*Welcome to Educator.com.*0000

*Today is our first lesson for the Algebra II series, and we are going to start out with some review of concepts from Algebra I.*0002

*If you need more detail about any of these concepts, please check out the Algebra I series here at Educator.*0008

*The first session is on expressions and formulas.*0014

*Recall the earlier concepts of variables and algebraic expressions:*0020

*starting out with some definitions, a ***variable** is a letter or symbol that is used to represent an unknown number.0025

*It could be any letter; frequently, x, y, and z are used, but you could choose n or s or w.*0034

**Algebraic expressions** means that terms using both variables and numbers are combined using arithmetic operations.0045

*Remember that a ***term** is a number, or a variable, or both.0057

*So, a term could be 4--that is a constant, and it is a term; it could be 2x; it could be y*^{2}.0065

*And when these are combined using arithmetic operations, then they are known as expressions.*0076

*And when variables are involved, then they are algebraic expressions.*0082

*For example, 4x*^{3}+2xy-1 would be an example of an algebraic expression.0085

*The rules specifying order of operations are very important; and they are used in order to evaluate algebraic expressions.*0097

*Recall the procedure to evaluate an expression using the order of operations.*0107

*First, evaluate expressions that are inside grouping symbols: examples would be parentheses, braces, and brackets.*0116

*The next thing, when you are evaluating an algebraic expression, is to evaluate powers.*0151

*So, if a term is raised to a power (such as 4*^{2} or 3^{4}), you need to evaluate that next; that is the second step.0160

*Next is to multiply and divide, going from left to right.*0175

*You start out at the left side of an expression; and if you hit something that needs to be divided, you do that.*0193

*And you proceed towards the right; if you see something that needs to be multiplied, you do that.*0198

*It is not "multiply all the way, and then go back and divide"; it is "start at the left; any multiplication or division--do it."*0203

*Move to the next step; move towards the right; multiply or divide...and so on, until all of that has been taken care of.*0208

*Finally, you do the same thing with addition and subtraction: you add and subtract from left to right.*0216

*And we will be illustrating these concepts in the examples.*0226

*One thing to recall is that a fraction bar can function as a grouping symbol.*0231

*For example, if I have something like 3x-2x+2, all over 4(x+3)+3, I would treat this as a grouping symbol.*0236

*And I would simplify this as far as I could, going through my four steps; and then I would simplify this;*0248

*and **then* I would divide this simplified expression on the top by the simplified expression on the bottom.0253

*And remember, the reason that we use order of operations is that, if we didn't, and everybody was just doing things their own way,*0259

*we couldn't really communicate using math, because people would write something down,*0265

*and somebody might do it in a different order and come up with a different answer.*0269

*So, this way, it is an agreed-upon set of rules that everyone follows.*0273

*Monomials: a ***monomial** is a product of a number and 0 or more variables.0281

*Again, refreshing your memory from Algebra I: examples of a monomial would be 5y, 6xy*^{2}, z, 5.0288

*So, it says it is a product of a number and zero or more variables.*0304

*Here, there aren't any variables, so that actually is simply a constant; but it is still called a monomial, also.*0307

*Here, I have 5 times one variable; here I have multiple variables.*0314

*This is examples of...these are all monomials.*0321

*A ***constant** is simply a number; so, it could be -3 or 6 or 14; those are constants.0328

*Coefficients: a ***coefficient** is the number in front of the variable.0346

*Up here, I said I had 5y and 6xy*^{2}: this is a coefficient: 5 is a coefficient, and 6 is a coefficient.0363

*When you see something like z, it does have a coefficient: it actually has a coefficient of 1.*0372

*However, by convention, we usually don't write the 1--we just write it as z, but it actually does have a coefficient of 1.*0378

*Next, degree--the degree of a monomial: the ***degree** of a monomial is the sum of the degrees of all of the variables.0386

*So, it is the sum of the degree of all of the variables.*0394

*For example, 3xy*^{2}z^{4}: if I want to find the degree, I am going to add the degree of each variable.0408

*This is x (but that really means x to the 1--the 1 is unstated) plus y*^{2} (the degree is 2), plus z^{4} (the degree is 4).0418

*Adding these up, the degree for this monomial is 7.*0429

*When we talk about powers: ***powers** refer to a number or variable being multiplied by itself n times, where n is the power.0435

*For example, if I say that I have 5*^{2}, what I am really saying is 5 times 5.0447

*So, 5 is being multiplied by itself twice, where n equals 2.*0455

*I could say I have y*^{4}: that equals y times y times y times y; and here, the power is 4.0461

*OK, continuing on with more concepts: a ***polynomial** is a monomial or a sum of monomials.0477

*Recall the concepts of term, like terms, binomial, and trinomial.*0485

*A polynomial is simply an expression in which the terms are monomials.*0492

*And we say "sum," but this applies to subtraction, as well--a polynomial can certainly involve subtraction.*0496

*For example, 4x*^{2}+x or 2y^{2}+3y+4: these are both polynomials.0504

*We also could say that 5z is a polynomial, but it is also a monomial; there is only one term, so it is a polynomial, but we usually just say it is a monomial.*0519

*OK, so looking at these other words: a ***binomial** is a polynomial that contains two terms.0532

*So, it is the sum of two monomials, whereas the ***trinomial** is the sum of three monomials.0544

*A monomial is simply a single monomial.*0552

*Recall that, as discussed, a ***term** is a number or a letter (which is a variable) or both, separated by a sign.0561

*Terms could be a number; they could be a variable; or they could be both.*0572

*3x-7+z: here, I have a number and a variable; here, I just have a number (I have a constant); here, I just have a variable.*0582

*And they are separated by signs--by a negative sign and a positive sign--so each one of these is a term.*0594

*The concept of ***like terms** is very important, because like terms can be added or subtracted.0602

*Like terms contain the same variables to the same powers.*0609

*For example, 1 and 6 are like terms; they don't contain any variables, so they are like terms.*0626

*3xy and 4xy are like terms; they both contain an x to the first power and a y to the first power.*0636

*2y*^{2} and 8y^{2} are also like terms: they both contain a y raised to the second power.0645

*And so, these can be combined: they can be added and subtracted.*0653

*A ***formula** is an equation involving several variables (2 or more), and it describes a relationship among the quantities represented by the variables.0661

*And we have worked with formulas previously: and just to review, one formula that we talked about is the Pythagorean Theorem.*0670

*That is a*^{2}+b^{2}=c^{2}, where c is the length0678

*of the hypotenuse of a right triangle, and a and b are the lengths of the two sides.*0683

*And this tells us the relationship among the three sides of the triangle.*0689

*And that is really what formulas are all about, and really what algebra is all about: describing relationships between various things.*0697

*And of course, during this course, we are going to be working with various formulas.*0706

*OK, in this example, we are asked to simplify or evaluate an algebraic expression.*0711

*5x*^{2}...and they are telling me x=3, y=-3; so I have some x terms and some y terms.0718

*My first step is to substitute: so, everywhere I have an x, I am putting in a 3; everywhere I have a y, I am putting in a -3.*0724

*So, here I have 3xy, so here it is going to be 3 times 3 times -3.*0736

*Recall the order of operations: the first thing I am going to do is to get rid of the grouping symbols.*0740

*Take care of the parentheses; and looking, I do have parentheses.*0746

*In here, I have a negative and a negative; so I am simplifying that just to positive 3.*0755

*OK, continuing to simplify inside the parentheses: 3 plus 3 is 6.*0766

*I completed my first step in the order of operations.*0775

*The next thing to do is evaluate powers; and I do have some terms that are raised to various powers.*0778

*3*^{2} is 9, minus 2 times 6^{3}; so, 6 times 6 is 36, times 6 is 216.0785

*That took care of my powers; and the next thing is going to be to multiply and divide.*0801

*And when we do that, we always proceed from left to right.*0807

*2 times 216 is 432; OK, now I have: 3 times 3 is 9; 9 times -3 is -27.*0812

*I am going to rewrite this as 9 minus 432 minus 27.*0825

*Finally, add and subtract; and this is going from left to right.*0830

*9 minus 432 gives me -423, minus 27 (so now I have another bit of subtraction to do--that is -423-27) gives me -450.*0839

*So again, the first step was substituting in 3 and -3 for x and y.*0854

*The next step was to get rid of my grouping symbols; evaluate the powers;*0860

*multiply and divide, going from left to right (and I just had multiplication here);*0867

