If you start thinking what numbers are less than 3 and also greater than 10, you will that you do not get any number.*0896

*You can not be both things at the same time.*0902

*They are both in completely different spots on a number line.*0906

*Let us make another one.*0913

* We can not shade in numbers that are both less than 3 and greater than 10.*0917

*What is that mean to our solution then?*0942

*I mean what can we put down?*0943

*This is where we say there is no solution.*0947

*Be careful on using those connectors and how it affects your solution.*0953

*Remember that when using the word or, it must satisfy one of the conditions.*0957

*If it does, go ahead and include it in your solution.*0961

*However, when using the word and, it must satisfy both of those conditions before you can include it in your solution.*0964

*Thanks for watching www.educator.com.*0971

*Welcome back to www.educator.com.*0000

*In this lesson we are going to go ahead and look at solving equations that have absolute values in them.*0002

*We only have one goal, how do you get down to the nuts and bolts of solving it when you see an absolute value.*0010

*When you have a number that is equal to an absolute value or any algebraic expression, then you can break it down into the following.*0021

*That expression is equal to the number or that expression is equal to the negative of that number.*0030

*What this is trying to say is that when you are dealing with this situation, absolute value equals number, is that you can break it down into 2 situations.*0035

*It takes care of 2 possibilities that whatever was inside the absolute value could have been positive *0045

*or maybe what is on the inside could have been negative.*0050

*The reason why we have 2 possibilities is the absolute values going to take away information about what the original sign was.*0053

*Maybe that original sign was positive, maybe it was negative, but we are not sure so we have to take them both into account.*0060

*To highlight how this works, let us look at some actual numbers.*0067

*Let us suppose that our algebraic expression was just x and we are going to look at the absolute value of x is equal to 2.*0073

*What would have to be inside that absolute value in order for this thing to work? *0081

*One possibility is that maybe x = 2, so we get this case over here.*0085

*What if the x was negative and we lost all information about that negative sign.*0091

*That looks like more of like the situation on the right, what we see is that x could be 2 or x could actually be -2.*0097

*The only difference between these two is I multiply both sides by -1.*0105

*What I like to do when handling these absolute values is immediately split them into 2 different problems.*0111

*I will handle what I called my positive case where I just keep everything the same and drop the absolute values.*0118

*Then I will take care of what I call the negative case, *0125

*I will put a negative sign on whatever was inside the absolute value and then continue solving from there.*0127

*When I’m all done, I will go ahead and take both of my solutions and check them in the original problem, just to make sure that they are correct.*0133

*Some other things to consider when solving an equation that have an absolute value is to make it fit the form.*0143

*You want to make sure that you isolate the entire absolute value first.*0150

*Do not worry about splitting it into 2 problems, until you have absolute value on one side and the one on all the numbers and other stuff on the other. *0155

*Once you do have that absolute value, go ahead and split it into 2 problems.*0165

*And what you are looking to do with each of those problems is you want one of them to handle what I said positive possibilities,*0172

*it will be exactly the same, just no absolute value.*0180

*And one to handle the negative possibilities, what would happen if there was a negative sign on the inside of those absolute values.*0183

*When you have two new problems then simply solve each of them separately as normal. *0191

*Follow all of your normal rules for algebra and you should be okay. *0196

*Let us give this a try. *0201

*The first one I'm looking at the absolute value x -3 -1 = 4. *0204

*The very first thing I want to do is to find those absolute values here and get them all alone onto one side of my equation. *0211

*I have the absolute value of x - 3 and isolate it then I will add one to both sides.*0220

*Once I get the absolute value completely isolated, this is the spot we are going to split this into 2 problems. *0230

*Get some space and get our two problems.*0237

*On the left side here, I'm going to consider this my positive possibility that whatever the absolute value was encapsulating it was positive.*0241

*It is one of the changes that I need to make.*0250

*The other possibility is that maybe the x -3 maybe that entire thing was negative.*0254

*Notice how I put a negative sign in there.*0258

*Now that we have two problems, we just go through each one separately and see how they turn out.*0262

*x - 3 = 5, add 3 to both sides and get that x = 8.*0268

*With the other one, it is okay if you want to distribute that negative sign first.*0276

*-x + 3 = 5, then we can subtract 3 and move that negative sign, x = -2.*0288

*I have two solutions to my equation that I started with.*0300

*It is a good idea to check these just to make sure they work out.*0306

*Let us take them individually and see what happens. *0309

*8 - 3 would give us 5.*0319

*Our absolute values make everything positive so the absolute value of 5 would be 5.*0326

*I can see that sure enough, 5 – 1 =4 and that solution checks out.*0333

*Onto the other side, we will put a -2 – 3 working with the inside of the absolute values,-2 - 3 would give us -5.*0340

*Notice, as we start to simplify that we can see what I was talking about how the sign got me positive so we do have to handle both possibilities.*0358

*Cancel the absolute value of -5 is 5, 5 - 1 = 4 and sure enough, we know that is true so our solution checks out, that x = 8 or x = -2.*0373

*In this next example, we will have to do a little bit more on getting the absolute value isolated, but again you will see this is not too bad.*0397

*To get rid of that fraction out front, I'm going to multiply both sides by 2, 2 over there and 2 over there.*0404

*It will give us the absolute value of 5x + 2 = 12.*0415

*We have the absolute value on one side of the equal sign.*0422

*Let us go ahead and split this into two different problems. *0427

*Here I will handle my positive possibility that whatever was inside absolute value it is positive.*0435

*Over here, maybe what was inside the absolute values was negative, so we will put that negative sign.*0443

*Let us solve both of these separately.*0453

*On the left, I will subtract 2 from both sides, and then we will go ahead and divide both sides by 5, I have x = 2.*0457

*On the other side, I’m going to do something a little bit different.*0471

*To move that negative sign at the very beginning, I will multiply both sides by -1. *0474

*It is important to know that you can also distribute through being your first step and then go through the solving process.*0481

*I’m just doing this so I can take care of the negative sign at the very beginning.*0487

*Negative × negative on the left side would give me 5x + 2 =-12.*0491

*Let us subtract 2 from both sides, and finally divide by 5.*0500

*Now I have two solutions to my absolute value problem.*0515

*Let us take a little bit of time just to check them and make sure they work out.*0521

*½ of 5 × 2 + 2 does that equals 6?*0527

*Working on the inside of the absolute value, I have 5 × 2 = 10 + 2 =12.*0536

*The absolute value of 12 is 12, 1/2 of 12 = 6, that one checks out.*0546

*On to the other side, this one is -14/5 + 2.*0556

*Let us see if this one equal 6 or not.*0570

*14/5 × 5 = -14, -14 + 2 =-12, it is very similar to what we got last time.*0573

*We are just dealing with the negative on the inside.*0593

*The absolute value will make that -12 into a positive 12 and sure ½ of 12 = 6.*0599

*Both of our solutions work out just fine. *0607

*There is still one more example to show you some of the things that might happen when going through the solving process.*0615

*This one is 3 times the absolute value of x + 2 + 3 = 0.*0620

*We will start off by trying to isolate that absolute value. *0626

*I'm going to subtract 3 from both sides.*0635

*To get rid of that 3 out in front of the absolute value, let us divide both sides by 3.*0639

*I have the absolute value of x + 2 .*0648

*Now following our steps, let us split this into 2 problems.*0651

*Here we have the problem exactly the same, just no absolute value and over here, -x + 2 = -1.*0660

*Let us solve each of them separately.*0670

*I will subtract 2 from both sides given the x = -3. *0674

*On the other side here, I will go ahead and distribute through with my negative sign and I will add 2 to both sides.*0679

*Lastly let us go ahead and multiply both sides by -1.*0693

*I have two possible solutions, x = -3 and x = -1.*0699

*Let us check them to see how they turn out.*0705

*Let us see, -3 + 2 = -1 then we will take the absolute value of -1, we get 1.*0716

*What I'm getting here is that 3 × 1 is 3 and when we add another 3, I get 6.*0730

*This shows that this one does not work out.*0738

*I'm going to get rid of that, it is not a solution.*0742

*Let us try it on the other side.*0747

*I will put in a -1.*0750

*I have 3, -1 + 2 = 1 and the absolute value of 1 is 1.*0758

*I have 3 × 1 + 3 I think that is the same situation as before or 6 = 0.*0772

*I think that does not work out as well so I have to throw away that solution. *0778

*What this shows is that if none of the work then we actually have no solution.*0784

*Let us highlight that it is a good idea to always check your answers.*0792

*I will also highlight that some funny things could happen when working with absolute values.*0796

*Remember that an absolute value takes whatever you put into it and makes it positive.*0800

*In fact, we can see a problem in this step when we are working. *0806

*I have that the absolute value of something is equal to a negative and that can not happen because our absolute value on one side will be positive.*0813

*On the other side we are setting equal to a negative and we can not get a positive and a negative, to agree.*0826

*That is why this one also has no solution.*0836

*If you catch it early on, it usually can save yourself a lot of work.*0839

*But if you do not, make sure your check your solutions also that you can definitely find these no solution problems.*0842

*Thank you for watching www.educator.com.*0850

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at inequalities with absolute values.*0002

*In our previous lessons, we did take care of inequality separately and we did take care of absolute value separately.*0009

*Our goal for this one is to mix the two together and see if we can solve some new types of problems.*0015

*When you have some number A greater than 0 and you have some algebraic expression x as packaged together in the following way.*0026

*The absolute value of x < A, or the absolute value of x > A then you can actually split these into two different problems. *0035

*Let us look at the first one, suppose that you have the absolute value of x < A, the way that we can put this is that the algebraic expression x > -A.*0043

*The algebraic expression of x could be less than A.*0054

*It is interesting to note that not only do we split it into two problems *0059

*and we will see that we have picked up our connector and it actually connected the two.*0063

*On the other side, when we are dealing with the absolute value of x > A.*0068

*This will still split into two problems but now we have the x < -A or x > A.*0073

*This is a little bit different and that now we have a different connector between the problems.*0084

*We will definitely use our tools on how to deal with these connectors and solving both problems separately*0091

*and figure out how their solutions should be connected in the end.*0096

*One thing that you may be looking at and be a little bit worried is keeping track of all of your signs.*0100

*It might start with less than and then these guys one is greater than and one is less than.*0107

*These guys over here, I have greater than and then I have less than or greater than.*0113

*Watch for my tips on how to deal with dealing with the absolute value and the absolute value separately so we can keep them both straight.*0118

*To understand why it splits into all of these different cases and why the signs look so funny, you have to recall two things that we covered earlier.*0131

*One has to do with those absolute values.*0140

*When we are dealing with absolute values this will end up being split into two different problems.*0145

*One that handles the positive possibilities, and one that handles the negative possibilities.*0157

*That is why we have two different things.*0162

*Also, you have to remember the rule that when you multiply or divide by negative that you will end up flipping your inequality sign.*0165

*That is why the direction of the inequality ends up changing in both of our cases.*0172

*Since one is that negative possibility, the sign ends up getting flipped.*0177

*To help you keep these both straight this is what I recommend.*0183

*When you are looking at your connectors, if you have the absolute value being less than a number then connect using and.*0187

*If you have the absolute value greater than a number, then use or to connect the two.*0198

*You will see that if you use each of these rules bit by bit, rather than trying to jumble them altogether, *0205

*it is not too bad being able to pick apart one of these absolute values and inequality problems. *0211

*Let us start out with some very small examples and work our way up.*0221

*That way we can keep track of everything we need to do.*0225

*This one simply says that the absolute value of x < 3.*0228

*Here is what I'm going to do first.*0232

*I’m going to use my techniques for our absolute values to simply just split this into two problems.*0234

*Remember before that we have our positive possibility that x < 3 or we have our negative possibility -x < 3.*0246

*I have not done anything with that inequality symbol, I have not even touched it yet I simply split it into two problems.*0258

*Because we have the absolute value less than a number, I will connect these using the word and.*0266

*It will become especially important when we get tour final solution.*0274

*Now I just have to solve these separately.*0278

*x < 3 is already done, -x < 3 I have to multiply both sides by -1.*0281

*Since I’m multiplying by a negative, we need to flip our sign.*0289

*Here I have two things that x must be a number that is less than 3 and x must be a number greater than 3.*0294

*Since we are dealing with and, we are looking for all numbers that satisfied both of the possibilities.*0302

*Let us look at our number line first to see if we can hunt down everything that does that.*0309

*All numbers less than 3 would be on this side of 3 and we will put in a big open circle.*0327

*They must be greater than -3 and I will put in another big open circle and -3.*0336

*We will see that it shades in everything greater than that value.*0341

*Our overall solution is all numbers between -3 and up to 3.*0346

*Use your techniques for splitting up into two problems.*0354

*Be careful on how you connect them and remember that you should flip the sign when you multiply or divide by negative.*0357

*Let us try another small one, and again watch for this process to happen.*0363

*This one says the absolute value of z = 5.*0369

*The first thing I'm going to do is just split this into two problems. *0373

*I’m doing this because I'm taking care of that absolute value.*0380

*z > or = 5, -z > or = 5 I have not touched my inequality symbol I just took care of my two cases.*0384

*Since I'm dealing with the absolute value being greater than or equal to a number over here and I will connect these using or.*0397

*This will become especially important when we get to our solution and we just solve each of them separately.*0408

*Z > or = 5 is already done and for the other one, I will multiply both sides by -1.*0413

*Now if I multiply it by a negative, this will flip its sign.*0423

*I have z > or = 5 or z < or = 5.*0428

*I think we are ready to start packaging things up here.*0434

*Since I’m dealing with the connection or, as long as it satisfies one of these possibilities I will include it in my final solution.*0442

*All the numbers greater than or equal to 5 that would be all of these numbers here on the number line.*0450

*All the ones less than or equal to 5, that would be these ones down here.*0460

*Now that I know more about the number line, let us go ahead and write this using our interval notation.*0465

*Negative infinity up to 5 brackets, because we are including from 5 to infinity, our little union symbol since it could be in either one of those intervals.*0470

*Now that we have the process down a little bit better for being able to split this up, let us look at ones that are just a little bit more complicated.*0487

*We want to solve the inequality involving our absolute value.*0498

*I want make sure that the absolute value is isolated on one side of my inequality sign and fortunately this one is.*0503

*Let us go ahead and split it into our two problems.*0510

*On this side we will have 3x + 2 < 5.*0519

*On the other side I have -3x + 2 < 5.*0525

*I will take note as to the direction of my inequality symbol, our inequality less than number.*0533

*Because of its direction I will connect these using and.*0541

*A little bit more work, let us go ahead and solve each of these.*0547

*For this one, I will subtract 2 from both sides and then divide both sides by 3 so x < 1.*0551

*On the other side, let us go ahead and distribute our negative sign and let us add 2 to both sides.*0563

*We can see that on this case I only did divide by -3 so because of that we will end up flipping our sign -7/3.*0576

*I need all numbers that are greater than -7/3 and they are also less than a 1.*0589

*On a number line I could see where -7/3 is, there is 0 and I can see where 1 is.*0600

*I’m just looking for all numbers in between. *0609

*Since those would be the only ones that are greater than -7/3 and also less than 1.*0614

*We have our interval -7/3 up to 1 and that would be our solution.*0620

*Let us get into another one. Solve the inequality using an absolute value.*0631

*In this one, we do a little bit of work to isolate that absolute value first.*0637

*Let us go ahead and add 1 to both sides, 5 - 2x > or = 1.*0642

*Now that my absolute value is isolated, we will split it into two problems.*0653

*This will handle our positive possibility, so no changes, *0660

*And the other one will take care of what happens when the inside part could have been negative.*0666

*Since I'm dealing with these absolute values and inequalities, let us look at the direction of everything.*0674

*I have the absolute value less than or equal to number.*0680

*This is that situation where we connect those using or.*0684

*I have two problems that are connected using or, let us continue solving.*0689

*On the left side, I will continue by subtracting 5 from both sides, then I can divide both sides by -2.*0696

*Since I'm dividing by negative, let us flip our sign, x < or = 2.*0710

*Onto the other side I'm going to distribute my negative sign in there first -5 + 2x > or = 1.*0719

*Let us get that x all alone and we are just a little bit closer being all alone by adding 5 to both sides.*0733

* And I will finally divide both sides by 2, x > or = 3.*0743

*I have what I’m looking for all numbers that are less or equal to 2 and all numbers that are greater than 3.*0753

*Let us see what that looks like on a number line.*0760

*All numbers less than or equal to 2 would include the 2 and everything less than that, so we will shade that in.*0767

*All numbers equal to 3 or greater than would be over here and so we have a big gap between 2 and 3 but that is okay. *0775

*Let us graph this using our interval notation.*0784

*From negative infinity up to 2 included and from 3 up to infinity and we are looking at the union of both of those two.*0787

*That is not too bad as long as you isolate that absolute value first before splitting it into two problems, you should be fine.*0799

*Let us look at one last one, and this highlights why you have to be careful on how you package up your solution using those connections and and or.*0807

*In this one, we already have our absolute value isolated to one side 3 - 12x all in absolute value.*0818

*I'm going to split this immediately into two problems.*0828

*3 - 12x < or = -3 and 3 - 12x < or = -3 because of the direction of our inequality symbol.*0838

*Let us go ahead and connect these using and.*0858

*Solving the one on the left 3 - 12x < or = 3.*0864

*Let us subtract 3 from both sides then we will divide both sides by -12.*0870

*I’m flipping my signs because I'm dividing by negative.*0883

*We can reduce this fraction as normal, just ½ so x > or = ½.*0889

*Working with the other one, I will distribute through with my negative sign and I will add 3 to both sides. *0897

*It looks like one last step and I can go ahead and divide both sides by 12, 0 ÷ 12 is simply 0.*0918

*Let us see exactly what this would include.*0933

*On a number line, I want to shade in all numbers that are greater than 1/2 and they must be less than 0.*0936

*Of course, that poses a huge problem.*0945

*There are lots of numbers greater than ½ and there is lots of numbers less than 0 but unfortunately I can not find numbers that satisfied both of them.*0951

*There are no numbers that satisfied both conditions.*0962

*In fact this one, I can say has no solution. *0966

*Now, you might have caught that earlier and if you did that is okay, you can definitely save yourself a lot of work.*0972

*If we look all the way back here at the original problem, we have a bit of an issue.*0980

*Remember that the absolute value makes everything positive and sitting on the other side of our inequality is a negative number.*0985

*We simply can not have something positive being less than something negative, that is not going to work. *0999

*That is another good reason why this one has no solution.*1005

*Be careful when going through these problems and make sure you split them up because of your absolute value.*1009

*Watch the direction of that absolute value so you know how to connect them using, say, either, and or or .*1013

*Remember to solve each of them separately and connect very carefully when you get to your solutions.*1020

*Thanks for watching www.educator.com.*1025

*Welcome back to www.educator.com.*0000

*In this lesson we are going to look at graphing inequalities using two variables.*0002

*In order to make this process work out , we will first briefly review how to just simply graph lines.*0009

*We will see how this works with our inequalities which involved entire regions of our graph.*0016

*The important parts for highlighting these regions will be shading in the proper part of the graph.*0022

*Watch for these things to come into play.*0028

*Recall it when you are dealing with any inequality it is not just usually an isolated point, it is usually a whole range of solutions.*0035

*When we are working with graphing, and we have more than one variable to worry about we are dealing with points like x and y.*0043

*If we are looking for the x and y that will make an inequality true, we are not talking about just one point on a graph,*0052

*we are actually talking about an entire region of the graph, and all the little points in that region.*0058

*The way we do this is when you start off by graphing the equation, corresponding with the inequality.*0065

*Let me show you what I mean by that.*0071

*I’m looking at the inequality y > or = 6/5x + 7 rather than trying to tackle on the entire inequality first, *0074

*I looked at the corresponding equation like y = 6/5x +7.*0086

*In doing so, I can look at my techniques for graphing just that equation. *0093

*Maybe find the slope and find the y intercept.*0098

*We will see how that equation connects back to our inequality.*0106

*What we are going to find is that equation is going to split our entire graph into two regions.*0115

*In one of the regions I will involve our solution and the other region will not.*0121

*I have cooked up another example, so you can see how this works.*0126

*We are going to start in graphing y = 2x -4.*0131

*The first thing we wanted to do is look at the corresponding equation y = 2x – 4 and focus on that first.*0141