*and then, add and subtract, going from left to right, to get -450.*0873

*In this second example, again, we are asked to evaluate an algebraic expression; and here, we have three variables: a, b, and c.*0879

*So, carefully substituting in each of these, a=-1...so -1*^{2}, minus 2, times b (b is 2), times c (c is 3), plus 3^{3}.0886

*Here, I have c*^{2} in the denominator; so that gives me 3^{2}, minus 2, times a (which is -1), times b (which is 2).0907

*Since there was a lot of substituting, it is a good idea to check your work.*0918

*a is -1 (that is -1*^{2}), minus 2, times b, times c, plus c^{3};0923

*all of that is divided by 3*^{2} (so that is c^{2}) minus 2, times a, times b.0934

*Everything looks good; now, the first thing I want to do is eliminate grouping symbols.*0941

*Recall that, in this type of a case, the fraction bar is functioning as a grouping symbol.*0945

*So, the whole numerator should be simplified, and the denominator should be simplified; and then I should divide one by the other.*0951

*Starting with the numerator: within the numerator, there are not any grouping symbols,*0958

*so I am going to go ahead and go to the next step, which is to take care of powers.*0964

*And -1 times -1 is 1; and then, I have 3*^{3}; that is 3 times 3 (is 9), times 3 (is 27).0971

*OK, I can do the same thing in the denominator; I can just do these both in parallel.*0991

*And so, I am going to evaluate the powers in the denominator.*0997

*3 times 3 is 9; and then, I don't have any more powers--OK.*1000

*So, I took care of that; my next step is going to be to multiply and divide.*1005

*OK, so I have, in the numerator, 1 minus 2 times 2 (is 4), and then 4 times 3 (is 12), plus 27.*1016

*So, that took care of that step; now, in the denominator, I have 9, and then I have minus 2, times -1.*1035

*So, 2 times -1 is going to give me -2; -2 times 2 is going to give me -4.*1046

*OK, so now, I have taken care of all of the multiplication and the division.*1063

*The next step is to add and subtract--once again, going from left to right.*1068

*So, starting up here, the next step is going to be 1 minus 12; 1 minus 12 is going to give me -11.*1073

*So, it is -11 plus 27; that is going to leave me with 16 in the numerator.*1086

*In the denominator, I have 9, minus -4; well, a negative and a negative gives me a positive,*1092

*so in the denominator, I actually have 9 plus 4, which gives me 13.*1099

*The result is 16 over 13.*1109

*Again, starting out by substituting values for a, b, and c...I have done that in this first step.*1112

*And then, I treat this fraction bar as a big grouping symbol, and then I take care of the numerator and the denominator separately.*1120

*You could have done them one at a time, or you can do steps at the same time.*1128

*So, first, evaluate the powers; I did that in the numerator; I did that in the denominator (I am treating them separately).*1132

*Multiplying and dividing: I did my multiplication here and in the denominator.*1138

*And finally, adding and subtracting to get 16/13.*1146

*Example 3: The formula for the area of a triangle is Area equals 1/2 bh.*1152

*So, this is actually that the area equals one-half the base times the height of the triangle.*1158

*Find the height if a is 32, and the base (b) is 8.*1164

*OK, here we are being asked to find the height, and we are given the other two variables.*1173

*So, let me rewrite the formula: area equals 1/2 base times height.*1179

*Now, I am going to substitute in what I was given.*1183

*I am given the area; I am given the base; and I need to find the height.*1185

*What I need to do is isolate h; so, first simplifying this: 32=...well, 1/2 of 8 is 4, so that gives me...4h.*1195

*Next, divide both sides of the equation by 4 (32/4 and 4h/4) in order to isolate that.*1206

*Well, 32 divided by 4 is 8; the 4's cancel out on the right; and then just rewriting this in a more standard form, with the variable on left, the height is 8.*1216

*So again, first just write out the formula; substitute in a and b (which I was given).*1228

*And then, solve for the height.*1237

*The temperature in Fahrenheit is F=9/5C+32, where C is the temperature in Celsius.*1242

*If the temperature is 78 degrees Fahrenheit, what is it in Celsius?*1250

*Rewrite the formula and substitute in what is given.*1255

*What is given is that the temperature in Fahrenheit is 78.*1263

*And I am looking for Celsius (I always keep in mind what I am looking for--what is my goal?).*1267

*And that is +32; my goal here is going to be to solve for C--to isolate that.*1272

*Subtracting 32 from both sides gives me 46=9/5C.*1281

*Now, in order to isolate the Celsius, I am going to multiply both sides by 5/9.*1290

*When I do that, I am going to get 5 times 46 (is 230), and that is divided by 9.*1302

*Here, that all cancels out; so, rewriting this, Celsius equals 230/9.*1310

*That is not usually how we talk about temperature; so simplifying that, if I took 230 and divided it by 9, I would get approximately 25.5 degrees Celsius.*1317

*So again, the formula for converting Fahrenheit into Celsius (or vice versa) is given.*1327

*I substituted in 78 degrees and figured this out: so, 78 degrees would be equal to approximately 25.5 degrees Celsius.*1335

*That concludes today's lesson on Educator.com; and I will see you again for the next lesson.*1346

*Welcome to Educator.com.*0000

*In today's lesson, we will be discussing properties of real numbers.*0002

*A ***real number** corresponds to a point on the number line, and real numbers may be either rational or irrational.0007

*So first, just talking about the number line: some examples of real numbers would be 0, 1, 1/2,*0016

*or (expressing a number as a decimal) it can be 2.8; negative numbers--maybe -2.18 or -4.*0032

*OK, these are all real numbers, **and* they are rational; so let's talk about the difference between rational and irrational numbers.0042

**Rational numbers** can be expressed as a fraction.0051

*So, rational numbers are expressed in the form a/b, where a and b are integers, and b does not equal 0,*0062

*because as you will recall, we cannot have 0 in the denominator, because that would result in an undefined expression.*0075

*When a rational number is expressed as a decimal, it will either be terminating or repeating.*0084

*Let me explain what I mean by this: if I have an example, such as 1/2, and I convert that to decimal form, it is going to terminate; it is .5.*0097

*I might have another number, 1/3, which is also rational; and it is going to end up being .3333, and on indefinitely.*0108

*This could also be written as .3 with a bar over it, indicating it is repeating.*0122

*So, this is also rational, because it repeats; the repeating pattern could be longer--it could be 2.387387387, so this is repeating.*0126

*The point is that they either terminate or repeat when expressed in decimal form.*0139

**Irrational numbers** cannot be expressed in the form a/b, so they are expressed as decimal form.0145

*But they neither terminate nor repeat in decimal form; they just go on indefinitely.*0156

*For example, I could have 4.871469837246, and on and on and on, with no pattern--no repeating and no ending.*0174

*Another example would be π; we often express π as 3.14, but it actually goes on and on indefinitely; it is just approximately equal to 3.14.*0189

*Therefore, we can add π up here (that is an irrational number) to represent some irrational numbers up here.*0201

*In addition, the square root of numbers, other than the square root of numbers that are perfect squares, are irrational numbers.*0209

*So, the square root of 3 is, or the square root of 2.*0217

*If it is a perfect square, that means it is the result of multiplying an integer by itself.*0221

*For example, 2 times 2 is 4, so that is a perfect square; and the square root of 4 is rational--in fact, it just equals 2.*0227

*All other square roots are irrational numbers; so again, both rational and irrational numbers are real numbers, but they have different properties.*0234

*Sometimes, we express the relationship between the various types of real numbers using a Venn diagram.*0244

*So, I am going to go ahead and break down the real number system into its subsets, and then show you how this works as a Venn diagram.*0252

*We have the real number system, and we have irrational numbers, and we also have rational numbers.*0261

*And a Venn diagram is a visual way of understanding the relationship between these and their various subsets.*0282

*You could use circles; I am using rectangles and squares...whichever.*0295

*Irrational numbers: this is sometimes just known as Q with a line over it, and some examples that I just discussed--*0300

*the square root of 2, π, the square root of 5--these are irrational numbers.*0313

*And then, I have rational numbers, sometimes expressed as Q; OK, this is the real number system.*0319

*Within rational numbers are some subsets; the first one is the ***natural numbers**.0333