*We can see that it has a y intercept and it has a slope.*0151

*I'm going to end up graphing this line, and I'm going to graph it using a nice solid line.*0158

*I will get into that y in just a bit.*0163

*y intercepts way down here at -4 and a slope of 2.*0166

*From there up 1, up to over 1 and now we have our line.*0172

*This line is going to split the entire graph in to two regions.*0190

*We had this region on the left side of it or the region on the right side of it.*0195

*One of those regions will be our solution for our inequality.*0201

*The question is, in which one should it be?*0205

*Since this is the entire region, we can pick out an entire or just one little point from that and see if that region should be included or not.*0208

*To get all points inside of the region is either solution or all points inside the region are not a solution.*0217

*We need one to the find out.*0223

*I'm going to borrow a point, let us take this guy right here.*0226

*I’m going to take a point and test it into my inequality, y =4 and x = -2.*0236

*I will put that little question mark in there because I want to see if this is true or not.*0249

*Is 4 > or = 2, 2 × -2 = -4 + 4?*0253

*That is 4 > or = -8, yes it is, 4 > or = -8.*0260

*I know that, that particular point should be in my solution region.*0269

*To highlight that it is in my solution I will go ahead and shade everything on that side of the line.*0275

*Everything on that side, shade it.*0297

*Everything that is shaded in red and including that line on my graph represents a solution to the original inequality.*0301

*When you have an inequality that has the or equals to it, we usually use a solid line to show that everything on that line is actually one of our solutions.*0311

*Even if I use to pick like a point right on that line it should satisfy the inequality.*0322

*If it was a strict then we will use a nice dotted line to show that the line should not be included.*0328

*Now that we have the basics to graphing out one of these regions, let us give it a few more try.*0338

*We know that our equation will split the region up in the two regions and that the equation will either have to be solid or dash.*0348

*It is determined by the type of inequality used.*0358

*If we are dealing with a strict inequality like strictly less than or strictly greater than, *0363

*feel free to use a dotted line because you should not include the points on that line.*0367

*If you have the or equals to in there and use a nice solid line to show that the line is included. *0373

*To determine the region, shade on one side or the other, use a test point to help you out,*0381

*that determines is that point and the region it is in included or not.*0388

*Pick something inside the region, you do not want to pick something on the line.*0395

*If the point satisfies your inequality, go ahead and shade that entire region that it is in.*0399

*If it does not satisfies then feel free to shade the other side, which it is not in.*0405

*One last thing if I could add to this, go ahead and pick something easy.*0410

*Potentially you could pick any point in one of these regions and you have some freedom, pick something that is nice and easy to evaluate.*0420

*I think we have our tools, let us go ahead and go to one of our examples.*0428

*We want to graph the inequality 3x -2y > 6.*0435

*First, I'm going to look at this as if it was a line.*0440

*Let us use our little equal sign in here and that is the corresponding equation.*0446

*The tools that I will use on this one is, maybe I will look at the intercepts.*0450

*What happens when x = 0 and y = 0.*0456

*Plugging in 0 for x, I would have 0 × 3 =0 and that term would drop away.*0459

*Then I divide both sides by -2 giving me -3.*0470

*I have one of my points, let us go ahead and put that on there 0, -3.*0479

*Let us put in a 0 for y.*0487

*It looks good -2 × 0 = 0 so that is gone and dividing both sides by 3 we will get 2.*0495

*I have another point I can go ahead and put on this graph 2, 0.*0508

*I’m going to use these points to graph the line but I’m going to be very careful to make it a dotted line since our inequality is a strict inequality.*0514

*I do not actually want to include this.*0522

*This looks good, nice big dotted line. *0541

*We have to figure out which region we should shade, either on the left side of this or on the right of this.*0545

*To help out, I'm going to borrow a test point just to see what is going on.*0551

*We could choose anything on here but I’m going to choose something nice and simple.*0556

*We are going to choose the origin since they are 0 for x and 0 for y.*0561

*Let us see if this works.*0566

*0 for x and 0 for y is that greater than 6?*0568

*0 × 3 and 0 × 2 both of those would cancel.*0576

*Is 0 > 6? I am afraid not, it is not greater than 6.*0580

*That tells me that origin is not in my solution region and I should shade the other side of the line.*0586

*Let us do that then.*0593

*This entire region would be considered as our solution region and we have graphed the inequality.*0604

*Let us try another one of these.*0619

*This one is to 2y > or = 4x -2.*0620

*Let us start off like before, looking at the corresponding equation and see if we can graph that .*0626

*2y = 4x -2 this one is almost in slope intercept form, I’m going to divide everything by 2 and put it into that form y = 2x – 1.*0635

*Now I can see that it has a y intercept at -1 and a slope of 2.*0649

*Let us put those points on them.*0656

*-1 and from there I will go up to over 1 and get a few more points.*0661

*I can definitely graph out my line and I will make it a solid line, since my inequality has the or equals to. *0669

*There we are, almost done.*0687

*We just have to figure out what side we should shade on.*0690

*Let us pick a nice easy point that we can evaluate and see where we go from there. *0695

*I already used the origin last time, so I'm going to pick something a little bit farther away.*0699

*Let us try this point right here, -2, 2.*0706

*When y = 2 is this greater than or equal to 4 × -2 -2.*0715

*I’m going back to the original.*0730

*I have put in the point into x and y.*0731

*2 × 2 = 4, 4 × -2 = -8 -2, is 4 > or = -10.*0735

*4 is positive and -10 is negative, positive number should be greater than negative numbers.*0747

*I know that point is in my solution region, I should shade everything on that side of the line.*0752

*This one we are not only including the entire region that I'm shading in right now but we are also including everything on that line since it is nice and solid.*0763

*There is our inequality all graphed out.*0782

*In this last example, let us go ahead and work in the other direction.*0786

*You will notice that I have an inequality I already graphed.*0790

*We are going to try and create what that inequality was.*0793

*When we are going and building these from the ground up, we focus on the equation associated with it first.*0798

*We are going to start in the same spot.*0805

*When you look at this dotted line and I can see that it has a y intercept up here at 2.*0807

*I think I can also interpret it slope.*0813

*To get back on the line I have to go down 2 and to the right 3 to get back on the line.*0816

*I definitely know its slope as well. *0823

*The equation of the line be y = -2/3x + 2, that worked out pretty good. *0826

*I can see that it is dotted so whatever inequality I’m going to deal with next should be nice and strict, either less than or greater than.*0838

*The question is which one should I include?*0848

*I want to make sure that it has the proper solution region.*0850

*In order to help me figure this out, I’m going to borrow a point from inside that solution region.*0854

*Let us borrow the point 2, 2.*0860

*I’m going to plug it into both sides of the equation and figure out what inequality symbol should go in between and make the rest of the statement true.*0863

*I have 2 for y and I will put in 2 for x.*0872

*Let us just go ahead and simplify this and see how this turns out.*0879

*2 is already simplified, over in this side I have -4/3 + 2, I have 2.*0883

*-4/3 + 2 I will have 2/3 left.*0892

*What is greater 2 or 2/3?*0899

*That does not take too long to figure out, 2 is definitely greater.*0903

*We will go back and put it into all of the spots all the way back up here and that way we can actually build our inequality y > -2/3x + 2.*0908

*That is the one represented by this graph.*0924

*You can notice how I’m using a strict inequality because it is a dotted line.*0927

*Thank you for watching www.educator.com.*0931

*Welcome back to www.educator.com. *0000

*In this lesson we are going to wrap things up with inequalities by looking at systems of inequalities.*0002

*We do not have a whole lot to cover but we will look at how you can satisfy a system of inequalities rather just individual inequalities.*0011

*That will bring us to looking at overlapping solution regions.*0019

*Recall it when we are usually dealing with the system, such as a system of equations,*0027

*we want to make sure that all of the equations in the system are satisfied.*0031

*When we are dealing with the system of inequalities, rather than equations, but it is the same goal. *0037

*We want to make sure that every single inequality in here gets satisfied.*0042

*This makes it a little unusual because rather than just looking for individual points that would satisfy inequality*0048

*remember that each of these is a region on our graph.*0054

*What we are looking for then is where these regions for each of them would end up overlapping.*0059

*That overlapping region would be where the entire system is satisfied.*0065

*Like before, after we are done identifying an overlapping region, we can use a test point to go ahead *0070

*and figure out whether that is the correct overlapping region.*0077

*It is not about process, we just have to get into figuring out what those regions are.*0081

*What we are going to do to basically figure out the overlapping region is, first we will graph each individual inequality in that system.*0090

*That breaks it down into figuring out the equation and figuring what side is shaded on that line.*0099

*Be very carefully that you are using either a dotted or dash line depending on the type of inequality in there.*0107

*We are using the dotted for our strict inequalities and we are using the nice solid line for our or equals to.*0113

* Since the line could be included as a solution.*0122

*Individually, we will make sure we shade on one side or the other and we can check that out using a test point.*0126

*Here is the big one that we have for our system, look for where the two regions overlap so we know where the solution to the system is.*0132

*That is a little bit of background, let us see these examples and see how this works in practice. *0145

*In my system of inequalities, I have 2x + y < or = 1 and I have x > 2y so individually let us take a look at these.*0150

*The corresponding equation to the first inequality would be 2x + y = 1.*0161

*If you want to graph that one out, you could get that one into slope intercept form by moving the 2x to the other side by subtracting a 2x.*0169

*From this I can see it has a y intercept of 1 and the slope of -2.*0181

*From that point I go down 2/1, down 2/1.*0186

*Looking at the original I can see it involves an or equals to so I'm going to use a nice solid line. *0192

*Now we have our equation, we need to figure out what side to shade it on.*0212

*We usually grab a test point to figure out what it should be and it is good to grab a nice test point that is to easy to evaluate.*0216

*I’m going to borrow the point 0, 0 here and see if it works in the original.*0224

*2x is 0, y is 0 is that less than or equal to 1, let us find out.*0231

*2 × 0 + = 0 and 0 < or = 1.*0240

*I know I should shade on that side of the line.*0248

*This entire region that we can see right now is only for the first inequality in our system.*0255

*Simply just going to go through the same process for the second inequality and figure out where both of their regions overlap.*0265

*Let us get some space and let us see this process for the second equation.*0278

*The second equation is x > 2y.*0289

*Maybe I will find this one in its slope intercept form so I will move the two together side by dividing everything by 2.*0293

*I could look at this as the equation y = ½ x, it has a y intercept of 0 and a slope of 1/2.*0305

*It goes through the origin 0 up 1/2, up 1/2, up 1/2, up ½.*0317

*Let us use a dotted line since the inequality is strict and see what we have.*0325

*We just need to shade on one side or the other.*0348

*We could test point for this particular line, let us just choose one way out here.*0353

*This one is at 3, 3.*0358

*Plug 3 in for y and plug 3 in for x and we will see if it is true not.*0363

*Is 3 < 3/2? That is not looking so good, that is not true.*0376

*What that means is we should actually shade on the other side of that line.*0386

*Let us shade everything below it.*0390

*Now comes the very important part of this.*0402

*We can see where each individual inequality is being satisfied but since we are interested in the region where the entire system is being satisfied,*0406

*we are looking at the overlapping regions.*0418

*I’m going to highlight that.*0421

*This region right down here is where the two overlap and I can see along one of the borders, it is dotted and along the other border it is solid.*0424

*This yellow area and its borders would be considered the solution to the system.*0440

*Let us try another of these and be careful as we walk through the process so we can see where that overlap is.*0450

*In this one we are dealing with the system 3x - 2y > 12 and 5x - y < 6. *0455

*Let us grab the first one and look at its corresponding equation.*0463

*The form of this ones in would be nice if we just use our intercepts to go ahead and see where it crosses the x and y axis.*0477

*I will put in a 0 for x and 0 for y.*0486

*If x = 0 what is y?*0491

*This term would go away for sure and I can divide both sides by -2 so y = -6.*0496

*Let us go ahead and put that on our graph, 0, -1, 2, 3, 4, 5, 6 right there.*0507

*We will go ahead and put in a 0 for y.*0515

*Let us see how this one turns out, 2 × 0 = 0, 3x = 12 divide both sides by 3 now and we will get x = 4.*0524

*4, 0 is our other point.*0536

*Now that we have 2 points, let us go ahead and connect it.*0541

*We will connect it using a dotted line and then we will figure out what side we should shade this on.*0544

*To determine what side to shade it on, let us borrow a nice, good test point.*0564

*Let us borrow the origin add 0, 0 and see if it is true or not.*0569

*(3 × 0) – (2 × 0) > 12? That would simplify to 0.*0574

*Is 0 > 12? No.*0583

*Let us shade the other side.*0587

*That entire side would be our solution region or least for the first one.*0602

*Let us get some space and do the same exact process for our second equation.*0608

*The corresponding equation would be 5x - y = 6. *0622

*This looks like maybe be easier to graph it if it was in slope intercept form.*0629

*I’m going to move the 5x to the other side and then multiply everything by -1. *0634

*Okay, my y intercept is way down there at -6 and it has a slope of 5 so up 1, 2, 3, 4, 5/1, up 1, 2, 3, 4, 5/1.*0649

*I will graph this one using a dotted line.*0669

*I have a few regions that it might be if we want to figure out until we shade one side of the blue line or the other. *0692

*Let us grab a test point for it.*0701

*I think origin 0, 0 is going to work out very nicely for this one as well.*0702

*(5 × 0) – 0 is that less than 6?*0708

*The entire left side simplifies to 0 and 0 < 6.*0713

*Let us go ahead and shade everything on that side of the line. *0720

*There is a lot of region over here to shade.*0734

*Now comes the important part where exactly do both of these overlap?*0744

*This is a little tough to tell, but if you look near the bottom you can see that there is the small triangle here where both of them overlap.*0751

*We would have that as our solution region.*0761

*As long as it is in the overlapping portion we are good to go.*0765

*For this last example I wanted something that was a little bit more like a word problem.*0773

*You could see how inequalities have a lot of good applications in them *0777

*and how some more problems lead to the rise of a system of inequalities rather than a system of equations. *0781

*What we want to do for this one is graph the solution region for the system and interpret what that means in the context of the problem. *0789

*What we are looking to do is recreate what it means when a person's heart rate, *0797

*or a person's maximum heart rate 220 – x, where x is the person’s age.*0802

*This one is good as long as it is for people between 20 and 70 years old.*0808

*When a person exercises and are trying to lose some weight, it is recommended that the person strive for heart rate *0815

*that is at least 60% of the maximum and at most 70% of the maximum.*0820

*There is a little bit of a range for that maximum heart rate that we are aiming for.*0826

*Let us see if we can interpret some of the things in here and create an inequality for what we have. *0832

*First of all, let us talk about a person's age.*0839

*Earlier it says that x is our person's age and that we are only interested when x is in between 20 and 70.*0844

*Let us say x > 20 and x < 70.*0856

*There are two inequalities that we can put in our system.*0864

*We also want to look at a person losing weight. *0868

*Also, it recommends that they strive for 60% of their maximum and no more than 70% of their maximum.*0873

*What exactly is their maximum?*0880

*It is way up here at this expression, 220 - x. *0883

*60% of their maximum 3.60 (220 – x) and we want to make sure their heart rate is greater than that.*0890

*You do not want to stress them too much, so we will make sure that it is less than .70 of the maximum heart rate. *0915

*What we now have here are 1, 2, 3, 4 different inequalities in our system that we will be looking to satisfy.*0924

*And we will graph this out so we can see the region it creates.*0934

*Let us end up rewriting this just so we have them handy.*0943

*x must be greater than 20, x is less than 70, *0947

*and we also have two more, H is greater than (.6 × 220) - x and H is less than (.7 × 220) – x.*0955

*If I'm going to graph these other two here I want to get them in a better form. *0974

*It is tough to see maybe what their slope is and what their y intercepts is.*0980

*Let us do a little bit of work with distributing that .6 and .7 in so we have a better idea.*0985

*This one will be H > .6 × 220 = 132 - .6x and for this other one we have 154 -.7x, that gives us a much better idea. *0995

*I think I’m going to write them again so I can put that x term first and see that is my slope.*1020

*We will end up graphing these two inequalities in our system as well as these two on our system.*1039

*Let us see how we can do that.*1046

*With the first two that have to deal with the age of the person, we will use that along our x axis.*1048

*We want the age of the person to be between 20 and 70.*1059

*Let us use a nice vertical line and I’m going to make this dotted because my inequality is strict.*1068

*Let us do the same thing for our 70 here.*1080

*If I was looking to just satisfy those first two then I would have to be between of these two lines. *1092

*Let us work in what we want their heart rate to be and see how that fits into this. *1104

*In the first one I can see it has a y intercept of 132 and the other one has a y intercept of 154. *1113

*My slope on the first one is a -.6x so since it is negative it should be going down, maybe something like this.*1127

*In the other one is very similar its slope is negative as well, but it is a little bit steeper since its .7 and we will put that on.*1153

*Now where should the shading for this one be?*1174

*We want our heart rate to be more than the .6x + 132.*1177

*We want to think of that as our minimum heart rate that we are aiming for in terms of exercise.*1183

*The other one would be a .7x + 157 we want them to be no greater than that value.*1189

*We are looking in between these 2 lines again. *1195

*From looking at these two solution regions, it is this little box looking thing where they end up overlapping.*1202

*We can interpret what that solution region means in terms of the problem. *1216

*If you we are to just pick a random point inside that box, what we are getting is maybe an age of a person, *1221

*let us say 65 and we are also getting what the heart rate should be when they exercise to try and lose weight. *1231

*As long as they stay inside that box they should do some good things to their heart.*1239

*If we try and fall outside of the solution region that means either the formula is for our maximum heart rate no longer apply*1245

*or we would end up overtaxing the heart and either one we definitely do not want.*1254

*When setting up a system of inequalities look at them individually and graph out each of the regions.*1261

*Then look at the overlap and see what would be included in both.*1267

*Thank you for watching www.educator.com.*1271

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at integer exponents.*0002

*What does it mean? You take a quantity and raise it to a power.*0005

*A lot of other things we will also be picking up along the way.*0010

*Let us get a good overview.*0013

*We will start off by interpreting what it means to raise something to a power when that power is a nice whole number.*0015

*We will see that we can often combine things using the product and the power rule because all these things have the same base.*0021

*We will see what it means to raise things to a 0 power and get into the quotient rule.*0028

*And get into understanding what happens when you raise something to a negative power.*0034

*Watch for all of these things to play a part.*0038

*One way that we can look at multiplication is that it is a packaging up of addition.*0046

*I have the value of x and we know that I have added together 5 times.*0053

*Rather than trying to string out to x + x + x over and over again I can write that same thing by just saying 5 multiplied by x.*0059

*It is a great way to take a long addition problem and package it all down into a nice single term. *0074

*Exponents are the things that we are interested in as long as we are dealing with whole numbers you can look at that as packaging up some multiplication. *0081

*Here I have those x’s once again, I have 5 of them all being multiplied together but rather writing all that, I'm going to use my exponents.*0090

*Specifically the bottom here will be our base. *0105

*Remember, that is what we are multiplying over and over again.*0108

*The exponent itself tells us how many times.*0111

*Let us get some practice with just understanding how exponents work before moving on. *0121

*With this, I simply want to write them into their exponential potential form.*0128

*If I have 5 × 5 × 5, I can see that it is the 5 being repeatedly multiplied and I did it 3 times.*0132

*I could consider this one like 5*^{3}.0139

*If I wanted to take this one little bit farther, this is equal to 125.*0143

*You have to focus on what is being multiplied over and over and over again to properly identify your base.*0150

* In this one, we have -2 × -2 × -2, we can see that it is being multiplied by itself 4 times.*0157

*We want the entire value of -2 multiplied by it self 4 times so we will use an exponent of 4.*0167

*If we had to simplify this one, I have -2 × -2 × -2 × 2 and that would be 16.*0177

*Keep an eye on the base, so you know what is being multiplied and the exponent will tell you how many times to do that.*0186

*In this next two, we just want to evaluate the exponential expression.*0196

*Let us see if we can identify the base here.*0200

*This is a 3*^{4} that happens to be negative sign out front and my next example, it is almost the same thing but it is -3^{4}.0203

*Watch how I treat both of these a little bit differently. *0212

*In this first one, the only thing that is considered the base is actually that number 3, 3 × 3 × 3 × 3 multiplied by itself 4 different times.*0215

*What should I do with that negative sign out front, it is still out front, but since it is not part of the base and it only shows up once.*0228