*And the natural numbers are the numbers that we use to count things (1, 2, 3, and on); they are the numbers we use in counting.*0344

*There is a slightly bigger set of numbers known as the ***whole numbers**.0354

*The whole numbers include the natural numbers (this square is around the natural numbers, because it includes all of those).*0362

*And it also includes 0; so, we add 0 to this set; and it does not include negative numbers.*0368

*Next are the ***integers**: the integers include natural numbers, whole numbers, and also negative numbers.0379

*So, 0 is included, and on.*0399

*OK, so then, you can come out to just rational numbers (that are not whole numbers, natural numbers, or integers).*0403

*And you can get fractions included, like -1/2, 0, 1, 2, 3/2, and on.*0411

*The real number system is broken down into rational and irrational; and rational numbers*0422

*are further broken down into the natural numbers, the whole numbers, and the integers.*0426

*In algebra, it is important to understand the properties of real numbers--what it is*0436

*that you are allowed to do, and not allowed to do, when working with real numbers.*0441

*So, we call the various properties commutative, associative, identity, inverse, and distributive.*0446

*Reviewing those: the ***commutative property** applies both to addition and to multiplication.0452

*And what the commutative property tells us is that two terms can be added in either order, or multiplied in either order.*0463

*So, if I have two numbers, a+b, I can change that order, b+a, and it is not going to change my result.*0471

*For multiplication, the commutative property under multiplication, I could say a times b equals b times a.*0478

*The ***associative property** also applies to both addition and multiplication.0487

*The associative property tells you that, when you are adding or multiplying, the terms can be grouped in any way, and the result will be the same.*0495

*Remember that we sometimes use grouping symbols with expressions and equations.*0504

*So, looking at this, I could group it as (a+b)+c, or I could group it as a+(b+c)--group those together.*0510

*Either way, they are equivalent; it is not going to change my result.*0522

*The same holds true for multiplication: if I have (ab)c, that equals a(bc), and it doesn't matter*0526

*if I decide to group it like this or like this--if I multiply these first or these first.*0536

*The ***identity property**: when I think of this, I just remember that the identity property tells me that the number maintains its identity.0544

*It doesn't change; so, for addition, what this says is that the sum of any number and 0 is the original number.*0553

*So, a+0 is still a; so, the identity of the number does not change just because you add 0 to it.*0565

*For multiplication, the product of any number and 1 is the original number; that is the identity property under multiplication.*0575

*So, a times 1 is still the original number, a.*0586

*The ***inverse property**, as applied to addition, says that, when you add the same number,0593

*but with the opposite sign (a + -a, or say, 3 + -3), the result is 0.*0607

*A number plus its additive inverse is equal to 0.*0617

*This property can also be applied to multiplication; and this only applies to real numbers other than 0, when applied to multiplication.*0625

*And you will see why.*0640

*What this says is that, if you multiply a number by its reciprocal with the same sign, the result will be 1.*0641

*We can't apply 0 here, because that would give us a 0 in the denominator, which is an undefined expression.*0650

*So, a number times its reciprocal gives you 1.*0657

*Finally, the ***distributive property**: you can review more about this in the Algebra I lectures.0663

*This is a very important property when working with equations, solving equations, and working with algebraic expressions.*0669

*Recall that a(b+c) equals ab + ac.*0677

*So, we go forward to multiply; and when we go in the reverse direction, recall that that is factoring.*0685

*This property also applies to multiplying a number by terms that are subtracting (ab-ac).*0691

*If you put the numbers you are adding first, it doesn't change the property--it still applies; you get ab + ac.*0704

*The same for subtraction.*0716

*And finally, recall that this property can be applied to several numbers.*0722

*You can have more than two numbers in the parentheses: this could be a(b + c + d).*0725

*And then, you just multiply each one out: ab + ac + ad.*0732

*And again, this is something we are going to be using a lot throughout the course.*0737

*All right, applying some of these concepts to the examples: Example 1: What sets of numbers do these belong to (starting with 6)?*0741

*Well, 6 is a real number; it is also a rational number (I can easily express this as a fraction: 6/1).*0750

*And so, it is a rational number; then, I think about the subsets.*0760

*Is it a natural number? Yes, it is a number that can be used in counting (1, 2, 3, 4, etc.), so it is a natural number.*0767

*And it is therefore also a whole number; the whole numbers encompass the natural numbers; and it is an integer.*0776

*6 belongs to all of these categories.*0786

*The square root of 20 is a real number; however, recall that, unless you are talking about the square root of a perfect square, it is irrational.*0789

*Square roots of perfect squares are rational; other square roots are irrational.*0799

*-4/5: this is expressed as a fraction, so...well, it is a real number; and it is also rational, because it can be expressed as a fraction.*0808

*It is not a natural number, because it is negative, and it is a fraction.*0821

*The same thing: it is not a whole number, and it is also not an integer.*0826

*So, this belongs to the two groups real and rational.*0830

*OK, Example 2 asks what properties are used: so, there is an expression here--a mathematical expression--and various steps are taken.*0838

*We need to determine which properties were used that allowed those steps to be taken.*0849

*Well, looking at what happened between here and here, we started out with 2, times 4 plus 3 plus 7.*0855

*The order of these was switched: 3+7 is still grouped together in parentheses, but it was put before the 4.*0863

*Remember that the commutative property is the property that says that, when adding,*0872

*you can change the order that you are adding terms in, and still get the same result; so this is the commutative property.*0877

*OK, the next step: I look at what happened, and this big set of parentheses is gone.*0884

*And the way it was removed is by use of the distributive property.*0891

*Recall that the distributive property says that a, times (b + c), equals ab + ac; and that is what was done here.*0894

*2 times the whole expression (3+7), and then 2 times 4; this is the distributive property.*0904

*In the next step, the order of these two numbers was changed; so that is commutative.*0919

*And a 0 was also added to 4; and remember that, according to the identity property,*0926

*you can add 0 to a number, and the original number is unchanged (4 + 0 is 4).*0931

*So, this is commutative and identity.*0937

*Finally, it is getting rid of the rest of these parentheses by using the distributive property: 2 times 7, plus 2 times 3, plus 2 times 4, plus 2 times 0.*0940

*So again, we are using the distributive property.*0951

*Example 3 asks what is the multiplicative inverse of -6 and 7/8.*0961

*Recall: ***multiplicative inverse**, the definition, is a number times the reciprocal of that number; and recall that that equals 1.0967

*OK, so the multiplicative inverse--I have -6 and 7/8, so I am going to change this from a mixed number to a fraction.*0979

*6 times 8 is 48, plus 7 is 55; and this is negative, so that is -55/8.*0987

*So, looking at it as a fraction makes it much simpler; and I just need to take the multiplicative inverse of that.*0996

*And so, I would change that to -8/55.*1007

*And it does satisfy this formula right here, because if I took -55/8 (which is my original number),*1012

*and I multiplied it by -8/55, two negatives (a negative times a negative) gives me a positive;*1020

*the 8's cancel; then, the 55's cancel to give me 1.*1028

*So, I was able to check my work by seeing that it satisfies this equation.*1033

*OK, in Example 4, we are asked to simplify and state the property used for each step of simplification.*1038

*First, I want to get rid of the parentheses; so I am going to use the distributive property.*1044

*Multiplying out, recall that the distributive property is: a(b+c)=ab+ac.*1051

*OK, this is 2(6x) + 2(3y + 4z).*1059

*So, right now, I am just removing these outer parentheses; I am keeping these intact--that will take a second round.*1068

*So, this is plus -3, times the entire expression in the parentheses, plus -3, times z.*1074

*OK, I am going to apply the distributive property again, in order to remove the remaining parentheses.*1084

*This gives me: 2(3y), plus 2(4z), plus -3(3x), plus -3(-y), plus -3(z).*1095

*Now, I am going to multiply these out: this is 12x + 6y + 8z - 9x + (a negative and a negative is a positive, so that gives me) 3y - (this is -3z).*1119

*OK, now I am going to group together like terms: and that is using the commutative property--I can change the order of these terms.*1145

*I have my x's, 12x-9x; I have my y's, and that is 6y and 3y; and then finally, z's: 8z and -3z.*1157

*All that is left to do is add like terms; so, 12x-9x is 3x; 6y and 3y gives me 9y; and 8z-3z is 5z.*1179