*I can take care of it from here, 3 × 3 × 3 × 3 a lot of 3’s in there, but that will equal 81.*0235

*Of course that negative sign is out front along for the right.*0244

*In the next one, these parentheses are highlighting that the entire -3 is what is in the base so I'll repeat the entire -3. *0248

*Look at this one, I'm dealing with a -3 multiplied by itself 4 times.*0267

*I know that a negative × negative would be positive and another negative × negative would be a positive, *0272

*the answer to this one action turns out to be 81.*0278

*Use those parentheses to help you know what is in the base and it is good for all of those negatives.*0283

*One of our first rules that we can throw into the mix is the product rule for exponents.*0293

*This will help us when we have two things and both of them are raised to an exponent.*0299

*It is a great way to actually put them together so they only have one thing raised to an exponent.*0306

*The way this rule works is, we want to make sure that both the bases are exactly the same.*0311

*I have a and a as my base, if they are the same then it says we simply need to take each of their exponents and add them together. *0318

*It seems like a little bit of an odd rule how you can always start with multiplication and end up with addition *0327

*but let me show you an example of why this works.*0333

*I’m going to base that off just what you know about exponents that it is repeated multiplication. *0336

*Here I have 2*^{4} power be multiplied by 2^{3}, let us pretend I knew nothing about the product rule.0342

*I can start interpreting this by using repeated multiplication, so 2*^{4} will be 2 × 2 × 2 × 2 until I done that 4 times in a row.0350

*I can do the same for 2*^{3} that would be 2 × 2 × 2.0364

*You can see what we have here, the whole bunch of 2s are all being multiplied together and there is even no real need to put on those parentheses.*0373

*I will just write 2 × 2 × 2 until I got all of them written down. *0381

*As long as I have all of these 2s written out, it was said earlier that exponents are just repeated multiplication, *0388

*I can actually package these all backup.*0394

*If you count how many are here, it looks like we have 7 total.*0398

*I can say that this is 2*^{7}.0404

*Now comes the important part, if you look at our original exponents that we have used the 4 and 3 and you add them together *0409

*you can see that sure enough, it does add to 7.*0418

*The reason why this is happening is because you are simply throwing in a few more numbers to multiply by and incrementing up that exponent.*0421

*That is why the product rule works.*0431

*Let us go ahead and practice with it now.*0434

*In our first one I have -7*^{5} × -7^{3}.0439

*The first thing you want to recognize is both of those bases are the same, so we will just focus on their exponents the 5 and 3.*0445

*This one as -7^(5+3) and then I can go through and add those 2 things together, -7*^{8}.0456

*You could simplify it further from there if you want to but I’m just going to leave it like that so you can see the product rule in action.*0472

*The next one will involve a little bit more work and that is why I put in the -4 and 3.*0479

*Before I can get too far, I’m going to start rearranging the location of the 4 and the p's raised to the exponents.*0488

*What allows me to do this is the commutative property for multiplication that the order of multiplication does not matter.*0495

*This is just helping me organize things a little better so that I have my numbers all in one spot and my exponents on one spot as well.*0504

*That looks good.*0513

*Put the -4 and 3, I will just straight multiply those two together and get -12.*0515

*With the other two, I can see that they are both the same base of p.*0521

*I will look at adding their exponents together so p^(5+8).*0526

*Here is -12 p*^{13} not bad.0535

*One last, let us see if we can use the product rule in that one.*0542

*Looking at this one you will know that they both have a base of 6.*0546

*One is raised to the 4th power and one is raised to the power of 2. *0549

*But there is a slight problem with this one.*0553

*This one is dealing with addition and the product rule, the one that we are interested in is for multiplication.*0557

*Since the product rule only applies to multiplication, it means we can not use the product rule here.*0570

*Even though that my bases are the same it simply not going to work here. *0584

*If I did want to continue this one out, I would not use the product rule. *0589

*I would just simply take 6*^{4} and see what value that is.0593

*I will get 1296 then I will get 6*^{2} = 36 and I will add the two together 1332.0597

*You can notice how we do not use the product rule since it has addition there on the bottom.*0608

*Let us try a few more examples and see if we can recognize our product rule in action.*0619

*I have thrown in a few more variables this time.*0623

*The first one is n × n*^{4}.0625

*It is tempting to say hey maybe this one is just the n*^{4}, but if you see missing powers like on that first one, assume that they are 1. 0629

*That way you actually have something to add.*0638

*This would be n1+ 4 and now that they actually have your exponents added together, we will get n*^{5}.0640

*Now you know if you strictly look at the product rule it only talks about putting two things together, you can even apply it for more than two things.*0655

*In this one, I have z*^{2} × z^{5} × z^{6} and we can apply the product rule as long as we do it two at a time.0664

*Let us just take the first two z’s here. *0673

*This would be z^(2+5), we will get to that other z in a bit.*0677

*I can see that I have a single z^(2+5) and z*^{6}, that simplifies into z^{7}.0686

*I can put these two together z^(7+6) and that would be z*^{13}.0696

*We could have even taken a much bigger shortcut if we simply just added all 3 of them together.*0707

*You can definitely do this as long as all of their bases are the same, and you are dealing with multiplication.*0716

*One more, this one is 4*^{2} multiplied by 3^{5}.0723

*We are definitely dealing with multiplication but again, this one has a little bit of a problem.*0730

*Notice how this one, the bases are not the same. *0737

*The condition for using that product rule is that we need the base is the same *0748

*and we need multiplication and so we simply can not use the product rule here.*0752

*If we are going to go any farther with this one, we have to evaluate these exponents.*0756

*4*^{2} would be 16 and then I will go ahead and multiply that by 3^{5} = 243 and then multiply the two together, so 3888.0762

*Be very careful that the conditions of the product rule are met before you try and use it.*0787

*Another good rule that we can use for combining things together is known as the power rule.*0797

*The way the power rule works is, we have something raised to an exponent and then we go ahead and re-raise all of that to another exponent.*0804

*It is like we have exponents of exponents.*0812

*The way the power rule handle these is we will take both of the exponents and we will actually multiply them together.*0816

*To convince you that this is the way it should work, we are going to deal with another example that has just numbers in it, *0826

*and show you that this is the way you should combine your exponents in this case.*0833

*We are going to take a look at 4*^{3} and all of that is being squared. 0838

*Say you knew nothing about the power rule, how could you end up interpreting this?*0845

*One way is you could use again repeated multiplication to interpret your exponents. *0849

*In the first exponent I'm going to interpret is this two right here.*0856

*I’m going to take my base, 4*^{3} and it will be multiplied by itself twice, you can see I have two of them.0860

*I want to go further with this, now I will interpret both of these 3s as 4 being multiplied by itself 3 times.*0873

*Now that I have expanded it out entirely, I’m going to work to go ahead and package it all up.*0891

*What I can see here is that I have a whole bunch of 2s and they are all going to be multiplied together.*0897

* In fact I have a total of 6, this is 4*^{6}.0905

*Go ahead and observe exactly what happened with the exponents.*0913

*Originally we started with 3 and 2 and if you multiply those together like the power rule says we should sure enough, you get 6.*0917

*If you fall along the process you can see why that happens. *0925

*Now that we know more about this rule, let us move on.*0929

*The power rule which we learned earlier can be applied to multiplication and division.*0937

*The way it does it is we can take two things that are being raised to a power and then that giving each of them that exponent.*0942

*When dealing with division, if we have things being divided, say, a ÷ b we can take that exponent and give it to each of them.*0953

*What I’m trying to say here is that if you have an exponential expression and it is raised to a power,*0964

*you are going to apply it to all the parts on the inside as long as those parts are being multiplied or divided.*0968

*This one helps us when we get to those larger expressions.*0973

*Let us try a few of these up, we simply want to use the power rule for each expression rated out.*0981

*The first one is 6*^{2} and all of that is being raised the fifth power.0988

*I'm not going to expand that out, I’m just going to go directly to the rule.*0993

*This would be 6 then we have 2 × 5.*0998

*Taking care of just the exponents this would be 6*^{10}.1004

*I could continue to try and figure out what number that is but I’m only interested in considering what the power rule is.*1010

*Let us try another one.*1017

*This is z*^{4}and all of that is being raised to the 5th.1019

*I can take my exponents and I will multiply them together. *1026

*This will be z*^{20}.1031

*Let us get into a few that are a little bit more complicated and see how we can tackle this.*1036

*In this next one I have 3 × a*^{2} × b^{4}.1042

*One of the first things I need use is my power rule and give that 5 to all of the parts.*1047

*Let us write that to each of these.*1057

*I’m going to do the 5*^{3}, I'm going to give it to the a^{2} and I'm going to give it to the b^{4}.1061

*It must go on to all of those parts and it is okay since all these parts were multiplied in the original.*1070

*Now that I have this, I will go ahead and start to simplify each of these.*1077

*My 3*^{5} is 243, then I will multiply these exponents together and get a^{10}.1086

*We will multiply the 4 and 5 together and would give us b*^{20}.1097

*This one is a completely condensed down far as I can go.*1104

*Let us see one that involves division.*1109

*I'm going to give the 3 to the top and bottom.*1116

*The top will be 3*^{3} and the bottom will be x^{3}.1121

*This will give us 27 all over x*^{3}.1133

*Let us try one more 4 + x*^{2}.1140

*What does the power rule say we should do with this one?*1145

*I threw this one in there because you should not use the power rule on this one. *1149

*Why not? Why can not we use the power rule here?*1156

*Notice the parts on the inside, they are being added 4 + x instead of being multiplied.*1159

*Since it is being added instead of multiplied, I can not simply give the 2 to all the parts.*1170

*This situation down here is completely different from this one because of that addition.*1175

*How could I take care of this?*1181

*What we will see in one of our future lessons is that we will handle this one using repeated multiplication*1185

*and we simply have to multiply those two terms together.*1193

*For now I want to point out is you should not use the power rule on that situation simply does not work.*1196

*All of these rules are extremely important and you should be very comfortable with them.*1207

*I recommend memorizing all of them.*1212

*Usually a lot of practice with them will get you comfortable with recognizing when you should and should not use them. *1214

*In addition, you should be comfortable with these that you can use more than one of them in a single problem.*1223

*Maybe you will end up using the power rule and then end up using the product rule all in the same one to condense and simplify an expression.*1229

*That is how comfortable with these rules you should be. *1237

*We can get a better handle and the least a little bit more comfortable with these we will give it a try.*1243

*We will try and use many of these rules together to simplify the following expressions.*1248

*Okay, this first one is (5k*^{3} / 3)^{2}.1253

*One of the very first rules that I'm going to use, is I'm going give that 2 to the numerator and to the denominator of that fraction.*1262

*You got to be careful in doing this, the top is 5k*^{3} and the bottom is just 3.1272

*We will give 2 to each of those.*1280

*Now that I have taken care of that rule I can see I have a little bit more of work to do on top. *1285

*I have multiple parts in there all being multiplied and I’m going to give the 2 to each of those parts. *1289

*Here is my 5k*^{3} so we will give the 2 to the 5 and we will give the 2 to the k^{3}.1296

*Now that we spread out our variable amongst all of the parts that are being multiplied and divided, let us see if we can continue.*1308

*5*^{2} that would be 25 and now I have k^{3} and all of that is being raised to the power of 2.1316

*What rules should we use for that?*1326

*As we only have a single base in there, this is that situation where we multiply the two together so this will be k*^{6}.1330

*Now I simply have 9 on the bottom from 3*^{2}.1341

*I do not see anything else that I can combine or any other rules that I can use, this is good to go.*1346

*Be careful in using multiple rules.*1353

*Let us take a peek at the second one here. *1357

*This one is (-3 × x × y*^{2})^{3}.1359

*In addition that is being multiplied by (x*^{2} × y)^{4}, lots and lots of different exponents flying around.1366

*Let us recognize that we do have some multiplication on the insides of those parentheses, *1376

*I will be able to spread out that exponent amongst all of its parts.*1381

*What parts do I have in there?*1386

*I have -3, x, y*^{2} so we will put 3 on each of those.1388

*Let us do the same thing for the other one.*1403

*I have x*^{2} and y^{4}, both of these looks like they need 4.1406

*(x*^{2})^{4} y^{4}, not bad.1413

*Now that I have applied that rule, let us go through and start cleaning other things up.*1421

*Starting at the very beginning I have -3*^{3}.1427

*That is -3 × -3 × -3 = -27.*1431

*Right now I have x*^{3} just as it is.1440

*It looks like this next part that is a good place where I can multiply the exponents together, y*^{6}.1445

*Here is a couple of more situations where again I can just multiply those exponents together x*^{8} y^{16}.1456

*It is important not to stop there because all of these pieces are still being multiplied together.*1470

*I can see that I have a couple of x and a couple of y.*1476

*Let us use our commutative property for multiplication to change the order of these parts.*1481

*x*^{3} x^{8} y^{6} y^{16}.1488

*The reason why I’m doing this is so we can better see that I need to apply one more of our rules that will be the product rule.*1494

*Both of these have the same base and I can simply add their exponents together, the 3 and 8.*1502

*-27 x^(3+8) is 11 and we can do the same thing down here with our y since they have the same base as well and are being multiplied.*1512

*y^(6+16) would be 22.*1531

*We finally got down to our answer and apply that to all the rules that we could.*1538

*You can be very comfortable with mixing and matching and putting these rules together.*1543

*Just be very careful that you do them correctly.*1548

*A very important rule that we will need is the 0 exponent rule.*1555

*What this rule says is that when you raise anything to the 0 power like a over here, that what you get is simply 1.*1561

*This is probably one of the most curious rules that you will come across.*1570

*After all, usually when you are dealing with 0, you are familiar with multiplying by 0 and getting 0 or even adding 0 and do not change anything.*1573

*Why is it that when we take something to the 0 power, it becomes 1?*1581

*One reason that we do this is if we want to be consistent with the rest of our rules. *1587

*After all, we have been building up a lot of other rules, combining things, our repeated multiplication, we wanted to the mesh well with those other rules.*1593

*Let us see how it actually does need to be defined as 1, so that it fits with everything else we are trying to do.*1600

*I’m going to look at 6*^{0} × 6^{2} and we will do this twice. 1609

*The first time that I go through this, I'm going to end up using my product rule.*1617

*The reason why I’m doing this is that I have the 6 and the 6 are both the same base and that rule says I need to add the exponents together. *1622

*This would be 6^(0+2) now the addition is not so bad 0+2 is 2 so this would end up being 36.*1631

*For this rule to stay consistent with that, I need it to also equal 36.*1648

*You can see that if I look at just 6*^{2} by itself, this guy over here that does equal 36.1656

*What should I call this thing to multiply by 36 so that my final result is 36.*1666

*You should not have to think on it to long, there is only one thing that I can call 6*^{0}.1676

*I must call it a 1 so that when it is multiplied by that 36, it still is 36 as the final result. *1683

*It is the only way that we can define something to the 0 power so that it mixes well with all the rest of our properties.*1691

*Let us play around this rule a little bit.*1701

*This new rule is the quotient rule.*1707

*In this one we are dealing with two things that have the same base so a and a and they are being divided.*1710

*When we divide like this, it looks like we can simply take their exponents and subtract them.*1720

*It is a lot like our product rule only that one dealt with multiplication and addition, this one has division and subtraction. *1726

*Some important things to know of course we must have the same base for this to work out.*1735

*We are dealing with division and subtraction.*1740

*This is a quick example to see why this might work.*1743

*Let us look at 5*^{4}/ 5^{2}.1747

*Suppose I knew nothing about the quotient rule, one way that I could just end up dealing with this, is looking at it as repeated multiplication 5 × 5 × 5.*1754

*Then I could look at the bottom and say okay what is 5*^{2}?1768

*That would be 5 × 5 and then I can go through canceling out my extra 5 in the top and in the bottom.*1774

*This would mean 5 × 5, which is the same as 5*^{2}. 1784

*If we look at the result of this, you know what happened to the original versus the answer in the end, *1792

*you can see that we also have to is subtract those exponents.*1798

*4-2 does equal 2.*1801

*This is the way that we will handle things that have the same base and we are dealing with division.*1804

*We will simply subtract their exponents. *1810

*Now, unfortunately this does lead to a very interesting situation and potentially it might give us some negative exponents.*1816

*To see why this might happen, let us use an example with numbers so *1825

*we get a good sense of some of the strange things that we want to be prepared for.*1829

*This one I have 2*^{4} ÷ 2^{5}.1833

*The first way I’m going to handle this, is I am going to use my rule for the quotient rule.*1838

*This will be 2^(4-5).*1845

*If you take 4 – 5, you will get 2*^{-1}.1852

*Now I can see that is what the product rule tells me to do, but you know what exactly does that mean to have 2*^{-1}.1858

*After all, when we are dealing with just whole numbers, then I would often look at this as repeated multiplication, but how can I multiply 2 by itself?*1867

*A -1 number of times and that just does not seem to make sense. *1875

*To get a handle of what that means, I'm going to look at the original problem as repeated multiplication.*1879

*This would be 2 × 2 × 2 × 2 all over 2 × 2 × 2 × 2 × 2.*1886

*I have 4 of them on top and 5 of them on the bottom.*1898

*Let us go through and cancel out all of our extra 2s, 4 from the top and 4 from the bottom.*1902

*You can see from doing this, there is only one left and it is on the bottom so this would be ½ .*1911

*Now comes the important part that if we look at each sides of our work out, we have used valid rules and we have done things correctly.*1918

*What I have here are two different expressions which are the same thing.*1927

*This gives us a good clue on what we should do with those negative exponents. *1934

*We want to interpret our negative exponents by putting the base in the denominator of our fraction.*1938

*It is how we will end up handling these things.*1945

*It will take a little bit more work to get comfortable with these, let us see what else we can do.*1949

*The rule that we have just developed is anytime we have a base raised to a negative exponent we will put it in the denominator of the fraction*1960

*and we will change that base, a rule change that exponent to a positive number. *1969

*Another way to say that is if you have an expression raised to a negative power it can be rewritten in the denominator with a positive power. *1975

*It leads to something very interesting. *1983

*If you have a negative in the top of your fraction and a negative in the bottom, *1986

*then what you can end up doing is changing the location of where those things end up.*1991

*I have a^-n and it was on the top but now it is in the bottom. *1997

*I had b^-n to in the bottom, now it is in the top.*2003

*A good rule of thumb that I give my students to keep track of what to do with that negative in the exponent *2010

*is think of it as changing the location of where something is.*2016

*If the negative exponent is in the numerator, think of the top, go ahead and move it to the bottom at your denominator and make it positive. *2020

*The other situation, if it is already in the bottom at your denominator, then move it to the numerator the top and make it positive.*2030

*It will help you handle these in a nice quick way.*2040

*Now that we know a lot more about our exponents and especially those negative ones, let us get into using them a bit more.*2047

*Here I have ¼*^{-3}.2055

*One of the first rules that I'm going to use is to spread out that -3 onto the top and bottom of my fraction.*2062

*I have 1*^{-3} / 4^{-3}.2070

*Here is where I’m going to handle that negative exponent.*2077

*That 1*^{-3} on the top, I’m going to move it to the bottom and make its exponent positive.2080

*I’m going to do the same thing with a 4.*2090

*4*^{-3} now it will go to the top and its exponent will now be positive as well.2091

*From here, I just go through simplify it, 4*^{3} = 64, 1^{3} = 1 so it is 64.2098

*Let us try the same thing with the next one and notice how things are changing location.*2116

*2*^{-3} will end up in the bottom as 2^{3}.2123

*3*^{-4} that one is going to go into the top as 3^{4}.2130

*I will go ahead and simplify from here.*2139

*There may be a few situations where some things will have positive exponents and some things will have negative exponents. *2144

*The only ones that will change locations will be the ones with negative exponents.*2155

*If I have an example like x*^{2} and y^{-3}, then the x^{2} will remain in the same spot but the y must go into the bottom since it had the negative exponent.2160

*Watch out for those.*2172

*Let us see if we can definitely combine many more of our rules together and simplify each of these expressions.*2177

*We got lots of rules to keep track of so we will just do this carefully, bit by bit. *2184

*In this first one I have the (x*^{2} / 2y^{3})^{-3}.2191

*One of the biggest features I can see here is probably that fraction.*2198

*I’m going to give the -3 to the top and the bottom of my fraction.*2204

*I have x*^{2} and will give it -3 and I will do the same thing with the bottom. 2209