*So, we are simplifying this, using first the distributive property (to remove the outer parentheses),*1193

*then the distributive property to remove these other sets of parentheses,*1198

*and the commutative property to re-order this to group like terms, and then simply adding or subtracting.*1203

*That concludes this lesson from Educator.com; see you next lesson.*1211

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be talking about solving equations.*0002

*Recall that verbal expressions can be translated into algebraic expressions, and vice versa.*0007

*So, you can go from verbal expressions to algebraic, and from algebraic back to verbal.*0014

*For example, if I were given a sentence (a verbal expression) such as, "The sum of two numbers is equal to 7,"*0019

*I could translate that into an algebraic expression.*0041

*And the technique is to first assign variables; so I am looking, and I see that I have two numbers.*0043

*And they don't tell me what the numbers are, so I need to assign variables; and I am going to assign x and y.*0049

*This is actually an equation, not just an expression; so if I see that I have an equal sign (like this is equal),*0057

*I note where that is, because that is going to divide it into the left and right sides of the equation.*0064

*So, I have an equal sign; I am going to first deal with one side of the equation, and it says "the sum of two numbers."*0068

*So, sum means adding; and my two numbers I am going to call x and y; "is equal to 7."*0075

*And then, double-check: the sum of two numbers (x + y--that works out) is equal to (I have that taken care of) 7.*0084

*Or another example: A number is 5 less than the square of another number.*0094

*And you need to be careful with this "is less than," because it is easy to get that backwards.*0113

*So, first I notice that I have two numbers; I have "a number" and "another number."*0118

*Once again, I am going to use x for one of the numbers; and for the other number, I am going to use y; so x and y are my variables.*0123

*Now, it says "a number is," so I have "is"--that tells me where the equal sign is.*0132

*So, the left side of the equation is just a number.*0141

*With "5 less than" or "something less than," the temptation is sometimes to say "5 minus."*0146

*But that is not correct; what this is saying is to look at what comes after (the square of another number) and subtract 5 from that.*0152

*So, a number is, and then 5 less than...what is it 5 less than? the square of another number.*0161

*I am calling my other number y, so y*^{2} - 5, *not* 5 - y^{2}.0168

*So, checking it: A number (x) is (=) 5 less than the square of another number (y*^{2} - 5).0176

*And later on in the examples, we will also see a situation where we are given an algebraic expression and asked to translate it into a verbal expression.*0188

*As you are solving equations, you need to keep in mind the properties of equality that you have used previously (but it is good to review).*0201

*The first property is the ***reflexive property**: the reflexive property states that, for every real number a, a is equal to a.0209

*So, a number is equal to itself.*0219

*The ***symmetric property** states that, for all real numbers, if a equals b, then b equals a.0222

*So, I can switch the left and right sides of the equation, and I am not going to change anything fundamental.*0233

*So, those are the reflexive and symmetric properties.*0239

*Another property is the ***transitive property**: the transitive property states that, if a equals b, and b equals c, then a equals c.0241

*Think of it this way: if a actually was 5 (let a equal 5), if I told you that a equals b, then the only way it is going to equal b is if b is also 5.*0257

*So, b must be 5; and then, if I went on and said, "OK, well, b equals c," if that is 5, then c must also be 5.*0270

*So, I go back; then 5 equals 5, or a equals c--it just follows through.*0277

*OK, that is the first three properties discussed up here.*0284

*But equality also satisfies the addition, subtraction, multiplication, and division properties.*0288

*And these are properties that you have used before in Algebra I; and again, you can review these in detail in the Algebra I lessons.*0293

*To refresh them here: the ***addition principle** says that, if the same number is added to both sides of the equation, the resulting equation is true.0299

*For example, if I have x - 5 = 10, and I want to solve for x, the way I am going to do that is: I am going to add 5 to both sides.*0309

*And these 5's cancel out; that will give me x = 15.*0325

*So, the resulting equation after I add 5, if I add 5 to both sides--this equation is also true.*0329

*The subtraction principle is the same idea, that if I have an equation (x + 3 = 5), I can subtract 3 from both sides;*0338

*I can subtract the same number from both sides, and the resulting equation is also true.*0357

*Multiplication principle: again, this is something you have used before to solve equations.*0363

*And it states that, if the same number is multiplied by both sides of the equation, the resulting equation is true.*0367

*So, I might have x/2 = 12; to solve that, I need to multiply both sides by 2; and that is allowable.*0376

*Division: the same idea--in this case, I may have 4x = 6; to solve that, I am going to divide both sides by 4.*0391

*So, the important thing is: if you do something to one side, you need to do the same thing to the other side of the equation.*0406

*OK, knowing those properties and the properties of real numbers (the properties of equality and the properties of real numbers), you can solve equations.*0417

*And we are going to use these properties frequently, even if you don't always know them by name.*0427

*You know how to do them, and you are going to be applying them throughout the course.*0432

*For example, if I have 19 = 3x + 4, since usually we have the variable on the left,*0437

*I can apply the symmetric property and just change this to 3x + 4 = 19.*0447

*That is the symmetric property that allows me to flip around the two sides of the equation.*0451

*Next, I need to isolate x, so I am going to subtract 4 from both sides.*0459

*And that utilizes the subtraction property, which is going to give me...subtracting 4 from both sides, I get 3x; the 4's cancel out, and I get 3x = 15.*0465

*Well, now I need to isolate x with one last step; and I am going to do that by using the division property.*0480

*If I divide both sides by 3, I am using the division property; and that is going to give me...the 3's will cancel out, and I will get x = 5.*0487

*So, by applying those principles, I was able to solve this simple equation.*0499

*Sometimes, in algebra, you will be asked to solve for a variable.*0505

*These properties can be used to solve a formula for a variable.*0509

*I may be given a formula such as 4x - 2y + z = 8.*0514

*And I might be asked to solve for z; so what I am going to do is solve for z in terms of x and y.*0521

*While I may not figure out the actual numerical value of z, I can solve for it; I can isolate it and solve for it in terms of the other variables.*0532

*So again, I am going to apply these same principles.*0541

*First, I am going to use the subtraction principle; and I can start out by subtracting 4x from both sides.*0545

*That is going to give me...these 4x's will cancel out; -2y + z = 8 - 4x.*0561

*Now, I need to add 2y to both sides, because I had -2y; so then, I am going to add 2y to both sides,*0570

*so that my 2y's cancel out to give me z = 8 - 4x + 2y.*0590

*And we usually write this so that we have the x's, then the y (in alphabetical order), and then the constant.*0600

*So, z equals -4x + 2y + 8; this is solving for z in terms of x and y.*0607

*First example: write an algebraic expression for the sum of three times the cube of one number and twice the square of another number.*0617

*OK, starting out, I am figuring out how many variables I have.*0627

*The sum of three times the cube of one number and twice the square of another number...x an y.*0632

*I am going to assign those as my variables, because I have two variables.*0640

*And there is no equal sign; this is an expression, not an equation, so I don't have an equal sign in it.*0643

*And it says "the sum," so I am doing addition.*0651

*"The sum of three times the cube of one number"--that is 3x*^{3},0656

*"and"--so that tells me that that is where the addition goes--"twice the square of another number";*0663

*so here is the other number: "twice the square of the other number."*0670

*Checking that: the sum of three times the cube of one number (3x*^{3}) and (+) twice--two times the square of another number (2y^{2}).0675

*Next, we are asked to do the opposite, which is to write a verbal expression for this algebraic expression.*0692

*OK, so looking at what I have: here, first, I have the difference of two squares;*0699

*and I am also looking and seeing that I have two variables.*0706

*So, the difference of the square of a number (x--I am just calling it "a number") and the square of another number (that is my y*^{2})...0709

*Next, I have an equal sign: so "is equal to"...*0737

*the sum of the square (we have sum--we are adding), and I want to make it clear*0743

*that we are talking about the same number here and here, so I am going to say "the square of the first number,"*0759

*and (I am adding again) 3 times (multiplying three times x times y) the product of*0770

*(my first number and my second number) the first number and the second number, plus the square of the second number.*0787

*OK, so this was quite long, but you can check it.*0820

*"The difference of the square of a number and the square of another number" (I have that right here)*0823

*"is equal to" (there is my equal sign) "the sum" (it's a sum) "of the square of the first number,*0827