*Now one thing I can see on the top is I have x*^{2} and that is being in turn raised to another power.2218

*Our rule for that says we need to multiply the exponents x*^{-6}.2225

*What to do with the bottom?*2233

*In the bottom we have some multiplication in there, so 2 × y*^{3}.2234

*I need to give that -3 to each of the pieces down here.*2239

*Moving on, bit by bit, I will deal with that negative on the x and eventually that negative on the 2.*2250

*Let us see what we need to do with the y*^{3} and y^{-3}.2258

*The rule says I need to multiply those things together and get -9.*2262

*I think I have used all of my product rules and power rules and quotient rule.*2269

*It is time to use that negative exponent rule.*2273

*Anything that has a negative exponent on it is on the top and I’m going to put in the bottom.*2277

*If it was on the bottom, it is going right to the top.*2281

*In the bottom, x*^{6} and that guy is done.2287

*In the top, 2*^{3} and y^{9}.2292

*One last thing to go ahead and clean this up, 2*^{3} is 8.2298

*I will just ahead and put it in there.*2303

*Our final simplified expression for this one is 8y*^{9} / x^{6}.2307

*It is quite a bit of work, but it is what happens.*2314

*Let us do another one. *2318

*This one is for 4h*^{-5} / m – (2 × k).2320

*I do not see a whole lot of rules in terms of spreading things out over multiplication, but one thing I do want to do is take care of those negative exponents.*2327

*I have one of them up here on the h and another one here on the m.*2336

*Let me first write down the things that do not have negative exponents, they will be in exactly the same spot. *2341

*The h*^{-5} I need to put that in the bottom as h^{5}.2348

*The m*^{-2} it now needs to go to the top as m^{2}. 2354

*Okay, I would continue simplifying from here if I could but I think I do not see anything else that has the same base.*2360

*I will leave it as it is and will call this one done.*2368

*It is time to put all of our rules in practice and see if we can do a much harder problem.*2376

*This one is (3*^{9} × x^{2} × y)^{-2} / 3^{3} × x^{-4} × y.2381

*One thing I want to point out with this one is that when you are applying the rules, you do have a little bit of freedom on which ones you do first. *2395

*Experiment with trying the rules in a different order and see if you come up with the same answer.*2403

*As long as you apply the rules carefully and correctly, it should work out just fine.*2408

*What shall we do with this one?*2414

*I'm going to go ahead and take that -2 and spread it out on my x*^{2} and y since both of those are being multiplied on the inside there.2417

*3*^{9} now have an (x^{2})^{-2}.2425

*I also have a y*^{-2} as well.2434

*I will put that -2 in both of them.*2438

*Let us go ahead and multiply our exponents here, 3*^{9} then 2 × -2 would be x^{-4} y^{-2}.2448

*Let us see where can I go from here. *2471

*I like dealing with those negative exponents and changing the location of things so I do not have to worry about negatives.*2472

*Let us do that first. *2479

*Everywhere I see a negative exponent that part will change its location.*2481

*I’m going to leave the 3*^{9} for now and leave my 3^{3} and let us take care of this one.2486

*x*^{-4} will be x^{4} in the bottom.2493

*This one which is x*^{-4}, let us put that in the top.2499

*y*^{-2} is in the bottom and this y already has a positive exponent, so no need to change that one.2505

*Continuing on, let us see what else we can do.*2516

*These 3s out here, they have exactly the same base so I can subtract their exponents, 9 – 3.*2519

*I can do the exact same thing with these x’s, subtract their exponents.*2529

*What should I do with those y’s on the bottom?*2541

*They have the same base, so I will add those exponents together.*2543

*Good way to start crunching things down.*2550

*This will be 3*^{6} x^{0} / y^{3}.2554

*3*^{6} = 729 x^{0}, remember that is one of our special ones is 1, all over y^{3}.2564

*This one crunches down quite a bit, but in the end we are left with 729 / y*^{3}.2581

*We do not have anything else to combine and so we will consider this one simplified.*2589

*Only one more to do this one, we want to make an expression that represents the area of the figure.*2597

*I want something that is a little bit more like a word problem, at least something that we have to drive out of.*2604

*We will simply use our rules along the way to crunch this down.*2609

*When I look at the area of a rectangle like this, it is formed by taking the width of that rectangle and multiplying it by the length.*2613

*I’m not sure what is the width and length here, since both of them are written as expressions.*2625

*I will write those in, area is the width 4x*^{2} and the length is 8x^{4}.2630

*That means I can take this expression for the area and put it together using some of my rules.*2640

*Let us rearrange that.*2646

*I’m going to put my numbers 4 and 8 together, and my variables and exponents together. *2648

*4 × 8 = 32 and I have x*^{2} and x^{4}.2657

*I can do these by adding them together since they have the same base giving me x*^{6}.2665

*This expression would be equal to the area of the rectangle.*2671

*It has quite a lot of rules to enjoy but again with a little bit of extra practice, they should become a little more familiar.*2677

*Watch for those special ones like raising something to the 0 power so that you know that is always 1.*2683

*Thank you for watching www.educator.com.*2689

*Welcome back to www.educator.com. *0000

*In this lesson we are going to take a look at how you can add and subtract polynomials.*0002

*Before we get too far though or have to say what polynomials are and how we can classify them.*0009

*I think we will finally get into adding and subtracting them.*0015

*To watch for along the way, I will cover how you can evaluate polynomials for several different values.*0018

*Recall some vocabulary that we had earlier.*0029

*A term is a piece that is either connected using addition or subtraction.*0033

*In this little example down below, this expression I have four terms. *0040

*A coefficient is the number in front of the variable.*0051

*The coefficient of this first term would be a 4 and I have a coefficient of 6, - 5 and 8 would be a coefficient.*0055

*We often like to organize things from the highest power to the smallest power.*0068

*The highest power here, it is a special name we will call this the leading coefficient.*0073

*This will be important terms you will often here me use later on when talking about polynomials.*0083

*Recall that things are like terms if they have the exact combination of variables with the same exponents. *0092

*In this giant list here, I have lots and lots of examples of like terms. *0100

*This first one is an example of like terms because they both have an m*^{3} and notice it has no difference what the coefficient out front is.0106

*I have 19, 14 but it does not matter that part. *0117

*What does matter is that I they both have an m*^{3}.0121

*The next two have both y*^{9} and if you have more than one variable in there then both of those variables better matchup.0127

*Both of them have a single x and they both have a y*^{2}.0136

*If we understand polynomials and understand terms then you can start understanding polynomials.*0144

*A polynomial is a term or a finite sum of terms in which all the variables have whole number exponents and no variables up here in the denominator.*0150

*To make this a little bit more clear, I have many different examples of polynomials and many different examples of things that are not polynomials.*0160

*I will pick these over and make sure that they fit the definition.*0168

*This first one here, I know that it is a polynomial as I can see that has a finite number of terms that means that stops eventually.*0174

*If I look at all the exponents present like the two, then all of those are nice whole numbers and I do not see anything in the denominator.*0182

*In fact, there are no fractions there and we do not have any variables in the denominator.*0191

*That is why that one is a polynomial. *0196

*I will put in this next example to highlight that you could have coefficient that are fractions*0201

*but the important part is that we do not have variables in the denominator, that would make it not a polynomial. *0206

*It is okay if we have more than one variable like start mixing around m and p, just as long as the exponents on those state nice whole numbers.*0216

*Even the next one is a good example of this. *0227

*I have y and x, but I have a finite number and eventually stops and it looks like I do not have any variables in the bottom. *0229

*Now a lot of things get to be a polynomial, even very small things. *0239

*For example, something like 4x is a type of polynomial, it is not a very big polynomial and has exactly one term, but is finite.*0245

*All of the exponents are nice whole powers and there are no variables in the denominator.*0253

*You can have even very, very short polynomials. *0259

*This one is just the number 5 has no variables, or even consider it x*^{0}.0261

*Compare these ones to things that are not polynomials and watch how they break the definition in some sort of way. *0267

*In this first one is not very big, it is 1 ÷ x and it is not polynomial because we are dividing by x.*0276

*We do not want that x in the bottom. *0282

*This next one is not a polynomial because of its exponents. *0286

*It has an exponent of ½ and another exponent of 1/3 and because of those exponents, it is not a polynomial. *0290

*The next one is a little tricky. *0299

*It looks like it should be a polynomial, I mean I have 1, 2x to the first and 3x*^{2}.0301

*The reason why this is not a polynomial has to do with this little…*0307

*That indicates that this keeps going on and on forever.*0311

*In order to be a polynomial, it should stop.*0315

*It should be finite somewhere.*0318

*That one is not a polynomial.*0319

*Other things that we want to watch out for is we do not have any of our variables and roots and none of those exponents should be negative.*0323

*That should give you a better idea of when something is a polynomial or not a polynomial.*0334

*We can start to classify the types of polynomials we have by looking at two aspects.*0342

*One is how many terms they have.*0348

*If it only has one term we would call that a monomial. *0351

*In example 3x, it only has a single term, it is a monomial. *0357

*If it has two terms then we can call that one a binomial, think of like a bicycle or something like that, two terms.*0363

*If I have 3 terms we will go ahead and call it a trinomial.*0374

*Usually if it has four or more terms you will get a little bit lazy we just usually call those as polynomials.*0384

*Technically, all of these are examples of polynomials but they are just a very specific type of polynomial. *0391

*Let me write that one with a bunch of different terms, 5x*^{4} - 3x^{3} + x^{2} - x +7 that would be a good example of a polynomial.0398

*In addition to talking about the monomial and binomial, that fun stuff, you can also talk about the degree of a polynomial.*0413

*What the degree is, it is the highest power of any nonzero terms.*0421

*You are looking for that biggest exponent.*0426

*In this first one here, you can see that the largest exponent is 2, we would say that this is a 2nd degree polynomial.*0429

*In the next one, just off to the right, the largest power in there is a 3, so this would be a 3rd degree polynomial. *0440

*Now you can combine these two schemes together and get specific on the types of polynomials you are talking about. *0452

*Not only for that first one can I say it is a 2nd degree because it has the largest power of 2 in there, *0459

*but I can say it is a 2nd degree binomial because it has two terms in it.*0465

*With the other one in addition to saying it is a 3rd degree polynomial, I can take a little further and say it is a 3rd degree trinomial.*0473

*That is because we have 1, 2, 3 terms present in a polynomial.*0484

*To evaluate a polynomial it is a lot like evaluating functions. *0494

*We simply take the value that we are given and we substitute it for all copies of the variables.*0498

*Let us give that a try with 2y*^{3} + 8y - 6 and we want to evaluate it for y = -1.0504

*I will go through and everywhere I see a copy of y, we will put in that -1.*0513

*Let us go ahead and work on simplifying this.*0527

*-1*^{3} would be -1 × -1× -1,that would simply be -1.0532

*8 × -1 would be -8 and then we have -6. *0539

*Continuing on, I have -2 – 8 – 6 = -16.*0546

*If you are evaluating it, just take that value and substitute it in for all copies of that variable.*0557

*Onto what we wanted to, adding and subtracting polynomials.*0568

*When we get into addition and subtraction, what we are looking to do is add or subtract the like terms present in the polynomial.*0573

*There are two ways you should go about this and both of them are perfectly valid so you will use whichever method you are more comfortable with.*0581

*Here I want to add the following two polynomials.*0590

*The way I’m going to do this is I'm going to simply highlight which terms are like terms.*0593

*Here is 4x*^{3} and 6x^{3} those are like terms - 3x^{2} 2x^{2}, those are like.0599

*2x and - 3x and so one by one we will take these like terms and simply put them together.*0610

*4x*^{3} and 6x^{3} = 10x^{3}.0619

*-3x*^{2} and 2x^{2} would be -1x^{2}.0626

*2x - 3x =- 1x.*0633

*You can see I have all of the parts there and it looks like I am left with a 3rd degree trinomial.*0640

*The other way that you can do this, if you are a little bit more comfortable with it, is you can take one polynomial over the other polynomial.*0645

*If you do it this way, you want to make sure you line up what are your like terms.*0660

*Put your x*^{3} on your x^{3}, your x^{2} on your x^{2} and you x on your x.0665

*It is essentially the same idea and that you go through adding all of the terms as you go along.*0672

*I will start over on the right side of this one, 2x and I'm adding - 3x, there is - x for that one.*0678

*Onto the next set of terms, - 3x*^{2} + 2x^{2} = - x^{2} and 4x^{3} and 6x^{3} = 10x^{3}.0688

*You can see you get exactly the same answer but just use whatever method you are more familiar with.*0701

*Let us try some examples.*0711

*In this one we want to determine if these are polynomials or not.*0712

*Let us try that first one.*0718

*I see I have 2x + 3x*^{2} – 8x^{3}.0719

*The things we are watching out for is, one does it stop.*0723

*I do not see any… out here, I know this is finite, that looks pretty good. *0727

*I do not see any variables in the denominator, so that is good.*0732

*There are no fractions with the x's in the bottom. *0736

*All of the exponents here, the 1, 2, and the 3 all of those are nice whole numbers.*0740

*This one is looking good. *0747

*I will say that this one is a polynomial.*0749

*2/x + 5/4x*^{2} – x^{3}/6.0757

*In this one I can think I can see a problem right away. *0762

*Notice how we have variables in the bottom and because of that I will say that this one is not a polynomial.*0765

*Be careful and watch out for that criteria. *0776

*This next example, we want to just go through and classify what types of polynomials these are.*0782

*We use two criteria for this, we look at its degree and we will see how many terms it has.*0788

*The first one, the largest power I can see in here is this 3.*0794

*I will say this is a 3rd degree.*0798

*It is a polynomial but let us be a little more specific. *0811

*It has 1, 2, 3 terms, so I will say this is a 3rd degree trinomial.*0814

*Let us try another one, the largest power here is 4, 4th degree.*0826

*It only has 1, 2 terms, so it will be our binomial.*0838

*One more to classify, this one has a bunch of different exponents, but of course the largest one is the only one we are interested in.*0847

*This is a 3rd degree and now we count up all of its terms, 1, 2, 3, 4.*0856

*Since it has four terms, I will just keep calling this one a polynomial.*0870

*Let us get into adding the following polynomials together.*0883

*The way I’m going to do this is I’m going to line them up, one on top of the other.*0886

*Starting with the first one, I have 3x*^{3}, I do not have any x^{2}, I have 4x and I have 1.0892

*All of that would represent my first polynomial there.*0905

*Below that I want to write the second polynomial but I want to lineup the terms, -3x*^{2} I will put it under the other x^{2}.0910

*I have a 6x I will put that under the other x and the 6 I will go ahead and put that with the 1.*0922

*You can now have things all nice and lined up.*0931

*Let us go ahead and add them one at a time.*0934

*1 + 6 = 7, -4 + 6x =2x, 0x*^{2} + -3x^{2} = -3x^{2}.0938

*One more, this has nothing to add to it so just 3x*^{2}.0961

*This would be our completed polynomial after adding the two together.*0967

*One last example is we are going to work on subtracting polynomials.*0977

*With these ones, what I suggest is being very careful with your signs. *0981

*You will see that I’m going to start with lining one on top of the other one.*0985

*But I’m going to end up distributing my negative signs, I will turn this into an addition problem.*0989

*Let us give it a try.*0995

*The polynomial on the left is 7y*^{2} -11y + 8 and right below that is -3y^{2} + 4y + 6.0996

*Now comes the important part, we are subtracting these polynomials so I will put a giant minus sign up front.*1015

*Before getting too far, I could go through and try and subtract this term by term, *1023

*but it is much easier to distribute my negative sign and just look at this like an addition problem. *1028

*My top polynomial will stay unchanged and we will leave that as it is.*1036

*After distributing the bottom, here is what we get negative × - 3y = 3y*^{2}, negative × 4y = -4y and for the very last one -6.1043

*We can take care of this as an addition problem and we know that the subtraction is taken care of because we put it into all of our terms.*1060

*8 + -6 = 2, -11y + -4y = -15y and then I have a 7y*^{2} + 3y^{2} = 10y^{2}.1071

*This final polynomial represents the two being subtracted.*1098

*Now you know a little bit more about polynomials and have put them together.*1105

*Remember that you are just combining your like terms. *1109

*Thank you for watching www.educator.com.*1112

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take care of multiplying polynomials.*0002

*Specifically we will look at the multiplication process in general, so you can apply that to many different situations.*0009

*We will look at some more specific things like how you multiply a monomial by any type of polynomial *0015

*and how can you multiply two binomials together.*0021

*I will also show you some special techniques like how you can organize all of this information into a nice handy table.*0026

*I will show you the nice way that you can multiply two binomials using the method of foil. *0033

*Watch for all of these things to play a part. *0038

*Now in order to multiply two polynomials together, what you are trying to make sure is that every term in the first polynomial gets multiplied*0044

*by every single term in the second polynomial.*0052

*That way you will know every single term gets multiplied by every other term. *0056

*If you have one of your polynomials being a monomial, it only has one term and this looks just like the distributive property.*0061

*Let us do one real quick so you can see that it is just the distributive property.*0071

*We are going to take to 2x*^{4} that monomial one term and multiplied by 3x^{2} + 2x – 5.0075

*We will take it and multiply it by all of these terms right here.*0083

*We will have 2x*^{4} × 3x^{2}, then we will have 2x^{4} × 2x, and 2x^{4} × -5.0091

*Each of these needs to be simplified, but it is not so bad.*0115

*You take the 2 and 3, multiply them together and get 6.*0119

*And then we will use our product rule to take care of the x*^{4} and x^{2} by adding their exponents together x^{6}.0124

*We will simply run down to all of the terms doing this one by one.*0134

* 2 × 2 =4, then we add the exponents on x*^{4} and x^{1} power = x^{5}.0137

*At the very end, 2 × -5 = -10 and we will just keep the x*^{4}.0147

*This represents our final polynomial after the two of them multiplied together.*0156

*Remember that we are looking so that every term in one is multiplied by every term in the other one. *0161

*What makes this a little bit more difficult is, of course, when you have more terms in your polynomials.*0171

*As long as you make sure that every term in one gets multiplied by every term in the other one should work out just fine. *0177

*Be very careful with this one and see how that turns out.*0183

*First, I’m going to take this first m*^{3} term and make sure it gets multiplied by all three of my other terms. 0187

*Let us put that out here.*0194

*I need to make sure m*^{3} gets multiplied by 2m^{2}.0198

*Then I will have m*^{3} × 4m and m^{3} × 3. 0206

*Now if I stop to there I would not quite have the entire multiplication process down. *0221

*We also want to take the -2m and multiply that by all 3.*0226

*Let us go ahead and put that in there as well.*0232

*We will take -2m × 2m*^{2}, then we will take another -2m × 4m and then -2m × 3.0234

*That is quite a bit but almost there. *0264

*Now you have to take the 1 and multiplied by all 3 as well. *0266

*1 × 2m*^{3}, 1 × 4m,1 × 3.0272

*That is a lot of work and we still have lots of simplifying to do, *0292

*but we made sure that everything got multiplied so now it is just matter of simplifying.*0295

*Let us take it bit by bit.*0301

*Starting way up here at the beginning I have m*^{3} × 2m^{2}, adding exponents that would be 2m^{5}.0302

*Now I have m*^{3} × 4m so add those exponents, and you will get 4m^{4}.0313

*Onto m*^{3} × 3 = 3m^{3} and now we continue down the list over here.0324

*-2m × 2m*^{2}, well 2 × -2 = -4 then add the exponents and I will get m^{3}, that will take care of that.0333

*We will go to this guy -2m × 4m = -8m*^{2}. 0345

*Now this -2m × 3 = -6m and that takes care of those.*0354

*Onto the last where we multiplied one by everything.,*0363

*Unfortunately, 1 × anything as itself we will have 2m*^{2}, 4m and 3.0366

*I have all of my terms and the resulting polynomial, but it still not done yet. *0376

*Now we have to combine our like terms.*0380

*Let us go through and see if we can highlight all the terms that are like.*0383

*I will start over here with 2m*^{5} and it looks like that is the only m^{5}, it has no other like terms to combine.0387