*and three times the product of the first number and the second number, plus the square of the second number."*0833

*So, it is pretty long, but we got everything covered clearly.*0839

*The third example is to solve: and we can use the properties that we discussed today in order to solve this.*0845

*First, I am going to need to get rid of these parentheses; and I can do that by using the distributive property.*0852

*So, -2 times x gives me -2x; plus -2, times -8, minus...actually, let's do plus -3 times 9, plus -3 times -2x, equals 4 times 2x plus 4 times -9.*0858

*So, just checking: I have -2x + -2(-8) + -3(9) + -3(-2x) = 4(2x) + 4(-9).*0892

*Multiplying everything out to simplify: -2x; and then -2 times -8 gives me +16.*0909

*And then, I have -3 times 9; that is actually a negative, -27.*0919

*-3 times -2x; that is positive, +6x.*0928

*That equals...4 times 2x is 8x; 4 times -9 is -36.*0934

*The next step is to combine like terms; and I have -2x and 6x--those can be combined; that is going to give me 4x.*0941

*I have constants: I have 16 - 27, which is going to give me -11.*0951

*I can't really do anything further with that right side.*0957

*Now that I am this far, what I want to do is isolate the x (isolate the variable); I can do that by using my addition property.*0961

*I am going to add 11 to both sides; the 11's cancel out--that is going to leave me with 4x = 8x, and -36 + 11 is -25.*0969

*Next, I am going to subtract 8x from both sides--again, trying to get like terms together and isolate the x.*0984

*4x - 8x is going to give me -4x; equals...these cancel; that is -25.*0994

*Now, I am going to divide both sides by -4 to get x = -25/-4.*1003

*And the negative and the negative is a positive, so x = 25/4; I can either leave it like that, or I can go on and say this equals 6 and 1/4.*1011

*So again, starting out, getting rid of the parentheses by using the distributive property and carefully multiplying out each section got me to here.*1020

*We're adding like terms to get to this step, then isolating the x through using the addition property,*1031

*the subtraction property, and the division property to get x = 6 1/4.*1039

*OK, in this example, we are asked to solve for h.*1047

*And we are given...this is actually the surface area of a cylinder; the surface area of a cylinder*1050

*equals 2π, times the radius, times the height, plus 2πr*^{2}.1055

*So, I need to solve for h in terms of these other variables.*1061

*And I am going to start out by using the symmetric property to rewrite this with the h on the left.*1069

*Let me rewrite this as 2πrh + 2πr*^{2} = s.1077

*To get the h by itself, I can start out by subtracting 2πr*^{2} from both sides.1083

*OK, once I have done that, then this is going to be gone from this side; and I will have 2πrh.*1095

*2πrh = s - 2πr*^{2}.1106

*In order to get the h by itself, I need to divide both sides by 2πr.*1111

*2πr will cancel out, leaving h isolated on the left, and s - 2πr*^{2} all divided by 2πr.1121

*And you really can't go any farther with this.*1130

*I started out by rewriting this with the value, the h, which we are trying to isolate on the left,*1132

*using the subtraction property and the division property to isolate the h.*1140

*That concludes this session of Educator.com, and I will see you next lesson.*1147

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be covering solving absolute value equations.*0002

*Recall that the absolute value x, written as |x| (this is the symbol for absolute value), is the distance from x to 0 on a number line.*0009

*For example, if you have the absolute value of x equals three, what you are saying is*0022

*that the absolute value of this number is 3 away from 0 on the number line (1, 2, 3).*0036

*OK, so looking at this: if I go from 0 to 3, this is 3 units away from 0 on the number line; therefore, x could equal 3,*0052

*because 3 is 3 away from 0 on the number line.*0070

*However, consider this: -3 is also 3 units away from 0 on the number line.*0074

*So, x could also equal -3; so, if the absolute value of x equals 3, x could be 3 (since the absolute value of 3 is 3),*0081

*and x could be -3 (since the absolute value of -3 is also 3).*0093

*Knowing this, and really understanding this definition, will allow you to solve equations involving absolute value.*0102

*Again, absolute value equations: you can solve equations containing absolute values, using the definition of absolute value.*0110

*For example, |9x + 2| = 29; well, I already said that, if the absolute value of x is 3, that means that x equals 3, or x equals -3.*0121

*So, you can apply that same concept right here: 9x + 2 = 29, or 9x + 2 = -29.*0134

*So, once you remove the absolute value bars (and you can do that by turning it into two related equations,*0144

*one where it equals the positive, and one where it equals the negative value)--once you have done that,*0151

*all you need to do is solve each equation.*0155

*So, I am going to solve, just using my usual techniques: subtract two from both sides, and that gives me 9x = 27.*0159

*Over here, if I subtract 2 from both sides, I am going to get 9x = -31.*0169

*Then, I am going to divide both sides by 9.*0176

*OK, so you handle the absolute value equations the same way, even if it is more complex.*0186

*The other thing to keep in mind is that sometimes, you have to first isolate the absolute value expression on the left side of the equation.*0193

*For example, if you were given 3 times |5x + 4| equals 6, the first step would be to divide both sides by 3,*0199

*because once you have done that, then you have the absolute value isolated,*0210

*and you can proceed by saying 5x + 4 = 2 or 5x + 4 = -2, and then solving both of those.*0214

*In some situations, some absolute value equations, where the absolute value equals c, if c is less than 0, these have no solution.*0227

*So, if it says that the absolute value of x is a negative number, then there is no solution; and we just say that the solution is the empty set.*0236

*And to make this more concrete: if I said that the absolute value of x equals -4...well, there is no situation*0249

*where an absolute value is going to be a negative number, because remember: the absolute value is defined*0258

*as the distance between that absolute value and 0 on the number line.*0266

*And you can't have negative distance; so there is no solution here.*0271

*And instead, we either just write the empty set as such, or like this, indicating that there is no solution.*0275

*We just talked about a situation where there is no solution.*0287

*An absolute value equation can have 0 solutions (which we just discussed--it is the empty set), 1 solution, or 2 solutions.*0290

*And it is important to check each answer to make sure that it is a valid solution.*0298

*For example, if I have |x + 4| = 9, I am going to go ahead and solve that,*0304

*using the usual technique of turning it into two related equations, x + 4 = 9 and x + 4 = -9.*0311

*I am going to solve that: x = 5; here, I am going to get x = -13.*0321

*So, let's check this by going back to the original, |x + 4| = 9.*0329

*If x = 5, then I am going to get |5 + 4| = 9; so, the absolute value of 9 equals 9--and that is true.*0335

*Since this is true, this is a valid solution; x = 5 is a valid solution.*0345

*OK, doing the same thing for my other solution: |x + 4| = 9...*0350

*trying this |-13 + 4| = 9, well, -13 + 4 is -9, so the absolute value of -9 is 9.*0358

*This is also a valid solution, since this is a true statement.*0368

*So here, I have two solutions: previously, we discussed that, if you end up with something like |x| = -6, there is no solution; it is the empty set.*0372

*The situation where you can get one solution is if you end up with |x| = 0.*0390

*So, if I went to solve this, I would say, "x equals +0, and x equals -0"; but that doesn't really just make sense; they are just 0.*0400

*Therefore, there is only one solution: x = 0; so here, I have one solution.*0408

*Three possibilities: no solution, one solution, or two solutions.*0414

*Or you may think you have two solutions, and then you go back and plug them in, and find out one is not valid.*0419

*And in that case, you could have something like this, that appears that it would end up with two solutions,*0425

*and it only ends up with one, or possibly even none.*0430

*OK, Example 1: Evaluate for x = -3; and this is multiple absolute value terms here--*0437

*substituting in -3 for x, solving for the absolute value of this would be 3 times -3...that is going to give me |-9 - 4|;*0447

*here, I have 2 times -3; that is |-6 + 3|; minus 3 times...the absolute value of a negative, times a negative, is a positive.*0469

*OK, this is |-9 - 4|; that is the absolute value of -13, plus |-6 + 3|; that is -3; minus 3...absolute value of 3.*0480

*Now, I just need to find the absolute value for each of these.*0493

*Well, the absolute value of -13 is 13; the absolute value of -3 is 3; the absolute value of 3 is 3;*0497

*but this time, we are multiplying it times a -3; this is -3 times |3| (which is 3).*0512