*We will go on to m*^{4}, let us see what do we got for that.0397

*I think that is the only one, so m*^{4}.0403

*3m*^{3} looks like I have a couple of m^{3}, I’m going to highlight those.0408

*I have some squares, I will highlight those.*0415

*Let us see what else do we have in here, it looks like we have single m’s.*0420

*There is that one and there is that one and there is a single 3 in the m.*0428

*We can combine all these bit by bit.*0433

*2m*^{5}, since it is the only one.0436

*4m*^{4}, since this the only one, let us check this after them.0440

*Now I have 2m*^{3}, 3 – 4=-1m^{3} and that will take care of those ones.0446

*-8m*^{2} + 2m = -6m^{2}.0459

*That one is done and that one is done. *0467

*-6m + 4m = -2m done and done.*0471

*And then we will just put our 3 in the end.*0478

*You can see it is quite a process when your polynomials get much bigger, *0483

*but it is possible to take every term and multiply it by every other term. *0487

*Now watch for later on how I will show you some special techniques to keep track of all of these terms that show up.*0492

*They will actually not be quite as bad as this one.*0498

*One way we can deal with much larger polynomials and keep track of all of those terms that multiplied together*0503

*is try and organize all of those terms in a useful way.*0508

*I’m going to show you two techniques that you can actually organize all that information. *0512

*One of them we will be using a table and another one we will look like more standard multiplication where you stack one on top of the other.*0517

*What I'm trying to with each of these methods is ensure that every term in one polynomial gets multiplied by every term in the other polynomials.*0524

*I’m not are changing the rule while we are doing a shortcut.*0531

*We are just organizing information in a better way.*0534

*No matter which method you use, make sure you do not forget to combine your like terms at the end so you can see the resulting polynomial.*0536

*Let us give it a try.*0543

*I want to multiply x*^{2} + 3x + 5 × x -4.0545

*The way I’m going to do this is first I’m going to write the first polynomial right on top of the second polynomial.*0550

*From there I’m going to start multiplying them term by term and I'm starting with that -4 in the bottom,*0566

*now multiply it by all the terms in that top polynomial.*0573

*Let us give it a try.*0580

*First I will do -4 × 5 = -20 then I will take a -4 × 3x = -12x and I have -4 × x*^{2} = -4x^{2}.0582

*That takes care of that -4 and make sure that it gets multiplied by all of the other terms.*0603

*We will do the same process with the x.*0609

*We will take it and we will multiply it by everything in that top polynomial.*0612

* x × 5 = 5x and I’m going to write that one right underneath the other x terms.*0618

*This will help me combine my like terms later.*0626

*x × 3x = 3x*^{2}.0629

*And one more x × x*^{2} = x^{3}.0635

*I have all of my terms it is a matter of adding them up and I will do it column by column. *0644

*This will ensure I get all of my like terms -20 - 7x - 1x*^{2} and at the very beginning x^{3}.0650

*That is my resulting polynomial. *0665

*Now another favorite way that I like to combine the terms of my polynomial is to use a table structure.*0669

*Watch how I set this one up.*0676

*First, along the top part of my table I'm going to write the terms of the first polynomial.*0679

*My terms are x*^{2}, 3x and 5. 0687

*Along the side of it I will write the terms of the other polynomials, so x, -4.*0697

*Now comes the fun part, we are going to fill in the boxes of this table by multiplying a row by a column.*0706

*In this first one we will take an x × x*^{2}.0713

*It feels like you are a completing some sort of word puzzle or something, only guesses would be a math puzzle.*0717

*Also x × x*^{2} = x^{3}.0722

*x × 3x = 3x*^{2} and x × 5 = 5x.0728

*It looks pretty good.*0738

*I will take the next row and do the same thing. *0739

*-4 × x*^{2} = -4x^{2}, - 4 × 3x = -12x and -4 × 5 = -20.0743

*You will get exactly the same terms that you do know using the other method in a different way of looking at them.*0757

*We need to go through and start combining our like terms.*0764

*Looking at my x*^{3} that is my only x^{3} so I will just write it all by itself.0768

*But I have a couple of x*^{2}'s so I will write both of those and combine them together, -4x^{2} + 3x =-x^{2}.0776

*Here I have -12x + 5x =- 7x and of course the last one -20.*0789

*Oftentimes you will find your like terms are diagonals from each other, but it is not always the case that seems to be very common.*0800

*A good important thing to recognize in the very end is that you get the same answer either way.*0807

*Use whichever method works the best for you, and that you are more comfortable with.*0813

*Now that we have some good methods and above, let us try multiplying these polynomials again and see how it is a little bit easier. *0821

*I will use my table method and we will take the terms of one polynomial write along the top. *0832

*I will take the terms of the second polynomial and write them alongside.*0843

*You will see this will go much quicker m*^{3} – 2m and 1, 2m^{2}, 4m and 3.0848

*Let us fill in the boxes.*0864

*2m*^{2} + m^{3} = 2m^{5}, 2 × -2 =-4m^{3}, 1 × 2m^{2} = 2m^{2}.0865

*On to the next row, 4m*^{4}, 4 × -2 = -8m^{2}, 4m × 1 = 4m.0880

*Last row, 3 × m*^{3} = 3m^{3}, 3 × -2 =-6m and 3 × 1 =3. 0896

*Let us go through and start combining everything.*0909

*I have a 2m*^{5} I will write that as our first term, 2m^{5}.0911

*I’m onto my 4m*^{4} and I think that is the only one I have floating around in there, 4m^{4}.0918

*We can call that one done.*0930

*3m*^{3} – 4m^{3}, two of those I need to combine, that will be -1m^{3}.0934

*I’m onto my squares, -8m*^{2} + 2m^{2} = -6m^{2}, -6m + 4m =-2m and the last number, 3.0949

*The great part is that it goes through and combines all of your like terms and I know I got them off because they are all circle.*0977

*I’m going to fix this -1, so it is just a - m*^{3} but other than that I will say that this is a good result right here.0984

*Some other nice techniques you can use to multiply polynomials together is if both of those polynomials happen to be binomials.*0995

*Remember that they have exactly two terms, this method is known as the method of foil.*1003

*That stands for a nice little saying it tells you to multiply the first terms together, the outside terms, the inside terms and the last terms.*1010

*It is a great way of helping you memorize and get all of those terms combined like they should.*1020

*It also saves you from creating a large structure like a table when you do not have to.*1027

*Let us see how it works with this one. *1032

*I have x -2 × x - 6 I’m going to take this bit by bit.*1035

*The first terms in each of these binomials would be the x and the other x.*1042

*Let us multiply those together and that would give us an x*^{2}.1047

*Then we will move on to the outside terms.*1056

*By outside that would be the x and -6 we will multiply those together, - 6x.*1060

*Continuing on, we are on inside terms, -2 and the x, they need to multiply together -2x.*1072

*And then our last terms -2 × -6 = 12. *1084

*We do get all of our terms by remembering first outside and inside last.*1096

*With this method, oftentimes your outside and inside terms will be like terms*1101

*and you will be able to combine them, and this one is no different. *1106

*They combined to be 8x.*1108

*Once you have all of your terms feel free to write them out. *1112

*This is x*^{2} - 8x – 4 + 12 and the more you use this method, it will come in handy for a factoring a little bit later on.1115

*Let us try out our foil method as we go through some of these examples.*1129

*Here I want to multiply the following binomials, 5x - 6 × 2y + 3.*1134

*First, I'm going to take the first terms together that will be the 5x and 2y, 5 × 2 = 10x × y.*1142

*That is as far as I can put those together since they are not like terms.*1154

*Outside terms that would be 5x and the 3 = 15x.*1159

*Onto inside terms, -6 × 2y = 12y and the last terms -6 × 3 = -18.*1168

*We got our first outside, inside, last and it looks like none of these are like terms.*1186

*I will just write them as they are 10xy + 15x -12y – 18 and we will call this one done. *1190

*Let us try another one, and in this one you will see it has few more things that we can combine.*1206

*We are going to multiply - 4y + x and all of that will be multiple by 2y -3x.*1211

*Starting off with our first terms let me highlight them.*1218

*- 4y × 2y = -8 and y × y = y*^{2}.1223

*That is the case here of our first terms.*1232

*Now we will do our outside terms, -4 × -3 = 12 and x × y.*1235

*Onto the inside terms, x × 2y = 2xy.*1250

*Of course our last terms, -3x*^{2}.1261

*Now that we have all of our terms notice how our outside and inside terms, they happen to be like terms so we will put them together.*1273

*That will give us our final polynomial, 8y*^{2} + 14xy - 3x^{2}.1280

*We can say that this one is done.*1294

*One more example and this one is a little bit larger one.*1303

*In fact, the second polynomial in here is a trinomial so we will not be able to use the method of foil.*1307

*That is okay, we will still be able to multiply it together, *1315

*but I will definitely use something like a table to help me organize my information a little bit better. *1318

*Okay, along the top of this table, let us go ahead and write our first polynomial, x – 5y.*1330

*Then along the rows we will put our second polynomial, I have an x*^{2} – 2xy and 3y^{2}.1340

*Here comes the fun part, just fill in all of those blanks by multiplying a row and a column.*1358

*x*^{2} × x =x^{3}, x^{2} × -5y = -5x^{2}y.1364

*Onto the next row, -2xy × x, the x’s we can put those together as an x*^{2} and the a y.1378

*The last part here -2xy × -5y, let us put the y’s together, -2 × -5 =10xy*^{2}.1389

*One more row, 3y*^{2} × x =3xy^{2}.1401

*I have 3y*^{2} × -5y -15y^{3}.1412

*We have all of our terms in there, now we need to combine the like terms.*1422

*Let us start here on the upper corner.*1427

*If we have any single x*^{3} that we can put with this one.1429

*It look like it is all by its lonesome, we will just say x*^{3}.1435

*We are looking for x*^{2}y, they must have x^{2} and they must have y, I think I see two of them, here is one and here is that other one.1441

*Let us put these together, -2 + -5 = -7 and they are x*^{2}y terms.1453

*Continuing on, I have an xy*^{2}.1464

*I have two of those so let us put them together, we will take this one and we will take that one, 10 + 3= 13xy*^{2}.1468

*That takes care of those terms.*1482

*One more -15y*^{3}, I will put it in -15y^{3}.1483

*Now I have the entire polynomial.*1491

*Remember, at its core when you multiply polynomials you just have to make sure that every term gets multiplied by every other term.*1493

*Use these techniques such as foil or a table to help you organize all of those terms.*1500

*Thank you for watching www.educator.com.*1506

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take care of dividing polynomials. *0003

*I like to break this down into two different parts. *0009

*First, we will look at the division process when you have a polynomial divided by a monomial.*0013

*We will look at what happens when you divide any two polynomials together.*0017

*In the very end, I will also show you a very special technique to make the division process nice and clean. *0022

*To get into the basics of understanding the division of polynomials, I like to take it back to looking at the division process for real-world fractions. *0031

*Suppose that when you were adding fractions together you know how that process will go.*0042

*One very important thing that you do when adding fractions is you would find a common denominator. *0047

*Once you have a common denominator, then you can go ahead and just combine the tops of those fractions together.*0053

*Think of a quick example like 2/3 and are looking to add 5/3.*0060

*Since they have exactly the same bottoms then you will only add the 2 and 5 together and get 7/3 as your result.*0067

*Since this has a giant equal sign in between it, it means you can also follow this process in the other direction.*0074

*This may look a little unfamiliar, but it does work out if you go the other way. *0080

*Suppose I had A + B and all of that was dividing by C.*0085

*The way I could look at this is that both the A and B are being divided by C separately.*0090

*This is the same equation I had earlier, I just turned it around. *0097

*The reason why I show this is this will help us understand what happens when we have a binomial divided by a monomial.*0102

*In fact, that is the example that I have written out.*0108

*The top is a binomial and the bottom is an example of a monomial.*0112

*You can see that the way we handle it is we split up that monomial under each of the different parts of the polynomial in the top.*0121

*Let us see this process with numbers and see what polynomials and you will get an idea of how this works.*0134

*Working the other direction, if I see (2 + 5) ÷ 3, I want to visualize that as the 2 ÷ 3 and also the 5 ÷ 3.*0140

*For our polynomials if I have something like (x + 3z) ÷ 2y, then I will put that 2y under both of the parts.*0152

*In addition, you will notice splitting up over both of the parts in the top, you always want to make sure that you simplify further, if possible. *0162

*This means if you use your quotient rule for exponents, go ahead and do reduce those powers as much as possible. *0172

*If you have just any two general polynomials, then you want to think of how the process works with numbers.*0183

*In fact we are going to go over a long division process so that we can actually keep track of all the parts of what goes into what.*0189

*To make this process easier, remember to write your polynomial in descending power.*0199

*Start with the largest power and write it all the way down to the smallest power. *0203

*An additional thing that will also help in the division process is to make sure you put placeholders for all the missing variables.*0214

*If I’m looking at a polynomial like 5x*^{2} + 1, then I will end up writing it with a placeholder for the missing x.0221

*It is not missing but it will help me keep track of where just my x terms go.*0233

*I'm going to show you this process a little bit later on*0239

*so you can see how it works with numbers and make some good parallels too doing this with polynomials.*0242

*Let us look at the division process for numbers.*0251

*Suppose I gave you 8494 and I told you to divide it by 3 and furthermore I said okay, let us see if you can do this by hand.*0254

*Some good news that you will probably tell me is that you do not have to take the 3 into that entire number all at once. *0263

*No, you will just take the 3 into 8494 bit by bit.*0269

*In fact, the first thing that you will look at is how many times this 3 go into the number 8.*0274

*That will be the only thing you are worried about.*0279

*How many times does the 3 go into 8?*0282

*It goes in there twice and you will write that number on the top. *0285

*Now after you have that number on the top, you do not leave it up there you go through a multiplication process, *0290

*2 × 3 and you will write the result right underneath the 8.*0296

*With the new number on the bottom, you will go ahead and subtract it away, so 8-6 would give you a 2. *0305

*That would be like one step of the whole division process.*0315

*You would continue on with the division process by bringing down more terms and doing the process again.*0319

*At this next stage, we will say okay how many times this 3 go into 24?*0329

*It goes in there 8 times.*0335

*Then we could multiply the 8 and 3 together and get a new number for that 3 × 8 = 24.*0338

*We can go ahead and subtract those away.*0352

*We will not stop there, keep bringing down our other terms and see how many times 3 goes into 9.*0359

*Write it onto the top and multiply it by your number out front.*0369

*Subtract it away and continue the process until you have exhausted the number you are trying to divide.*0377

*Let us see.*0389

*3 goes into 4, it looks like it goes in there once.*0390

*I will get 3, subtract them away and I have remainder of 1.*0396

*That is a lengthy process but notice the Q components in there.*0401

*You are only dividing the number bit by bit.*0405

*You do not have to take care of it all at once.*0408

*The way you take care of it is you are saying how many times 3 goes into that leading number.*0410

*You write it on the top, you go through the multiplication process and then you subtract it away from the number.*0415

*You will see all of those same components when we get into polynomials.*0423

*We have our answer and I could say it in many different ways but I’m going to write it out.*0429

*8494 if we divide this by 3 is equal to 2831 and it has a remainder down here of 1.*0434

*We could say +1 and still being divided by 3.*0446

*We have many different parts in here that are flying around, and you want to keep track of what these parts are.*0451

*The part underneath your division bar is the dividend. *0458

*What you are throwing in there this is your divisor.*0466

*Your answer would be your quotient.*0472

*This guy down here is our remainder. *0477

*Notice how those same parts actually show up in our answer.*0483

*Dividend, divisor, quotient, and remainder.*0488

*We put the remainder over the divisor because it is still being divided.*0506

*Now that we brushed up on the process with numbers, let us take a look at how we do this with polynomials. *0510

*I want to divide 2x*^{2} + 10x + 12 and divide that by x +3.0519

*The good news is we do not have to take care of the entire polynomial all at once.*0526

*We are going to take it in bits and pieces.*0530

*We will first going to look at x and see how many times it will go into 2x*^{2}.0532

*In fact one thing that I can do to help out the process is think to myself what would I have to multiply x by in order to get a 2x*^{2}.0537

*1 × what would equal to 2? That would have to be 2.*0548

*What would I have to multiply x by to get an x*^{2}?0552

*I have to multiply it by x.*0555

*I will put that on top, just like I did with numbers.*0557

*After I do that, we will run through a multiplication process.*0563

*We will take the 2x to multiply it by x, I will multiply it by 3.*0568

*We will record this new polynomial right underneath the other one.*0573

*2x × x = 2x*^{2}.0577

*2x × 3 = 6x.*0581

*Now comes a very important step.*0589

*Now that we have this new one, we want to subtract it away from the original.*0592

*Notice how I put those parentheses on there, that will help me remember that I need to subtract away both parts and keep my signs straight.*0599

*Starting over here, I have 10x – 6x = 4x.*0606

*Then I have 2x*^{2} – 2x^{2}, that will cancel out and will give me 0x^{2}.0616

*If you do this process correctly, these should always cancel out.*0622

*If they do not cancel out, it means we need to choose a new number up here.*0628

*That is just one step of the division process, let us bring down our other terms and try this one more time.*0635

*I want to figure out how many does x goes into 4x?*0648

*What would I have to multiply x by in order to get 4x?*0652

*I think I have to multiply it by 4 that is the only way it is going to work out.*0658

*Now we have the 4, go ahead and multiply it by the numbers out front.*0663

*4 × x = 4x and 4 × 3 = 12.*0669

*Once you have them, put on a giant pair of parenthesis and we will go ahead and subtract it away.*0678

*12 – 12 =0 and 4x – 4x = 0.*0687

*What that shows is that there is no remainder and that it went evenly.*0693

*Let us write this out.*0699

*When I had 2x*^{2} + 10x + 12 and I divided it by x + 3, the result was 2x + 4.0699

*There was no remainder.*0714

*We can label these parts as well.*0717

*We have our dividend, divisor, quotient, and if I did have a remainder I will probably put it out here.*0720

*Here is our dividend.*0739

*Here is our divisor and our quotient.*0744

*Now that we know a lot more about dividing polynomials, let us look at a bunch of examples.*0758

*Some of them will take a polynomial divided by a monomial.*0764

*Some of them will take two polynomials and divide them.*0767

*We will approach both of those cases in two different ways.*0770

*Example 1, divide a polynomial by a monomial.*0774

*I will take (50m*^{4} -30m^{3} + 20m) ÷ 10m^{3}.0777

*Since we are dividing by a monomial, I will take each of my terms and put them over what I’m dividing them by.*0784

*50m*^{4} ÷ 10m^{3}, 30m^{3} ÷10m^{3} and 20m ÷ 10m^{3}.0792

*Now that I have done that, I will go through and simplify these one at a time.*0811

*50 ÷ 10 = 5, m*^{4} ÷ m^{3} = I can use my quotient rule and simply subtract the exponents and get m^{1}.0816

*Continuing on 30 ÷ 10 = 3, if I subtract my exponents for m*^{3} and m^{3}, I have m^{0}.0830

*Onto the last one, 20m ÷ 10m*^{3}, there is 2.0844

*Let us see, if I subtract the exponents I will m*^{-2}.0852

*Then I can go through and just clean this up a little bit.*0857

*5m – anything to the 0 power is 1, 1 × 3 + and I will write this using positive exponents, 2/m*^{2}.0860

*This will be the final result of dividing my polynomial by a monomial.*0872

*Let us try this scheme with something a little bit more complicated. *0881

*This one is (45x*^{4} y^{3} + 30x^{3} y^{2} - 60x^{2} y) ÷ 50x^{2} y.0884

*We have to put our thinking caps for this one.*0894

*We will start off by taking all of our terms in the top polynomial and putting them over our monomial, only one term.*0897

*Once we have this all written out we simply have to simplify them one at a time.*0917

*Let us start at the very beginning.*0925

*15 goes into 45 3 times, now I will simplify each of my variables using the quotient rule.*0929

*4 – 2 =2 and 3 – 1= y*^{2}.0939

*There is my first term, moving on.*0946

*15 goes into 30 twice.*0950

*Using my quotient rule on the x, 3 – 2 = 1 and y*^{2} – y^{1} = y^{1}.0955

*Both of these have an exponent of 1 and I do not need to write it.*0964

*One more, 15 goes into 60 four times.*0969

*I have x*^{2}/x^{2} which will be x^{0} and y/y will be y^{0}.0975

*Anything to the 0 power is 1.*0983

*I can end up rewriting that term 3x2 y*^{2} + 2xy – 4.0987

*There is my resulting polynomial.*1000

*In this next example, we are going to take one polynomial divided by another polynomial.*1007

*I would not be able to split it up quite like I did before, now we will have to go through that long division process.*1012

*Be careful you want to make sure that you line up your terms in descending order. *1018

*If you look at the powers of the top polynomial you would not mix up.*1024

*I want to start with that 3rd power then go to the 2nd power, then go to the 1st power, just to make sure I have it all lined up.*1027

*Let us write it out.*1034

*2x*^{3} + x^{2} + 5x + 13, now it is in a much better order to take care of.1035

*We will take all of that and we will divide it by 2x + 3.*1048

*That looks good.*1055

*It is time to get into the division process.*1057

*Our first terms there, and what would I need to multiply 2x by in order to get 2x*^{3}?1060

*The only thing that will work would be an x*^{2}.1068

*Once we find our numbers up top, we will multiply them by the polynomial out front.*1072

*2x × x*^{2} = 2x^{3}, that is a good sign, it is the same as the number above it.1079

*X*^{2} × 3 =3x^{2}, now comes the part that is tricky to remember.1088

*Always subtract this away and do not be afraid to use this parenthesis to help you remember that.*1097

*x*^{2} – 3x^{2}, 1 – 3 = -2x^{2}.1106

*Then I have 2x*^{3} – 2x^{3} and those will be gone, cancel out like they should.1116

*The first iteration of this thing looks pretty good. *1122

*Let us try another one.*1126

*I want to figure out how may times does 2x go into -2x*^{2}?1129

*I have to multiply it by –x.*1137

*Let us bring down some more terms.*1146

*I will get onto our multiplication process.*1149

*-x × 2x =-2x*^{2}.1153

*-x × 3 =-3x.*1158

*We have our terms in there, it is time to subtract it away and be careful with all of our signs.*1164

*5x - -3x when we subtract a negative that is the same as addition.*1171

*I’m looking at 5x + 3x = 8x.*1179

*-2x*^{2} - -2x^{2} = that is a lot of minus signs.1188

*That would be the same as -2x + 2x*^{2} and they do cancel out like they should.1193

*That was pretty tricky keeping track of all those signs but we did just fine.*1201

*We have 8x + 13 and I’m trying to figure out what would I have to multiply 2x by in order to get that 8x.*1207

*There is only one thing I can do I need to multiply by 4.*1218

*We will multiply everything through and see what we get.*1224

*4 × 2x = 8x, 4 × 3 =12.*1228

*I can subtract this away and let us see what we get.*1238

*13 – 12=1, 8x – 8x = 0.*1245

*Here I have a remainder of 1.*1250

*I could end up writing my quotient and I can put my remainder over the divisor.*1257

*There is our answer.*1269

*In some polynomials you want to make sure you put in those placeholders.*1275

*That way everything lines up and works out good.*1280

*It is especially what we will have to do for some of those missing powers in this one. *1283

*I’m missing an x*^{2} and a single x.1287

*Let us write this side and put those in.*1291

*x*^{3}, I have no x^{2}, no x – 8.1294

*All of that is being divided by x – 2.*1305

*I’m going to go through and let us see what I need to multiply x by in order to get x*^{3}.1310

*I think I’m going to need x*^{2}.1318

*Now that I found it, I will go ahead and multiply it through.*1323

*x*^{2} × x =x^{3} and x^{2} × -2 = -2x^{2}.1327

*We have our terms, let us go ahead and subtract it away.*1336

*0x*^{2} – 2x^{2}, this is one of those situations where if we subtract a negative is the same as adding.1345

*0x*^{2} + 2x^{2} =2x^{2}.1354

*x*^{3} – x^{3}, that is completely gone.1360

*Bring down our other term here and we will keep going.*1366

*What would I have to multiply x by in order to get 2x*^{2}?1374

*I’m going to need 2x.*1381

*Let us multiply that through, 2x × x =2x*^{2}.1386

*2x × -2 =-4x.*1392

*Now we found that, let us subtract that away.*1399

*Starting on the end, 0 – -4, that is the same as 0 + 4x = 4x.*1405

*Then 2x*^{2} – 2x^{2}, they are completely gone, you do not have to worry about it.1415

*We will bring down our -8 and continue.*1422

*What would I have to multiply x by in order to get 4x?*1428

*That will have to be 4.*1433

*4 × x = 4x and 4 × -2 = -8.*1437

*It looks like it is exactly the same as the polynomial above it.*1447

*I know when I subtract, I would get 0.*1452

*There is no remainder for this one.*1454

*I will take (x*^{3} – 8) ÷ x -2 and the result is x^{2} + 2x + 4.1457

*Division process can take a bit but as long as you do the steps very carefully, you should turn out okay.*1467

*Let us try this giant one.*1476

*This is (2m*^{5} + m^{4} + 6m^{3} – 3m^{2} – 18) ÷ m^{2} + 3.1478

*There are a lot of things to consider in here.*1489

*One thing that I will be careful of is putting those placeholders for this guy down here.*1493

*Notice that it is missing an m, let us give it a try.*1498

*I will have m*^{2} + 0m + 3 and all of that is going into our other polynomial 2m^{5} + m^{4} + 6m^{3} – 3m^{2} – 18.1501

*Lots of things to keep track of but I think we will be okay.*1523

*What would I have to multiply the m*^{2} by in order to get a 2m^{5}?1527

*That would be 2m*^{3}.1535

*I can run through the multiplication and write it here.*1539

*2m*^{3} × m^{2} = 2m^{5}.1543

*Now we can multiply it by our 0 placeholder but anything times 0 will give us 0 so 0m*^{4}.1549

*By my last one, let us see 2m*^{3}× 3 + 6m^{3}.1558

*Let us subtract that away.*1568

*6m*^{3} – 6m^{3} those are gone.1572

*I have m*^{4} – 0m^{4} I still have m^{4}.1575

*2m*^{5} – 2m^{5} those are gone.1580

*I dropped away quite a bit of terms.*1584

*Let me go ahead and write in my 0m*^{3} as one of those placeholders so I can keep track of it.1587

*Let us try this again.*1598

*m*^{2} goes into m^{4}, if I multiply it by another m^{2}.1600

*Multiplying through I have m*^{2} × m^{2} = m^{4}.1609

*m*^{2} × 0m = 0m^{3} and m^{2} × 3 = 3m^{2}.1614

*We will take that and subtract it away.*1626

*I need to go ahead and subtract these.*1636

*Be very careful on the signs of this one.*1638

*I have –m*^{2} and I’m subtracting -3m^{2}, the result here will be -6m^{2}.1641

*The reason why it is happening is because of that negative sign out there.*1652

*Now I have 0m*^{3} – 0m^{3}, 0 – 0 =0, m^{4} – m^{4} = 0.1659

*It looks like I forgot an extra placeholder.*1669

*I need 0m and then I need my 18.*1676

*Let us bring down both of these.*1684

*I need to figure out what I have to what would I have to multiply m*^{2} by in order to get -6m^{2}.1691

*-6 will do it, -6m*^{2} 6 × 0 = 0m and -6 × 3 =-18.1699

*It is exactly the same as the polynomial above it.*1718

*Since they are exactly the same and I’m subtracting one from the other one, 0 is the answer.*1723

*There is no remainder, it went evenly.*1729

*The quotient for this one would be 2m*^{3} + m^{2} – 6.1732

*In this polynomial, I have (3x*^{3} + 7x^{2} + 7x + 11) ÷ 3x + 6.1744

*The reason why I put this one is because it can get a little bit difficult figuring out what you need to multiply to get into that second polynomial.*1753

*I’m going to warn you, this involves a few fractions.*1762

*Let us give it a shot.*1767

*(3x*^{3} + 7x^{2} + 7x + 11) ÷ 3x + 6.1770

*Let us start off at the very beginning.*1788

*What do I need to multiply my 3x by in order to get 3x*^{3}?1790

*The only thing that will work will be an x*^{2}.1797

*I will go through and I will multiply and get the result.*1804

*3x*^{3} + 6x^{2} and let us subtract that away.1809

*7x*^{2} – 6x^{2} = 1x^{2}.1821

*Now comes the tricky part, I need to figure out what I need to multiply 3x by in order to get x*^{2}.1837

*If I’m looking at just the variable part of this, I have to multiply and x by another x in order to get an x*^{2}.1845

*We will go ahead and put that as part of our quotient.*1851

*What do I have to multiply 3 by in order to get 1 out front?*1854

*That is a little bit trickier.*1859

*3 × what = 1.*1861

*That is almost like an equation onto itself.*1865

*What we see is that x would have 1/3, a fraction.*1869

*It is okay, we can use fractions and end up multiplying by those.*1874

*3x × 1/3x = 1x*^{2}.1881

*Let us multiply that through.*1887

*1/3x × 3x = 1x*^{2} I will write it down and 1/3x × 6 = 2x.1888

*Now we can take that and subtract it away.*1900

*7x – 2x = 5x.*1904

*Bringing down our extra terms and I think this one is almost done.*1913

*What would I have to multiply 3x to get 5x?*1917

*Let us see.*1922

*he only way I can get an x into another x is to multiply by 1, but I’m going to think of how do I get 3 and turn it into 5?*1924

*Let us do a little bit of scratch work on this one.*1933

*3 × what = 5?*1939

*If we divide both sides by 3 I think we can figure out it is 5/3.*1945

*That I can write on top 5/3.*1951

*We can go through multiplying.*1956

*5/3 × 3 = 5x.*1958

*5/3 × 6 =10.*1966

*We will go ahead and subtract this away.*1977

*11 – 10 = 1 and 5x – 5x = 0.*1980

*We have a remainder of 1.*1985

*Now that we have all of the quotient and the remainder, let us go ahead and write it out.*1993

*We have (x*^{2} + 1/3x + 5/3 + 1) ÷ 3x + 6.1998

*Definitely do not be afraid some of those fractions to make sure it goes into that second polynomial.*2010

*This process can get a little messy as you can definitely see from those examples.*2018

*I have a nice clean way that you can go through the division process known as synthetic division.*2022

*This is a much cleaner way for the division process so that you can keep track of all the variables.*2027

*It is much clean but be very careful in how you approach this. *2033

*It works good when dividing by polynomials of the form x + or – number.*2037

* It will work especially with my little example right here (5x*^{3} - 6x^{2} +8) ÷ x -4.2043

*First watch how I will set this up. *2051

*I'm going to create like a little upside down division bar and that is where I will end up putting the polynomial that I'm dividing.*2054

*But I will not put the entire thing I’m only going to put the coefficients of all of the terms.*2062

*The coefficient of the x*^{3} is a 5. 2069

*The coefficient of my x*^{2} is -6.2073

*I will put a 0 placeholder in for my missing x and then my last coefficient will be 8.*2077

*Once I have all of those I will put another little line. *2086

*I want to put in the value of x that would make this entire polynomial 0, if x was 4 that would be 0.*2090

*I’m going to write 4 out here.*2105

*That turns to be a tricky issue and many students remember what to put out over here because it will be the opposite of this one.*2109

*If you see x – 4 put in a 4.*2118

*If you see something like x + 7 then put in -7.*2121

*We got that all set let us go through this synthetic division process.*2127

*It tends to be quick watch very carefully how this works.*2131

*The very first thing that you do in the synthetic division process is you take the first number here and you simply copy it down below.*2137

*This will be a 5.*2146

*Once you get that new number on the bottom, go ahead and multiply it by your number out front, 4×5 = 20.*2149

*That is one step of the synthetic division process.*2161

*To continue from there simply add the column -6 + 20 and get the result. *2165

*This would be a 14.*2173

*Once you have that feel free to multiply it out front again.*2177

*14 × 4 = 56.*2181

* There are 2 steps now we will take the 56 and we will add 0, 56.*2191

*When we get our new number on the bottom, go ahead and multiply it right out front.*2200

*4 × 50 = 200.*2205

*4 × 6 = 24, 224.*2212

*One last part to this we got to do some addition.*2222

*8 + 224 = 232.*2225

*It does not look like we did much of any type of division. *2234

*We did a lot of adding and we did a lot of multiplying but do you know what these new numbers stand for on the bottom.*2236

*That is the neat part, these new numbers I have here in green stand for the coefficients of our result.*2243

*You know what happens after the division.*2249

*The way you interpret these is the last number in this list will always be your remainder.*2252

*I know that my remainder is 232.*2260

*As for the rest of the values, the 5, 14, and 56, those are the coefficients on our variables. *2264

*What should they be? Let me show you how you can figure that out.*2270

*Originally we had x*^{3} as the polynomial that we are dividing and these new ones will be exactly one less in power.2274

*That 5 goes with 5x*^{2} and the 14 goes with the 14x and 56 has no x on it. 2282

*The result for this one is 5x*^{2} + 14x + 56 with a remainder of 232 which you can write over x – 4.2293

*Since it is still being divided*2312

*It is a much cleaner and faster method for division.*2315

*Let us go ahead and practice it a few times just to make sure got it down.*2318

*We will go ahead and do (10x*^{4} - 50x^{3} – 800) ÷ x – 6.2325

*It is quite a large problem.*2332

*We will definitely remember to put in some of those placeholders to keep track of everything.*2334

*First I will write in all of the coefficients of my original polynomial.*2339

*I have a 10x*^{4} - 50x^{3} I need to put in a placeholder for my x^{2} and another placeholder for my x.2344

*We will go ahead and put in that -800.*2356

*What shall we put on the other side?*2364

*Since I'm dividing by x – 6, the value of 6 will be divided that would make that 0.*2367

*I will use 6 and now onto the synthetic division process.*2376

*The first part we will drop down to 10, just as it is.*2381

*Then we will multiply by the 6 out front 60.*2387

*Now that we have that, let us add the -50 and the 60 together, 10 again.*2396

*We will multiply this out front.*2405

*That result will be 60.*2409

*We will go ahead and add 0 + 60 = 60 and multiply 60 × 6= 360.*2417

*Now we will add 0 + 360 = 360.*2430

*I have to take 360 × 6.*2435

* think I have to do a little bit of scratch work for that one.*2442

*6 × 0 = 6 × 6 = 36 and then 3 × 6 = 18 + 3 = 21.*2443

*I have 2160, let us put it in.*2455

*Only one last thing to do is we need to add -800 to the 2160 and then let us do a little bit of scratch work to take care of that.*2465

*0 – 0= 0, 6 – 0= 6, 20 and we will breakdown and will say 11 – 8 =3.*2478

*I have 1, I have 1360.*2489

*Now comes the fun part, we have to interpret exactly what this means.*2493

*Keep in mind that this last one out here that is our remainder.*2498

*Originally our polynomial was x*^{4} so we will start with x^{3} in our result.2505

*10 x*^{3} + 10x^{2} + 60x + 360 and then we have our remainder 1360.2513

*It is all still being divided by an x – 6.*2532

*That was quite a bit of work, but it was a lot clean than going through the long division process. *2536

*Let us see this one more time.*2540

*In this last one we will use (5 - 3x + 2x*^{2} – x^{3}) ÷ x + 1.2545

*We will first go ahead and put the coefficients of our top polynomial in descending order.*2555

*Be very careful as you set this one up.*2560

*On that side you can write out the polynomial first in descending order and then go ahead and grab its coefficients.*2564

*-x*^{3} would be the largest, then I have 2x^{2} and then I have 3x, and 5.2570

*I need -1, 2, -3 and 5.*2579

*I’m dividing by x + 1 so the number I will use off to the left will be -1.*2589

*I think we have it all set up now let us run through that process.*2597

*The first I’m going to bring down is -1 then we will go ahead and multiply.*2600

*Negative × negative is positive.*2608

*2 + 1 =3, 3 × -1= -3.*2615

*-3 + -3 = - 6, -6 × -1 = 6 and 5 + 6 =11.*2630

*Now we have our remainder, we can finally write down the resulting polynomial.*2647

*We started with x*^{3} so I know this would be x^{2}.2657

*-x*^{2} + 3x – 6 and I still have my 11 being divided by x + 1.2662

*Now you know pretty much everything that there is to know about dividing polynomials.*2676

*If you divide by a monomial, make sure you split it out among all the terms.*2680

*If you divide a monomial by a polynomial, you can go through the long division process *2685

*or use this synthetic division process to make it nice and clean.*2690

*Thank you for watching www.educator.com.*2694

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at the greatest common factor and the technique of factor by grouping.*0002

*First we will learn how to recognize a greatest common factor for a list of different terms*0012

*and how we can actually factor out that greatest common factor when you have something like a polynomial. *0017

*This will lead directly into the technique of factor by grouping, a good way of breaking down a large polynomial.*0023

*You are going to hear me use that word factor quite a bit in this entire section.*0032

*You are probably a little curious about what exactly factoring means.*0036

*Technically it means to write a quantity as a product.*0043

*We will be breaking down into pieces that are multiplied.*0047

*In a more practical sense, the way I like to think of factoring is that you are ripping things apart.*0051

*It is almost like doing the opposite of multiplication. *0056

*For example if I have 2 × 3 I could put them together and get 6. *0059

*When I have the number 6 then I can factor it down into those individual pieces again 2 × 3.*0066

*It is like multiplication but we are going in the reverse direction. *0072

*Now some expressions and numbers could have many different factors.*0077

*Let us take the number 12. *0082

*If you break that one down, you could look at it as 2× 6 and you could continue in breaking down the 6 and 2 × 3.*0085

*You have factors like 2 would go in there, 4 would go into 12, 3 would be a factor, 6 would be factor.*0095

*All those things could go into 12.*0103

*The greatest common factor will be the largest number that will divide into all the numbers present.*0110

*I will show you this to you twice just to get a better feel for it.*0116

*Suppose I'm looking at 50 and 75, the greatest common factor would be the largest number that could divide into both of those evenly.*0120

*5 could divide it into both of those evenly and that would be a pretty good choice, but is not the largest thing that could go in there.*0128

*The largest thing that could go in would be the number 25.*0136

*You will see that if I do take them both and divide them by 25, true enough they both go in there evenly.*0140

*25 is my greatest common factor.*0150

*We may only do this for pairs but you could do it with an entire list of numbers.*0155

*In this one I have 12, 18, 26, 32, and again we are looking for the largest thing that divides into all of that. *0162

*Sometimes a good way to find the largest thing that goes into all of them is just find something that works, *0169

*maybe something like 2 and end up reducing them bit by bit.*0177

*If I divide everything in here by 2 I would get 6, 9, 13 and 16.*0183

*Nothing else will go into all of those since 13 is a prime number.*0191

*My greatest common factor is 2 in this case.*0197

*It is not a very big number but it happens to be the largest in there that will go into all for these numbers evenly.*0201

*In algebra we are not interested in just single numbers we also want to see this process work out when we have lots of variables.*0209

*Let us see how this works out. *0216

*Maybe I’m looking at the numbers over the terms -16r*^{9}, -10r^{15}, 8r^{12}.0218

*To find the greatest common factor, I’m first going to look at those numbers and see if there is a large number that could go into all of them. *0226

*I could divide all of them by 2, that would work but actually I think 2 is the largest one.*0234

*I will say that 2 will divide it into all of these evenly.*0242

*Divided by 2, divided by 2, divided by 2 and I would get -8, -5, 4.*0248

*We can also do this thinking about our variables. *0258

*What was the largest variable raise to the power that we can divide into all of them?*0261

*Think of using your quotient rule to help out.*0266

*With this one r*^{9} is the greatest thing that I can divide all of them by.0272

*I would not have to deal with negative exponents or anything like that.*0279

*Let us write that over here, r*^{9}.0284

*r*^{9} ÷ r^{9} would be a single r^{0} or 1.0288

*r*^{15} ÷ r^{9} = r^{6} and r^{12} ÷ r^{9} = r^{3}.0293

*Our greatest common factor would be 2r*^{9}.0303

*Be careful when you are dealing with those variables and exponents.*0307

*Sometimes you will hear greatest common factor in your mind and you think you should go after the greatest exponent.*0310

*But as we can see in this example, it is not the case. *0315

*We actually took the smallest exponent because it was simply the greatest thing that could go into all of them.*0319

*Put that in mind and let us try this again.*0327

*In this one I have nothing but variables.*0330

*h*^{4} k^{6} h^{3} k^{6} and h^{9} k^{2}.0332

*What is the largest power of h that could go into all of them?*0340

*It is going to be h*^{3}.0345

*I picked up on that because I can see that looking in the middle one it is the smallest power.*0349