*So here, it's 13 + 3...-3 times 3 is -9, so 13 + 3 is 16, minus 9 is 7.*0523

*OK, so we are solving this by substituting in -3 for x, finding the absolute values for these three, and then this one is multiplied by -3, and then simply adding.*0536

*Example 2: I have an absolute value expression on the left, but it is not isolated.*0550

*So, the first step is to isolate the absolute value, and then find the two related equations and solve them.*0555

*Divide both sides by 4 to get that isolated.*0562

*Now, recall that the absolute value of x equals 2, for example: this means that x could equal 2,*0570

*or x could equal -2--to just illustrate the definition of absolute value.*0578

*In order to get rid of these absolute value symbols, I am going to create two related equations,*0585

*3x + 4 = 12, or 3x + 4 = -12: then I am going to solve both, and check to make sure that they are valid solutions.*0590

*3x = 8; so divide both sides...I had 3x + 4, so I subtracted 4 from both sides.*0602

*And then, over here, I have 3x = -16, subtracting 4 from both sides--I am just doing these together.*0619

*Now, dividing both sides by 3 is going to give me x = 8/3; dividing here, I get -16/3.*0631

*Now, check each of these in the original: that is this 4 times the absolute value of 3x + 4 equals 48.*0641

*First checking this one: 4, and this is 3 times 8/3 + 4, equals 48...so the 3's cancel out, and that gives me 8 + 4...equals 48,*0652

*so 4 times the absolute value of 12 equals 48; the absolute value of 12 is 12, so 4 times 12 equals 48; and it does, so this is valid.*0672

*This first solution is valid.*0685

*Now, substituting in -16/3, again, in this original: that is 4 times |3(-16/3) +4| = 48.*0688

*The 3's cancel out; that gives me 4|16 + 4| = 48.*0704

*-16 +4 is -12...equals 48...the absolute value of -12 is 12; so again, I come up with 4 times 12 equals 48, and that is valid.*0710

*That is a true statement; so both of these solutions are valid.*0724

*So this time, I had two solutions, 8/3 and -16/3, that satisfy this absolute value equation.*0729

*Example 3: again, my first step is to isolate the absolute value expression on the left side of the equation.*0740

*And I am going so start out by subtracting 15 from both sides, and that is going to give me -3 on the right.*0748

*Now, I want to divide both sides by 3, and that is going to give me |2x - 4| = -1.*0756

*And I don't even need to go any farther, because what this is saying is that*0766

*the absolute value of whatever this expression ends up being (once I solve for x)--the absolute value of this equals -1.*0769

*Well, that is not valid; you cannot have an absolute value equal a negative number, because it violates the definition of absolute value.*0778

*Therefore, I don't even need to go any farther; I can just say that there is no solution, or that it is the empty set.*0789

*So, there is no solution to this absolute value equation.*0795

*The important thing is to just look carefully at your work and make sure you didn't make any mistakes.*0802

*And if you did all the math correctly and handled this correctly, and you come up with something like this, then you didn't do anything wrong.*0806

*It is just that there is no solution.*0813

*OK, another absolute value expression: this time, the absolute value is already isolated on the left.*0817

*So, looking at this, it looks more complex; but we just use the same logic that we did with the simple case.*0824

*If the absolute value of x is 2, x equals 2, or x equals -2.*0834

*So, I do the same thing here: I get rid of the absolute value bars, and this is my positive permutation.*0838

*And then, I also have 2x - 7 = -(3x + 8).*0846

*OK, solving each of these: I am going to add 7 to both sides; that is going to give me 2x = 3x +15.*0854

*Then, I am going to subtract 3x from both sides, which is going to give me -x = 15.*0867

*I am going to multiply both sides by -1 to get x = -15.*0876

*So, that is my first solution: solving this...this is 2x - 7 = -3x - 8.*0884

*Adding 7 to both sides is 2x = -3x -1.*0893

*Adding 3x to both sides: 5x = -1; divide both sides by 5: x = -1/5.*0900

*OK, check: with absolute value equations, you always have to check your solutions.*0909

*So, checking this back in the original equation: 2 times -15, minus 7--the absolute value of that--equals 3 times -15, plus 8.*0917

*OK, so I have 2(-15), which is -30, minus 7, equals 3(-15), which is -45, plus 8.*0931

*This gives me |-37| equals...well, -45 + 8 is -37; the absolute value of -37 is 37.*0945

*Well, 37 does not equal -37, so this is not a valid solution; this is not true--this did not satisfy this equation.*0959

*So, x = -15 is not a valid solution; let's try this one, x = -1/5, substituting it in here.*0970

*|2(-1/5) - 7| = 3(-1/5) + 8: that is going to give me |(-2/5)-7| = (-3/5) + 8.*0977

*So then, adding these two together, I am going to get |-7 2/5| = 8 - 3/5...is 7 2/5.*0999

*OK, the absolute value of -7 2/5 is 7 2/5, equals 7 2/5; and that is true; this is a valid solution.*1010

*So, I actually have only one solution to this equation, and it is x = -1/5.*1022

*We are starting out by breaking this into two related equations and removing the absolute bar,*1030

*solving each, getting two solutions, and then checking and finding out that the first one is not valid, and that the second one is valid.*1036

*That concludes this session for Educator.com, and I will see you for the next Algebra II lesson.*1046

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be discussing solving inequalities.*0002

*Recall the properties of inequalities: these are needed to solve inequalities.*0008

*The first one is that, if a is greater than b, and you add the same number to both sides, this inequality has the same solution set as the original inequality.*0015

*And you can think of this using numbers to make it clearer.*0030

*If I said that 6 is greater than 5, and then I decided to add 2 to both sides, 8 is greater than 7; so this still holds up.*0035

*You are allowed to add the same number to both sides of an inequality; and that is true for both greater than and less than.*0048

*And it is also true for greater than or equal to and less than or equal to.*0056

*The subtraction property states that, if a is greater than b, and I subtract the same number from both sides,*0064

*then the resulting inequality has the same solution set as the original inequality.*0072

*The same idea as up here: if 6 is greater than 5, and I subtract 3 from both sides, I am going to end up with 3 > 2; that still holds up.*0079

*And that also is valid for less than, and for greater than or equal to and less than or equal to.*0093

*With multiplication properties, if a is greater than b, and a positive number (c > 0), the same positive number,*0104

*is multiplied by both sides of the inequality, then the solution set of the resulting inequality is the same as that of the original inequality.*0113

*If I have 3 > 2, and I multiply both sides by 4, I am going to end up with 12 > 8, which is still valid.*0125

*The same for less than: a < b--you are allowed to multiply both sides of the inequality by the same positive number, without changing the solution set.*0142

*Again, this applies to greater than or equal to and less than or equal to, as well as to strict inequalities.*0152

*The case is different for negative numbers: if c is less than 0 (if you were multiplying both sides of an inequality*0161

*by a negative number), you must reverse the inequality sign.*0167

*If both sides of an inequality are multiplied by the same negative number, the direction of the inequality must be reversed.*0172

*And if you do that, then the resulting inequality has the same solution set as the original inequality.*0179

*If you don't reverse them, the solution set will not necessarily be the same.*0184

*For example, if I have 8 > 4, and I multiply both sides by -2, let's say I didn't reverse the sign:*0190

*then, I am going to end up with -16 > -8; and this is clearly not true.*0205

*So, as soon as I multiply by a negative number, this is incorrect; I immediately have to reverse the direction of the inequality.*0212

*And that will give me -16 < -8, which is valid; and again, the same for greater than or less than or equal to.*0226

*So, multiplying by a positive number, the solution set of the resulting inequality is the same as the original.*0235

*Multiplying by a negative number, you need to reverse the inequality sign in order for the solution set for this inequality to be the same as for the original.*0241

*A similar idea with division: if you are dividing both sides of an inequality by a positive number,*0251

*the resulting inequality has the same solution set as the original.*0259

*That is only if it is for a positive number; so 15 > 20--if I wanted to divide both sides by 5, I could do that.*0264

*And I don't need to do anything to the inequality; I just keep it the same.*0275

*And that is going to give me 3 >...well, I need to start out with an inequality that is valid!*0279

*So, if 20 > 15, and then I divide the same number by both sides...so I am starting out with 20 > 15;*0288

*20 divided by 5 is greater than 15 divided by 5, which is going to give me 4 > 3, which is true.*0305