*Looking at all the k’s, the largest exponent of k that could go into all them would be k*^{2}.0354

*My greatest common factor will be h*^{3} k^{2}.0364

*Let us see what this will end up reducing down to if I take that greatest common factor.*0373

*h*^{1} k^{4} h^{3} and h^{3} will cancel each other out.0378

*k*^{4} I have h^{6} and I have no k that will end up cancelling each other out.0388

*Being able to recognize the greatest common factor will help us in the next process.*0396

*Make sure it is pretty solid. *0400

*If you can recognize the greatest common factor you can often factor it out of a polynomial. *0405

*What I mean by factoring out is we are going to place it out front of a pair of parentheses *0413

*and put the reduced terms inside of that parentheses .*0418

*3m + 12 this is like looking at 3m and looking at 12, what is the greatest common factor of those two terms?*0423

*What could potentially go into both of them evenly and what is the largest thing that will do that?*0431

*The number 3 will go into 3m and the number 3 will also go into 12.*0438

*3 will be my greatest common factor.*0446

*I’m going to write that out front on the outside of the parenthesis.*0449

*What I write on the inside, what happens when I take 3m ÷ 3?*0453

*I will get just single m and when I take 12 ÷ 3 I will get 4.*0458

*Here I have taken a polynomial and essentially factored it into a 3 and into another polynomial m + 4.*0467

*The neat part about this is since factoring is like the opposite of multiplication you can check this by running through the distribution process.*0475

*If you take 3 and you put it back in here do you get the original answer?*0487

*You will see that in fact you do get 3m + 12.*0492

*That is how you know this has been factored correctly.*0496

*In some instances, just like with numbers when you are looking for the greatest common factor*0501

*it turns out you would not be able to find greatest common factor to pull from all of the terms or the largest thing will actually be 1 or itself.*0506

*In cases like that, we say that the polynomial is prime.*0513

*In other words it does not break down into any other pieces.*0517

*Let me give you a quick example of a prime polynomial 5x + 7.*0520

*There is not a number that goes into 5 and into 7 so I would consider that one prime.*0526

*How about 8x + 9y?*0534

*Individually those numbers are divisible, but there is not a number that goes in the 8 and 9 it would be a prime polynomial.*0542

*Let us work on factoring out the greatest common factor from various different polynomials, *0555

*just so we get lots of good practice with it.*0560

*Starting with the first one I have 6x*^{4} + 12x^{2}.0562

*Let us hunt down those numbers first. *0567

*The largest number that would go into both 6 and 12 would be 6.*0570

*Let us look at those x.*0578

*What is the largest power of x that would divide into both of them?*0581

*That have to be x*^{2}.0586

*We will write in the leftovers inside our parentheses 6x*^{4} ÷ 6x^{2}=x^{2} and 12x^{2} ÷ 6x^{2} = 2.0592

*I factored out the greatest common factor for that polynomial.*0611

*Let us give this another shot with something has a bunch of terms to it.*0615

*30x*^{6} 25x^{5} 10x^{4}.0619

*Looking at the numbers 30, 25, and 10, what is the largest thing that could go into all of them, 5 will do it.*0625

*Looking at our variables the largest variable that could go into all them would be x*^{4}.0634

*It is time to write down what I left over 30 ÷ 5 = 6, x*^{6} ÷ x^{4} = x^{2}.0645

*25x*^{5} ÷ 5 = 5, x^{5} ÷ x^{4} =x and 10x^{4} ÷5x^{4} = 2.0659

*That one has been factored out.*0675

*Continuing on, let us get into some of the trickier ones. *0678

*This one has 3 terms and has multiple variables in there.*0681

*It has e and t.*0685

*Let us look at our numbers.*0688

*What could go into 8, 12 and 16?*0689

*2 would definitely go into all of them.*0696

*4 is larger and it would also go into all of them and I think that is largest thing, so we will go with 4.*0698

*Let us take care of these variables, one at a time.*0707

*Looking at the e’s, what is the largest power of e that could go into all of them?*0710

*This will be e*^{4}.0715

*Onto the t’s, the largest exponent of t that would go into all of them is t*^{2}.0721

*Now that we have properly identified our greatest common factor, let us write down what is left over.*0732

*8 ÷ 4 =2, e*^{5} ÷ e^{4} =e and t^{2} ÷ t^{2} =1.0737

*16 ÷ 4 = 4, e*^{6} ÷ e^{4} = e^{2} and t^{3} ÷ t^{2} =t.0752

*Onto the last one, -12 ÷ 4 = -3, e*^{4} ÷ e^{4} = 1 and t^{7} ÷ t^{2} = t^{5}.0766

*We have our factor polynomial.*0790

*One last one, I threw this one in so we can deal some fractions.*0794

*We will first think of what fraction could divide into ¼ and into ¾ ?*0800

*¼ is the only thing that will do it.*0808

*What could go into y*^{9} and y^{2}?0812

*Look for the small exponent that usually helps out.*0817

*That will be our greatest common factor, let us write what is left over.*0823

*¼ ÷ ¼ = 1 and y*^{9} ÷ y^{2} = y^{7}.0828

*¾ ÷ ¼ = 3 and y*^{2} ÷ y^{2} = 1.0839

*At all of these instances you identify your greatest common factor first and simply write your leftovers inside the parentheses. *0849

*One quick thing that can help out, you can double check your work by simply multiplying your greatest common factor*0856

*back into those parentheses using your distribution property.*0861

*If it turns out to be the original problem, you will know that you have done it correctly.*0865

*In some terms, there is a much larger piece in common that you can go ahead and pull out.*0873

*Even though it is larger, feel free to still factor it as normal.*0880

*That is the hard part.*0884

*To demonstrate this I have 6 × (p + q) –r (p + q).*0887

*It has that p + q piece again and both of its parts.*0894

*When I'm looking at the greatest common factor of this first piece and the second piece, it is that p + q.*0899

*I’m going to take out the p + q as an entire piece and that is my greatest common factor.*0910

*Inside the set of parentheses I will write what is left over.*0919

*From the first part there is a 6 that was multiplied by p + q so I will put that in there and there was a -r on the second piece.*0924

*This looks a little strange, I mean it seems weird that we can take out such a large piece, but unusually that it does work.*0933

*To convince you I will go ahead and multiply these things back together using foil.*0941

*What I have here is my first terms would be 6p, outside terms would be –pr.*0947

*Inside terms 6q and my last terms –qr.*0955

*Compare this after I use my distribution property on the original.*0962

*(6p + 6q) – (rp + rq).*0970

*What you see is that it does match up with the original. *0984

*There is my 6p, I have my – pr and – rp, there are just in a slightly different order. *0988

*I have 6q and 6q -qr and I forgot to distribute my negative sign.*0995

*This should be a -rq and so it is the same term.*1007

*What you find is that you can pull out that large chunk of p + q and that leads to what is known as factor by grouping.*1013

*The idea behind factor by grouping is you try and take out the greatest common factor from a few terms at a time.*1024

*Rather than looking at all of them, just take them in parts.*1031

*This grades usually a very large piece and you can collect it into two groups.*1034

*Then you factor within those groups, you take out one of those large pieces. *1040

*This will allow you to factor the entire polynomial. *1045

*Sometimes factor by grouping does not seem to work or usually you end up struggling with a little bit.*1048

*If that is the case, try rearranging the terms and try to factor by grouping one more time. *1053

*It is a neat process and let us give it a try, when you factor by grouping on p × q + 5q + 2p +10.*1060

*Watch how this works.*1071

*Rather than looking at all the terms at once, I’m going to take them two at a time.*1073

*A part of my motivation for doing that is if I were to look at all the terms all at once, they do not have anything in common.*1078

*Let us take these first two.*1086

*What would be the greatest common factor for p × q + 5q.*1088

*q is in both of the terms so that is my greatest common factor for both of those.*1094

*What is left over? There is still a p in there and there is still 5.*1101

*Those are done. *1109

*Let us look at the next two terms.*1111

*What is the greatest common factor of just those two. *1115

*I can see a 2 goes into both of them and then let us see, the only thing left over would be a p + 5.*1119

*I will factor them into those little individual groups and now notice how I only have two things and they both have a p + 5 in common.*1131

* I'm going to take out the p + 5 as my greatest common factor of those two terms.*1142

*What will be left behind will be the q + 2.*1151

*This will represent my factor polynomials.*1156

*Let us try this again and see how it could work. *1162

*I have 2xy + 3y + 2x + 3.*1164

*Looking at the first two.*1170

*These have a y in common, let us go ahead and take that out.*1174

*2x will still be left when I divide 2xy by y and I still have a 3 over here when I divide 3y by y.*1187

*Looking at the next two terms, it looks like these ones do not have anything in common.*1198

*I might consider that the only thing that you do have in common is just a 1.*1205

*I could divide them both by 1.*1209

*I still have a 2x and I still have a 3.*1213

*Notice how we have that common chunk in there 2x + 3.*1218

*We will take that out 2x + 3 and then we will write what is left over just the y +1.*1223

*We can consider this one factored.*1237

*In this next example, be on the watch out for some negative signs which could show up as we factor out that greatest common factor.*1242

*I’m going to take this two at a time.*1251

*Looking at the first two, they do not have a number in common, but they both looks like can be divided by x*^{2}.1255

*Let us take that out.*1262

*After taking our x*^{2} we would have left over a single x and a 3.1268

*Let us look at the next two terms.*1280

*I can see that if I would go into -5 and 5 would divide into a -15 and they both have the negative sign. *1284

*That is important because I’m thinking ahead and try to think how I also have this common x + 3 piece.*1292

*Especially if I’m trying to factor by grouping.*1300

*The only way that is going to work out is if I take out a greatest common factor that is a -5.*1303

*What would that does to our left over pieces now?*1310

*-5x ÷ -5 = x and -15 ÷ -5 = 3.*1314

*I get those left over pieces like I need to, that same x +3 on the other side. *1324

*Now I have this I can take out my common piece of x + 3 and I can go ahead and write the leftovers x*^{2} – 5.1332

*Be on the watch out for certain situations where you need to rearrange things first.*1352

*In this one I have 6w*^{2} - 20x + 15w - 8wx.1357

*And it is tempting to just go ahead and jump in there and try and factor.*1364

*Watch what happens if you do so. *1367

*First we take the first two and we would look for something common.*1370

*The only thing I can see that would be common is a 2 goes into 6 and into 20 and they do not have the same variables.*1375

*That is all I can do.*1385

*I have 3w*^{2} and -10x.1386

*Looking at the other terms the only thing that I see common over there is they have a w in common.*1393

*What would be left over? I still have a 15 and have 8x.*1406

*That is definitely a problem because these pieces are not the same. *1413

*I need to do some rearrangement and retry this factor by grouping one more time.*1419

*Let us go ahead and rewrite this.*1433

*I will rewrite it as 6w*^{2} + 15w - 8wx and we will put that -20x on the very end. 1435

*Here is a little bit of my motivation for putting in that order. *1452

*When looking at the w's, notice how this one is a w*^{2} and this one is a single w.1456

* I have the one is one more in power after the left.*1460

* And looking at the same w's, here is my w*^{1}, this is like w^{0}.1464

*It is like I have lined things up according to the w power.*1470

*Watch how I factor by grouping works out much better.*1475

*Taking the first two they both have a w in common and I can pull that out.*1481

*It also looks like I can pullout 3.*1485

*3w will be like our greatest common factor.*1488

*What would be left over on the inside?*1494

*6w*^{2} ÷ 3w = 2w and 15w ÷ 3w = 5.1499

*I have -8wx -20x, both of those are negative, I think I will go ahead and pull out a -4, that will go into both.*1510

*They both contain x so we will also take out an x.*1523

*-8wx ÷ -4x.*1532

*There are still 2w left in there.*1536

*-20x ÷ -4 = 5.*1540

*I can see there is that common piece that I wanted the first time.*1546

*We can go ahead and factor that out front.*1552

*2w + 5 and 3w - 4x and now this is completely factored.*1556

*Let us do one last one, and this is the one where we might have to do a little bit of rearranging just make sure it works out.*1575

*9xy – 4 + 12x – 3y.*1581

*What to do here?*1587

*Let us go ahead and put the things that have x together.*1590

*9xy + 12x - 3y - 4 and I have just rearranged it.*1606

*We will look at these first two terms here and take out their greatest common factor.*1612

*You can see that 3 goes into both of them and they both contain x.*1618

*It will take both of those out.*1624

*9x ÷ 3x, 9 ÷ 3 = 3, x ÷ x =1 and we still have a y sitting in there.*1628

*12 ÷ 3 = 4, x ÷ x = 1.*1639

*Those two would be gone.*1643

*That looks pretty nice and it is actually starting to match what I have over here. *1646

*Notice that the only difference is a negative.*1651

*I'm going to take out a -1 from both of the terms that not should be able to flip my signs and make it just fine.*1655

*-3y ÷ -1 = 3y and -4 ÷ -1 = 4.*1663

*I have that nice common piece that I need. *1672

*We will go ahead and factor that out .*1677

*3y + 4 and 3x - 1.*1681

*Now this is completely factored.*1690

*When using the technique of factor by grouping take it two at a time and factor those first.*1692

*Look for your common piece and factor that out and that should factor your entire polynomial completely.*1698

*Thank you for watching www.educator.com.*1705

*Welcome back to www.educator.com. *0000

*In this lesson we are going to work on factoring trinomials.*0003

*We are not going to tackle up all types of trinomials just yet.*0006

*For the first part we are going to focus on ones where the squared term has a coefficient of 1.*0010

*We will also look at polynomials where we can factor out a greatest common factor.*0015

*In future lessons we will look at the more complicated trinomials.*0020

*One of our first techniques we have to dig back in our brains and recall how we used foil in order to multiply two binomials together.*0028

*For example, what did we do when we are looking at x -3 × x +1.*0037

*Using the method of foil we would multiply our first terms together and get something like x*^{2}.0042

*We would multiply our outside terms together 1x then we would multiply our inside terms.*0049

*And finally we would multiply our last terms together. *0060

*Sometimes we had to do a little bit of work to clean this up.*0067

*As long as we made sure that everything got multiplied by everything else. *0070

*We where assured that we could multiply these two binomials out.*0075

*Since we are working with factoring and breaking things down into a product you want to think of this process, but do it in reverse. *0079

*If you had a trinomial to begin with, how could you then break this down into two binomials?*0087

*Since this is the exact same one as before I will simply write down the two binomials that it will break up into.*0096

*This is the process that we are after of taking a trinomial and breaking it down into two binomials.*0105

*In doing so it is not quite a straightforward process.*0114

*The way we are going to attack this is to think of that foil process in our minds.*0119

*This will help us determine our first and last terms in those binomials. *0124

*Now, after we have chosen something for those first and last terms *0129

*we will have to check to make sure that the outside and inside terms combined to be our middle term.*0133

*Sometimes we will have to do some double checking just to make sure that it does combine and give us that middle term.*0140

*Sometimes there are lots of different options.*0146

*We may have to do this more than once until we find just the right values that make it work. *0149

*Watch how that works with this trinomial.*0154

*I have x*^{2} + 2x -8 and what we are looking to do is break this down into two binomials.0157

*I’m going to go ahead and write down the parentheses just to get it started.*0165

*I first want to determine what should my first terms be in order to get that x*^{2}.0169

*We are looking to multiply two things together and get x*^{2}.0177

*The only two things that will work is x and x.*0180

*We have a good chance that those are our first terms.*0184

*Rather than worrying about the outside and inside just yet, we jump all the way to the last terms.*0191

*That will be here and here and I'm looking to multiply them and get -8.*0197

*Here is the thing there are lots of different options that we could have, it could be 1 and 8, 2 and 4.*0203

*We could also look at the other order of these maybe 4 and 2, 8 and 1.*0211

*We want to choose the proper pair that when combined together will actually give us that 2x in the middle.*0217

*Let us go ahead and try something.*0224

*Watch how this process works. *0226

*Suppose I just tried the first thing on the list, this 1 and 8.*0228

*1, 8, with this combination I can be sure that my first terms work out and that my last terms work out.*0233

*I'm not completely confident until I check those outside and inside terms to make sure that they work out.*0242

*I will do some quick calculations.*0250

*Let us check our outside terms, 8 × x = 8x and inside terms 1 and x, and these would combine to give us a 9x.*0252

*If you compare that to the original it is not the same.*0265

*What that is indicating is that pair of 1 and 8, those are not the ones we want to use.*0270

*We can backup a little bit and try another pair of numbers.*0277

*Let me try something different.*0287

*I'm going to try 4 and -2.*0290

*Now when I do my outside and inside terms I will get -2x on the outside, *0298

*4x on the inside and those combine to give me 2x, which is the same as my middle term.*0305

*I know that this is how it should be factored.*0313

*If you want you can go through the entire foil process, just to double check that all the rest of terms work out.*0319

*In fact it is not a bad idea when you are done with the factoring process, just to make sure it is okay. *0324

*Now, there are a few tips to help you along the way when doing this reverse foil method.*0333

*It works good, as long as you are leading coefficient is 1 and you can take a look at the signs of the other two coefficients. *0339

*In general, here is what you are trying to do.*0351

*You are looking for two integers whose product will give you c.*0354

*They multiply and give you c but whose sum is b.*0358

*It is what we did in that last example. *0363

*Now you can get a little bit more information if you look at the signs of b and c.*0367

*If b and c are both positive then those two integers you are looking for must also be positive. *0371

*One situation that might happen is you know both integers that you are looking for will be negative if c is positive, *0383

*that is this one in the end and d is negative.*0391

*Watch for that to happen. *0396

*Of course one last thing, you will know that the integers you are looking for are different in sign, one positive and one negative if c is negative.*0400

*That is the only way they could multiply together and give you a negative number here on the end.*0410

*Watch for me to use these shortcuts here in just a little bit.*0416

*We want to use this reverse foil method in order to factor the following polynomial y*^{2} + 12y + 20.0426

*I’m going to start off by writing set of parentheses this will break down into some binomials.*0435

*Let us start off for those first terms.*0444

*What times what would give us a y*^{2}?0447

*One thing that will do it is just y and y.*0451

*Let us look for two values that would multiply and give us a 20.*0457

*It is okay for me to write down some different possibilities like 1 and 20, 2 and 10, 4 and 5.*0462

*You can imagine the same values just flipped around.*0470

*Let us see if we can use any information to help us out. *0474

*Notice how this last term out here is positive and so is my middle term, both of them are positive. *0479

*Now what is that telling me about my signs, I know that the two numbers I'm looking for will both be positive.*0488

*That actually is quite a bit of information because now when we look at our list I can pick two things that will add to be the middle term 12. *0496

*I know they multiply between.*0504

*Let us drop those in there, 2 and 10.*0507

*I have factored the trinomial.*0511

*Let us quickly go through the foil process just to make sure that this is the one we are looking for. *0514

*We are looking to make sure that this matches up with the original. *0520

*First terms would be a y*^{2}, outside terms 10y, inside terms 2y and last term is 20.0524

*These middle ones would combined giving us y*^{2} + 12y + 20 and that shows that our factorization checks out.0534

*We know that this is the proper way to factor it.*0544

*Let us try another one, this one is x*^{2} - 9x – 22.0551

*Let us start off in much the same way.*0557

* Let us write down a set of parentheses and see if we can fill in the blanks.*0559

*We need two numbers that when multiplied together will give us x*^{2}.0566

*That must be an x and another x.*0571

*Now we need a pair of numbers that will multiply and give us a -22.*0577

*Let us write down some possibilities like 1 and 22, 2 and 11, I think that is it.*0581

*Let us get some information about the signs of these numbers. *0589

*I’m looking at this last number here and notice how it is negative,*0594

*I know that these two numbers I’m looking for on my last terms, one of them must be positive and one of them must be negative. *0598

*The question is which one is positive and which one is negative?*0607

*We will look to our middle term to help out. *0613

*I need the larger term to be negative, so that I will get a negative in the middle, that -9.*0617

*We have plenty of information it should be pretty clear that it is actually the 2 and 11 off my list that will work.*0624

*2 and 11.*0632

*Let us just check it real quick by foiling things out to make sure that this is how it should factor.*0637

*First terms x*^{2}, outside terms -11x, inside terms 2x, and last terms -22.0643

*Since I have a positive × negative, these middle terms combined I will get x*^{2} - 9x -22.0656

*That is the same as my original, so I know that I have factored it correctly.*0666

* Let us use the reverse foil method for this polynomial.*0679

*It is not very big, it is r*^{2} + r + 2.0682

*We are going to start off by setting down those two parentheses.*0686

*We are hunting for two first terms that will multiply to give us r*^{2}.0691

*There is only one choice that will do that, just r and r.*0697

*We turn our attention to the last terms and we need them to multiply to be 2.*0704

*Unfortunately, there is only one possibility for that, 1 and 2.*0711

*Let us hunt down our signs.*0717

*The last term is positive, the middle term is positive that says both of the terms that we are looking for must both be positive.*0720

*Let us put them in and now we can go and check this using the foil process.*0728

*r × r =r*^{2}, the outside terms should be 2r, inside terms 1r, and the last term is 2.0735

*Combining the two middle terms here, I get r*^{2} + 3r + 2.0746

*Something very interesting is happening with this one, let us take a closer look.*0755

*If we look at the resulting polynomial that we got after foiling out and we compare that with the original, they are not the same.*0759

*That tells us something. It tells us that this is not the correct factorization. *0769

*Now if this is not how it should be factored then what other possibilities do we have?*0783

*If we look at all of our possibilities for those last terms, it must contain 1 and 2 if it is going to multiply and give us 2.*0789

*Since that did not work and I have no other possibilities, it tells us that this does not factor into two binomials using 1 and 2.*0798

*This is an indication that our original is actually prime.*0807

*Watch out for ones like this where when you try and factor it, it simply does not factor into those two binomials.*0813

*Let us try something with a few more variables in it.*0824

*This one is t*^{2} – 6tu + 8u^{2}.0827

*Even though we have a few more variables in here, you will see that this process works out the same as before. *0833

*Starting off with those first terms, something × something will give us t*^{2}.0841

*That must be t and another t.*0848

*I need two things to multiply and give us 8u*^{2}.0853

*Well I'm not quite sure about that 8 yet because it could be 1 and 8, 2 and 4, or could be those reversed.*0860

*I did know about the u, you better have a u and another u in order to get that u*^{2}.0867

*Let us just focus on numbers for bit and see what information we can get from there.*0874

*Looking at the sign of my last term it is positive, but my middle term is negative.*0881

*The information I’m getting from there is that both of my numbers I’m looking for must both be negative.*0887

*I think we can pick it out from our list now and it looks like we must use the 2 and 4.*0896

*Finally let us check that to make sure this one works.*0906

*Let us be careful since we have both t’s and u’s.*0909

*First terms t*^{2}, outside terms -4tu, inside terms -2tu and my last terms – 2 × -4 = 8u^{2}.0912

*We can combine our middle terms giving us t*^{2} – 6tu + 8u^{2}.0931

*And now we can see that yes, this has been factored correctly since the resulting polynomial is the same as my original.*0940

*This one is good.*0947

*One thing to watch out for is to make sure you pull out any common factors at the very beginning. *0953

*That is what we will definitely need to do with this example before we even start the forming process.*0959

*Always check out for a good common factor. *0966

*This is also good idea because it can potentially make your number smaller, so that you do not have to think of as many possibilities.*0970

*We are going to quickly factor 3x*^{4} -15x^{3} + 18x^{2}.0978

*Look with this one, everything here is divisible by 3 and I can pull out an x*^{2} from all of the variables.0985

*Let us take it out of the very beginning, I have 3x*^{2} and let us write down what is left.0996

*3x*^{4} ÷ 3x^{2} = x^{2}, =15 ÷ 3 = -5x, 18 ÷ 3 + 6 and I think that is all my leftover parts.1004

*With this one now, I will go and do the reverse foil process on that.*1024

*I need two binomials, let us break it down. *1030

*What should my first terms be in order to get my x*^{2}?1035

*That must be an x and an x.*1042

*I have to look at my last terms and then multiply together to give us a 6.*1047

*1 and 6, possibly 2 and 3, but you can use the signs to help you out. *1052

*The last term is positive, the middle term is negative, so both of these will be negative.*1058

*It looks like I need to use that 2 and 3.*1066

*Some quick checking to make sure this is the correct factorization.*1077

*I have x*^{2} - 3x - 2x + 6 and looks like my outside and inside terms do combine and give us that - 5x.1080

*This is the correct factorization and let us go head and write out that very first 3x*^{2} that we took out at the very beginning.1093

*Now that we have all the pieces, we can say that this is the correct factorization.*1103

* Always look for a common factor that you could pull out from the very beginning before starting the foil method. *1109

*Sometimes you can pull something out, sometimes you can not but it will make your life easier if you can find something.*1116

*Keep that in mind for this next example. *1125

*This one is 2x*^{3} - 18x^{2} - 44x.1128

*Let us come at this over.*1134

*It looks like everything is divisible by 2 and they all have an x in common.*1136

*Let us take out a 2x at the very beginning. *1142

*What do we have left?*1148

*2x*^{3} ÷ 2x = x^{2}, -18 ÷ 2 = -9x and -44x ÷ 2x = -22.1150

*Now we want to factor that into some binomials.*1168

*Let us go ahead and copy over this 2x just we can keep track of it.*1174

*I need my first terms to multiply together and get an x*^{2}.1180

*That will be an x and another x.*1186

*I need to look at my last terms so that they multiply together to give me a -22.*1190

*Some of our possibilities are 1 and 22, 2 and 11.*1197

*Since the last one is negative and my middle term is negative, I know that these will be different in sign.*1204

*Let us take the 2 and 11 off of our list, those are the ones we need, -11 and 2.*1214

*Let us quickly combine things together and make sure that it is the correct factorization.*1224

* x*^{2} - 11x + 2x – 22.1229

*Combining these middle guys x*^{2} - 9x -22, so that definitely checks with this polynomial right here. 1236

*Now if you want to go ahead and put in the 2x as well, this will take you back all the way to the original one.*1249

*Remember to use your distribution property so you can see how that will work out.*1258

*2x*^{3} - 18x^{2} - 44x and sure enough that is the same as the original.1264

*Everything checks out. I know that this is the correct factorization for our polynomial.*1275

*Just a few things, make sure that when you are using this method, always check for common factor to pull it out. *1286

*Make sure you set down your first terms and then your last terms *1292

*and definitely check those signs to help you eliminate some of your possibilities. *1297

*Thank you for watching www.educator.com.*1303

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at factoring some trinomials using a method known as AC method. *0002

*We have already seen factoring trinomials once before but these ones are going to be a little bit more complicated*0010

*and that our squared term will be something other than the number 1.*0015

*This is why we are going to pick up the AC method, we have a little bit more algebraic way to approach these types of problems.*0019

*Recall some of the earlier trinomials that we have been factoring so far.*0029

*The squared term in front has always been 1 and that made life pretty easy on us *0033

*because when we went searching for those two binomials to break it down into.*0039

*We did not have a lot of options in order to get that first term.*0044

*It is probably something like y and y or x and x.*0048

*There is not a whole lot of other things it could be.*0051

*The reason why this made things a little bit easier is we only have to focus on the last term *0054

*and making sure our outside and inside terms combined to give us the middle term. *0059

*Now, we do not want to necessarily stick with those types of trinomials.*0064

*We want to go ahead and factor things where the initial term is something other than 1. *0068

*Now this will end up making things a little bit more difficult.*0073

*The good news is with these ones you can still use something like the reverse foil method.*0080

*You have not only possibilities for your first term, but now you also have possibilities for your first and last terms, both of those.*0086

*This will make it a little bit more difficult when we are checking to make sure that the outside and inside terms combine to give us that middle term.*0096

*Let us do a reverse foil example, so you can see that we have to track down many more possibilities. *0108

*This one is 2x*^{2} + 7x + 6.0113

*It is not that big but we will look at the two binomials that we are looking to break this down into.*0117

*Like before, I will be looking for two terms that multiplied to give us the 2x*^{2}.0124

*Since that number is there I have to look at possible things that will give us 2.*0129

*This one is not too bad, it has to be 1 and 2 to get that x*^{2}, 1x and 2x.0136

*We will look at the possibilities that will give us our second term.*0149

* 6 could be 1 and 6, could be 2 and 3, or it could actually be those values flipped around.*0154

*The question is what should it be? What things are we looking for here?*0163

*I know that these two numbers whatever they are looks like they both better be positive since these are both positive.*0170

*It could be 1 and 6, could be 2 and 3.*0177

*Let us go ahead and put one of those in there just to see what happens. *0182

*Let us suppose I'm trying out 1and I'm trying out 6. *0187

*My first terms multiply out just fine. *0192

*My last terms multiply out just fine.*0194

*Let us check out our outside and inside terms.*0197

*The outside would give us 6x and the inside would give us 2x.*0202

*When you combine those together, you get 8x which unfortunately is not the same as that middle term.*0212

*I need to come up with some other choice for those last terms.*0219

*Let us see, if it is not 1 and 6, I guess we have to try 2 and 3.*0226

*I will put those in there and we will double check our outside and inside terms.*0232

*Outside would be 3x, inside would be 4x and sure enough those do combine to give us that 7x.*0240

*The important part to recognize is that if your initial term is something other than 1, *0250

*you have to look at your possibilities for your last term and your first term *0256

*and play around with how they are ordered in order to get your proper factorization.*0259

*Let us try another one that has a few more possibilities for that first term just to make things a little bit more interesting.*0270

*Looking at the first term, we need something that will multiply and give us 6y*^{2}.0283

*That could be 1 and 6, could be 2 and 3, but they both will definitely contain y because of that y*^{2}.0290

*Looking at our last terms, many different things could multiply and give us the 10.*0300

*1 and 10, 2 and 5 or possibly those just reversed around.*0307

*I have lots of different options that I can end up packaging this together. *0313

*Maybe this is 1 and 6 with the 1 and 10 or maybe I should use the 2 and 3 with the 1 and 10, *0319

*or the 1 and 6 with the 2 and 5, or the 2 and 3 with the 2 and 5.*0329

*There are many different ways and I can also do these in different orders to make sure that they combine.*0333

*The key for figuring out which combination should you use is looking at those outside and inside terms.*0340

*In this one, we want them to combine to give us that 19.*0349

*We definitely have to do a little bit of work to figure out what that is.*0352

*Let us see for the current setup my outside would be 10y and my inside would be 6y,*0355

*which unfortunately does not combine enough to give us that 19y in the middle.*0360

*I know that my 1 and 6, and my 1 and 10 I need to change something around this one.*0366

*This one is just not going to work that way.*0370

*Let us play around with our first term.*0374

*Let us try something else for the beginning here.*0378

*Let us try 2 and 3.*0381

*What will that give us?*0386

*I can see the outside is 20y, the inside is 3y but that is a little too much, 23y not going to work out.*0389

*The good news is I did do this one earlier so I do have a combination that will actually factor.*0406

*What we are looking for this one is 2, 3, 5 and 2.*0414

*Let us check the outside and inside terms for this guy.*0423

*The outside would be 4y and the inside would be 15y and sure enough those combine to give us 19y.*0427

*I know that this is the proper factorization.*0437

*Notice what this highlight is that when your initial term is something other than 1, and you have lots and lots of possibilities to run through, *0441

*it can be very difficult to find out just the right combination of numbers to use in order to make it all work out correctly.*0449

*If my numbers were even bigger I would have even more possibilities to run through.*0456

*This is a problem.*0462

*To fix this problem where our leading term is something other than 1, and our numbers could get fairly large,*0467

*we do not necessarily want to use the reverse foil method.*0474

*There are simply too many possibilities to consider for some problems and it gets too difficult. *0478

*This is why I run to pick up something known as the AC method. *0483

*This is a little bit more of an algebraic method that we can use and hunt down some of the possibilities we need for breaking it down. *0486

*Let me quickly run you through how the AC method works, and then I will give you some quick tips on using it.*0493

*The very first thing that you want to do when using AC method is just to see if everything has a common factor or not.*0499

*And if it does have a common factor, go ahead and factor that out before beginning any other type of factoring process.*0507

*If they have anything in common, pull that out.*0513

*Then I will multiply the first and the last terms together, this is known as the A and C terms.*0516

*This is where the method gets its name. *0521

*Once you get that new number, you will be looking for two numbers that multiply to give you AC and they actually add to get you B. *0524

*This will have the feel of looking for those two integers, but it will be a little bit more straightforward than what you see in the past.*0532

*I do have a nice way to organize that step to keep track of the two numbers are looking for.*0538

*Here comes the interesting part, when you find those two numbers we will actually split up your middle term into two new numbers *0543

*and then you will have four terms total and you will actually use factor by grouping to move from there.*0550

*The AC method is a way of splitting up your middle term and using a different approach that factor by grouping to handle it instead.*0556

*Let me give you some tips on using the AC method.*0565

*When you use the AC method you want to organize where your AC and your B terms go.*0572

*What I recommend is that you draw a small x.*0578

*In the top part of that x you put the values of A× C and in the bottom part of that actually put the value of B.*0585

*What you are looking to do is you want to fill out the rest of this x by putting in these two numbers.*0595

*The numbers that will go there, they must multiply to give you that top number and they must add to give you the bottom number.*0604

*You will feel like you are filling out a very small crossword. *0617

*Be very careful in doing this and make sure that the signs matchup.*0621

*If you need to add to give you a negative number then make sure the two side numbers will add to give you that negative number.*0625

*Always be careful on your signs with this one, they should matchup.*0633

*You have heard a lot about the AC method and have not seen it yet, let us go ahead and do an example see can see it all in action. *0637

*We want to factor 10q*^{2} – 23q +12.0647

*This is a good example of one that you want use the AC method on.*0652

*If you try to factor directly you have lots of possibilities for the first term, like 1 and 10, 2 and 5. *0656

*You have lots of possibilities for the second term, 1 and 12, 2 and 6, 3 and 4.*0662

*Of course all of those reversed.*0667

*Let us tackle this using the AC method.*0671

* In step one, check all of your numbers to see if they have a common factor.*0675

*I get 10, 23 and 12 it looks like they do not have anything in common. *0679

*Unfortunately, means I can not pull anything out and make the number smaller.*0685

*In to step two, I want to go ahead and multiply my a term and my c term together.*0689

* 10 × 12 = 120.*0697

*I’m looking for two numbers that will multiply to give me 120 and add to be -23. *0702

*This is where little box will come in handy.*0709

*Let me just put in our few little notes.*0718

*We want them to multiply that gives us our top number. *0720

*We want them to add to give us that -23. *0725

*To help us better find the numbers that will go ahead and do this, *0732

*I’m going to start listing out all the pairs of numbers that will multiply to give us 120.*0736

*1 and 120, 2 and 60, I have 3 and 40, it just keep continuing making this list until you get as many numbers as possible.*0742

*I got 4 and 30, 5 and 24, 6 and 20, 8 and 15.*0757

*Now that we have a bunch of numbers on this list, let us see how we can use it. *0770

*We want these two numbers to multiply to be 120 and when we built those list that all should multiply to give us 120.*0775

*But they must add to give us -23. *0784

*The only way you want to add to get a negative number and multiply to get a positive number *0788

*is if both of these new numbers here were negative.*0792

*Let us say if you are going to pick two things off this list that will give us -23 when added together.*0797

*I think it is going to be that last two, 8 and 15.*0804

*Notice how those will multiply negative × negative will give us that positive 120 and will definitely add to be -23.*0812

*Those are the two numbers we want.*0820

*Now comes the interesting thing.*0824

*What I'm going to do with those two numbers is end up splitting up my original middle term.*0827

*I'm writing down the numbers of my original polynomial but I'm not writing that -23 in there. *0833

*This is where I split it into two terms.*0841

*This will be -8q – 15q so I have not changed my polynomial. *0845

*I just take a look at it in a different way.*0855

*You will notice how this new polynomial has four terms 1, 2, 3, 4.*0858

*I'm going to now attack it using factor by grouping, which means I will take these terms two at a time.*0863

*Let us do the first two and see what they have in common.*0871

*Both are divisible by 2q.*0875

*Let us see what we got left over in here.*0887

*2q, I have 5q - 4 and let us see what is in common with the next two.*0890

*It looks like I can pull out -3 from both of them.*0909

*That would leave us 5q - 4 and notice how the signs do match up, -3 × 5 =-15.*0923

*-3 × -4 =12.*0933

*We can go ahead and wrapped this one up. *0939

*They both have a 5q - 4 in common, I will write that for my first binomial with the leftover pieces being 2q and – 3.*0940

*It is quite a journey to get to those final two binomials, but notice how it is a little bit more methodically, *0956

*not necessarily guessing or picking things out of here.*0962

*You have a better hunting way of going about it. *0964

*Let us quickly check to make sure that this is the correct factor polynomial just by running through the foil process.*0969

*5q × 2q = 10q*^{2}, outside terms -15q, inside terms – 8q, last terms +12. 0974

*It is already starting to look pretty good since it looks like that one.*0987

*Just combine my little terms here and I get 10q*^{2} – 23q + 12, which is exactly the same as I had originally.0991

*I know that this one is factored correctly using that AC method.*1005

*Let us see the AC method again just we can get more familiar with it.*1012

*In this one we want to factor 5t*^{2} + 13t – 6.1016

*I want to make sure that they have a greatest common factor of only one, *1025

*which means do they have anything in common that I can factor out at the very beginning.*1028

*5, 13, and 6 do not have anything in common.*1032

*Let us move on to multiplying the A and C terms together.*1035

*5 × -6 = 30.*1041

*Two numbers that will multiply to give us 30, but add to give us a 13.*1046

*We will use our box to help us out.*1051

*It need to multiply to be 30 and add to be 13.*1057

*The two numbers that we put in here, they must multiply to give us 30 and they will add to give us 13.*1064

*To help out with the search, we will list down all the things that multiply to give us 30.*1077

*1 and 30, 2 and 15, 3 and 10, 4 and 15.*1083

*The only two things that are going to work from this list or the least that I can see I think will be our 3 and 10.*1099

*Let us go ahead and put those in.*1109

*Hold on, I think we forgot one of our signs here should be -30 on top.*1122

*We will try this again. *1131

*We need two numbers from our list that will multiply to give us -30, but add to be 13.*1133

*I think the 2 and the 15 will have to be the one to do it because they have to be different in sign to multiply to give us that -30. *1141

*How about 15 and -2?*1150

*We are ready to split up our middle term.*1156

*I have written the first term and the last term now we will write out that middle term split using these two new numbers.*1159

*– 2t + 15t looking pretty good.*1169

*Now that we have this, we want to factor by grouping.*1177

*We will take these two at a time. *1181

*5t – 2t they only have one thing in common and that would be t.*1186

*Let us see what is left over.*1199

*5t - 2 looking at the next two terms 15t – 6 they have a 3 in common.*1200

*5t – 2.*1219

*We have the 5t -2 common piece, we will go ahead and take it out of both of them and write out our leftover pieces, t + 3.*1224

*This one did take quite a bit of work let us quickly check it again by our formula.*1236

*5t*^{2} + 15t – 2t -6. 1242

*These two little terms combine giving us 13t - 6, which is the same as the original.*1251

*I know that this is the correct factorization.*1260

*Let us go ahead and try to factor this one using the AC method. *1269

*This one has a few more variables in it so notice I have x*^{2} and y^{2}.1273

*Do not worry too much about those x and y, what you will see just focus mainly on those numbers and hunt down what those need to be in.*1278

*Is there anything I can take out at the very beginning?*1288

*Do they have anything in common?*1291

*It looks like the 11 is going to make it where they do not have anything in common.*1295

*I will multiply the A and C terms together.*1301

*6 × -10 = -60.*1304

*We need two numbers that will multiply to be -60, but add to be 11.*1312

*Let us try our box to help us out.*1317

*-60 and 11. *1322

*They need to multiply and give us -60 and add to be the 11.*1327

*Starting off with writing down all the possibilities to give is that 60.*1337

*1 and 60 would do it, 2 and 30, 3 and 20, 4 and 15.*1341

*The two numbers that we use must multiply to be -60.*1353

*We want one of these numbers to be positive and the other one to be negative.*1357

*Since we are adding to be 11, the larger number must be positive.*1362

*I think I see a good option on this list, the 4 and the 15.*1368

*The 15 is the larger one, so it must be positive and the 4 is the smaller one, so we will make it negative.*1372

*We will write down our polynomial and split up the middle term into two new terms.*1380

*We will use 15xy and - 4xy.*1391

*Notice that we are using that xy here because those are what is on the middle t*