*The same holds up for less than, and greater than or equal to/less than or equal to.*0314

*Now, if we are dealing with dividing by a negative number, you have to reverse the inequality sign.*0320

*And then, that resulting inequality is the same solution set as the original inequality.*0325

*So, if I have 4 < 6, and I divide both sides by -2, I need to immediately reverse this inequality symbol.*0330

*And then, I am going to get 6 divided by -2, and I am going to end up with -2 > -3; and that is valid.*0344

*If I hadn't reversed this, I would have gotten something that is not valid, or does not have the same solution set as the original.*0354

*So, you need to be very careful, as soon as you multiply or divide by a negative number,*0360

*when you are working with inequalities, that you reverse the inequality symbol.*0366

*There are multiple ways of describing the solution set of an inequality.*0371

*You can either express it as a graph, a set, or simply as an inequality.*0379

*For example, if I came up with the solution set x ≥ 2, I could just leave it as an inequality like that; that is a little less formal.*0386

*I could describe it as a set, using set builder notation.*0396

*And I would put it as follows: what this is saying is "the set of all x, such that x is greater than or equal to 2."*0400

*So, this is set builder notation: the set of all x, such that x is greater than or equal to 2.*0414

*You will see the same things, sometimes, with two dots here; this is more common, but you will see this sometimes; and it means the same thing.*0420

*You can also use a graph on the number line; you can express it as a graph.*0429

*And remember that, if you are saying "greater than or equal to," you are including this number 2 in the solution set.*0436

*So, in that case, you would want to use a closed circle, and then continue on to the right.*0443

*Now, let's say I was going to say x is less than 3.*0450

*In this case, it is a strict inequality; and 3 is not part of the solution set, so I am going to use an open circle.*0458

*So, you can express an inequality's solution set either using set builder notation, using a graph, or less formally, just as a simple inequality.*0465

*Looking at Example 1, we have -3x < -27, so I need to solve that.*0478

*I need to isolate this x; and in order to do that, I am going to need to divide by -3.*0489

*But as soon as I start thinking about dividing by a negative number, I have to reverse the inequality symbol.*0494

*So, instead of less than, this becomes greater than.*0503

*This is going to give me x > 9.*0507

*I can leave it just as an inequality; I can use set notation to show this; I can also graph it on the number line.*0511

*OK, and graphing it on the number line, I am going to use an open circle (since it is a strict inequality)*0532

*and show that x is greater than 9, but that 9 is not part of the solution set.*0538

*Solve -4x - 7 ≥ 9: the first step is to add 7 to both sides, to get -4x ≥ 16.*0547

*Now, to isolate x, I need to divide both sides by -4; and as soon as I do that, I am going to change this direction to less than or equal to.*0561

*16 divided by -4 is -4, so the solution is x ≤ -4.*0570

*Using set notation, and graphing on the number line--multiple ways...-1, -2, -3, -4.*0578

*And this is saying "less than or equal to"; therefore, -4 is going to be included in the solution set, as indicated by a closed circle.*0596

*OK, in Example 3, it is a little bit more complicated; but you just handle it using the same principles.*0611

*When you are dealing with fractions, the best thing to do is to get rid of them first, because they are difficult to deal with.*0621

*So, I am going to multiply both sides of the equation by -9, and (since that is a negative number)*0628

*immediately reverse the inequality symbol, so that you don't forget to do that.*0633

*This is going to be -9 times -2 times 3x - 6; I'll move this over a bit...and that is +(-9), times 4x, divided by -9.*0637

*It is greater than -3 times -9.*0657

*OK, so -9 times -2 is 18, times 3x - 6...these -9's cancel out, and that is going to give me...+4x, is greater than (-3 times -9 is) 27.*0660

*Using the distributive property, I am going to multiply this out; and 18 times 3x is 54x; 18 times -6 is -108; plus 4x, is greater than 27.*0677

*Now, I need to isolate this x; so first, I am going to combine like terms.*0696

*I have 54x + 4x, is 58x, -108 is greater than 27.*0702

*Adding 108 to both sides will give me 58x > 27 + 108, so that is 58x > 135.*0712

*So, x is greater than 135/58; I can leave it like that, or I can use set notation.*0729

*To graph this, you need to figure out...if you divide 135/58, it is a little bit larger than 2; it is approximately 2.3.*0746

*So, I could go ahead and graph that out, as well.*0754

*2 is here; 2.5 is about there; 2.3 is about right here; open circle at 2.3, or actually right over here;*0762

*1 is here; 2 is here; 2.5 is about here; so, 2.3 is going to be right about here; and that is open circle, like that.*0771

*There are three ways to express the solution.*0788

*OK, in Example 4, again, it is a little bit more complex, and there is a fraction, so we are going to get rid of that first.*0794

*Start out by multiplying both sides by -7, which tells me that I immediately need to reverse the sign.*0803

*-7 times -3 times (x - 4), over -7, plus -7 times 3 times (9 - 2x); reverse the inequality symbol--greater than or equal to -7 times (4 - x).*0811

*OK, this cancels; that gives me -3 times (x - 4), and this is -7 times 3, so that is -21, times (9 - 2x), is greater than or equal to...*0843

*-7 times 4; that is -28; -7 times -x is plus 7x.*0860

*So now, I just need to do some more simplification.*0868

*-3 times x is -3x; -3 times -4 is +12; -21 times 9 is -189; -21 times -2x is +42x.*0872

*And this is -28 + 7x; I can't really do anything with that.*0888

*Now, I am going to combine like terms (and I do have some like terms).*0905

*I have a -3x and 42x; combining those, I am going to get 39x; -189 +12 is -177.*0908

*The next step is to isolate the x, so I am going to add 177 to both sides; that is going to give me 177 - 28 + 7x.*0921

*Subtract 7x from both sides: 39x - 7x ≥ 177 - 28.*0934

*Combine like terms to get (39x - 7x is) 32x ≥ 149.*0945

*Now, I am going to divide both sides by 32; and it is a positive number, so it just becomes x ≥ 149/32.*0956

*Using set notation, x is greater than or equal to 149/32.*0970

*You could also graph this; this is approximately equal to 4.7, so 0, 1, 2, 3, 4, and 5;*0977

*it is going to be a little over halfway between 4 and 5; closed circle; and graph it.*0989

*OK, so the first step was to eliminate the fraction by multiplying both sides of the equation by -7 and reversing the inequality symbol.*0996

*Once that was done, we are using the distributive property to multiply everything out and get rid of the parentheses.*1005

*Then, we are using the addition and subtraction principles and combining like terms to simplify this inequality.*1012

*And the result was x ≥ 149/32.*1021

*That concludes this lesson on solving inequalities at Educator.com; see you again!*1030

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be solving compound and absolute value inequalities.*0002

*A compound inequality consists of two or more inequalities combined by either "and" or "or."*0009

*To solve a compound inequality, solve each part.*0016

*For example, if you are given 4x - 2 < 10, and x - 1 ≥ 4, it is connected by the word and, so it is a compound inequality.*0022

*So, I am going to solve each part of that.*0040

*4x...add 2 to both sides; that would be less than 12; divide both sides by 4, so x is less than 3.*0044

*And if I add 1 to both sides, I am going to get that x is greater than or equal to 4.*0054

*We will talk, in a second, about how the solution set works for this; but the first step is to just solve both parts of the inequality.*0063

*The same idea applies if an inequality is a compound inequality connected by the word or.*0074

*For example, you might be given 3x > 6, or 4x < 4.*0080

*Again, solve each part: x > 2, or x < 1.*0090

*Breaking this down and talking about the solution sets for each: if "and" is used, the solution set is the intersection of the solution sets of the individual inequalities.*0103

*So, I am going to review this here--the idea of intersection--and this is also covered in detail in the Algebra I series.*0113

**Intersection** means the common members of both solution sets.0121

*Just looking at numbers: if I have 2, 4, 6, 8, and 10, and that is my first set; and then I have a second set*0127

*that is 6, 8, 10, 12, and 14; the intersection would be the elements common to both.*0136

*So, I see that I have 6 in both; 8; and 10; so the intersection would be 6, 8, and 10.*0149

*Apply that to inequalities connected by the word and: for example, x - 4 < 10, and 2x ≥ 4.*0158

*OK, solve each part as discussed; so add 4 to both sides...this means that x is less than 14;*0173

*and (both conditions must hold) if I divide both sides by 2, I get x ≥ 2.*0179

*To visualize this, look at the number line: if I have a 0 right here, and then 2, 4, 6, 8, 10, 12, 14,*0187

*this is telling me that x is less than 14; and that would just go on and on and on.*0200

*This is telling me x is greater than or equal to 2, which is going to start here and go on and on and on and on.*0207

*But I am looking for the intersection, or the common elements; and the common elements would be greater than or equal to 2, and less than 14.*0215

*So, the intersection would come out as 0, 2, 4, 6, 8, 10, 12, 14--greater than or equal to 2, and less than 14.*0225

*And you could write that out as an inequality, or using set notation, that x is greater than or equal to 2, and less than 14.*0240

*So, this is actually just a more efficient way of writing, instead of two separate sections here with the word and;*0255

*you can just write it out more efficiently like this; x is greater than or equal to 2 and less than 14; or showing it on the number line like this.*0262

*So, that is for inequalities joined by the word and.*0271

*If or is used in a compound inequality, the solution set is the union of the solution sets of the individual inequalities.*0276

*Now, reviewing what a union is, just using numbers: if you have a set, such as 4, 5, 6, 7, 8; and then you have another set,*0285

*-2, -1, 2, 3, 4, 5; and you are asked to find the union; well, the union is any elements that are in either one of these, or both.*0298

*So, if it is in one; if it is in the other; or if it is in both; that would be the union.*0315

*Now, I am looking, and I have 4, and that is in both; I don't need to write it twice; 5; 6 is just in this one, but it is included;*0321

*7, just in that one; 8; and then, I already covered 4 and 5, but I also have to include -2, -1, 2, and 3.*0332

*And I could rewrite this in ascending order, but it is all included here.*0344

*The union means that it includes the elements that are in either one of these, or both.*0348

*Applying this concept to inequalities: 3x + 2 > 8, or 4x - 3 < 1 is a compound inequality joined by the word or.*0353

*First, solve each inequality: 3x >...subtract 2 from both sides...6; divide both sides by 3 to get x > 2.*0366

*Or: solve this one also--if I add 3 to both sides, I am going to get 4x < 4; I am going to divide both sides by 4 to get x < 1.*0380

*Now, looking at this on the number line, what this is saying is that...0, 1, 2, 3, 4...x is greater than 2.*0391

*This is saying x is less than 1; and there is no overlap here, but that is OK, because the union*0403

*means that, if something is in either one of these or both, it is included.*0409

*This is saying that the solution set is that x is greater than 2, or x is less than 1.*0414

*And this can be written in set notation; so, I would write it as x is greater than 2, or x is less than 1.*0420

*So, when you use "and," it is the intersection of the two solution sets.*0429

*When "or" joins a compound inequality, the solution set is the union of the two solution sets.*0432

*Now, we have worked with absolute value equations; and this time, today, we are working with absolute value inequalities.*0440

*To solve an absolute value inequality, you need to use the definition of absolute value.*0448

*And recall that the definition of absolute value is the distance that a value is from 0 on the number line.*0453

*So, the absolute value of 3 is 3, because the distance between this and 0 on the number line is 3.*0461

*The absolute value of -3 is also 3, because the distance between -3 and 0 on the number line is 3.*0481

*So, the absolute value of 3 is 3, because it is 1, 2, 3 away from 0 on the number line.*0489

*The absolute value of -3 is 3, because -3 is also 3 units from 0 on the number line.*0494

*For more complicated problems, you may need to rewrite the inequality as a compound inequality.*0505

*Talking about this in a little bit more detail: you may see two different forms.*0511

*You may see inequalities in the form |x| < n.*0516

*If you see that, rewrite the inequality as a compound inequality, as follows.*0523

*Looking at this, I have an example: the absolute value of x is less than 4.*0532

*Think about what this is saying; it is saying that the absolute value of x is less than 4 units from 0.*0541

*OK, so anything that is less than 4...it could be 3, 2, 1, but it is less than 4 away from 0.*0552

*That could be satisfied by anything between 0 and 4; but it can also be satisfied by anything between 0 and -4,*0562

*because the distance between -4 and 0 on the number line is less than 4.*0580

*-3: the distance between that and 0 is less than 4.*0585

*So, anything in this range is going to satisfy it; so what this is really saying is that x is less than 4, and x is greater than -4.*0589

*So, in order for the answers to fall in this range, they can't just be less than 4 and go all the way down;*0605

*because then you will get numbers way over here, that have an absolute value that is much bigger than 4.*0610

*So, it is in this range, where x is greater than -4, but less than 4.*0614

*So, in general, when you see an absolute value in this form, where it is less than something, you rewrite it*0621

*as two related inequalities: x < 4, and x > -4.*0630

*We are just generalizing out: |x| < n can be rewritten as x < n, and x > -n.*0637

*OK, the other possibility is that you have inequalities that are in the form |x| > n.*0647

*For example, |x| > 3; well, what that is saying is that the absolute value of the number is more than 3 away from 0 on the number line.*0660

*So, anything bigger than 3 is going to have an absolute value of greater than 3.*0673

*4 has an absolute value greater than 3; and on up.*0680

*In addition, however, if I look over here on the left, any number smaller than -3 is also going to have an absolute value that is greater than 3.*0685

*If I took -4, the absolute value of that is 4, which is greater than 3.*0695

*So, this would translate to x > 3, or x < -3.*0700

*You need to memorize these two forms and be familiar with them.*0709

*And generalizing this out, this would say that |x| > n could be rewritten as x > n, or x < -n.*0712

*So, you need to keep these in mind: when you see |x| < n, you rewrite it as x < n, and x > -n.*0722

*When you see |x| > n, then x > n, or x < -n.*0731

*And you can apply these to more complex inequalities, such as |4x + 1| ≥ 12.*0741

*I would recognize that it is in this form, and I would rewrite it as 4x + 1 ≥ 12, or 4x + 1 ≤ -12.*0752

*So, just follow this pattern; and we will work on this in the examples.*0770

*OK, starting with Example 1: this is a shorthand way of a compound inequality that is joined by the word "and."*0775

*Instead of writing out 7 < 14x - 42, and 14x - 42 ≤ 35, they just combine the same term, put it in the middle, and left these on the outside.*0785

*But this is really a compound inequality that is joined by the word "and."*0805

*And remember that, in order to solve these, you solve each inequality; and then you find the intersection of their solution sets.*0810

*So, for absolute inequalities joined by the word "and," we are going to need to find the intersection of the solution sets.*0818

*All right, let's solve each of these; that is the first step.*0824

*I am going to rewrite this with the x on the left; 14x - 42 ≥ 7--more standard.*0828

*It is saying the same thing: 14x - 42 ≥ 7; I just reversed the sides of the inequality.*0845

*Next, add 42 to both sides; and that is going to give me 49; 14x ≥ 49.*0854

*Now, divide both sides by 14; and since that is a positive number, I don't need to reverse the inequality symbol.*0869

*I can simplify this, because they have a common factor of 7; so to simplify this, remove the common factor; that is going to give me 7/2.*0878

*I just pulled out the 7 from both the numerator and the denominator--factored it out.*0890

*OK, solving this inequality--my second inequality: adding 42 to both sides is going to give me 14x ≤ 77.*0895

*Again, dividing both sides by 14 is going to give me 77/14; again, I have a common factor of 7, so this is going to give me 11/2.*0909

*Now, the intersection of these, graphing this out: well, this is (2, 4, 6)...about 3.5, 3 and 1/2; and this is equal to 5 and 1/2.*0922

*OK, to help me graph it out, I am going to write it in decimal form: 1, 2, 3, 4, 5...a little more room....6.*0945

*OK, so what this is saying is x ≥ 3.5, which is right here; and it includes that point 3.5, so I am going to put a closed circle.*0961

*Here, the other restriction is that x ≤ 5.5; so this is greater than--it is going up this way;*0975

*but I have to stop when I get to here, because the intersection of the solution set has to meet both conditions.*0983

*It has to be greater than or equal to 3.5, and less than or equal to 5.5.*0993

*So, I can write it like this as an inequality; I can graph it here, or do set notation: where x is greater than or equal to 7/2 and less than or equal to 11/2.*0998

*OK, again, recognize that this is really a compound inequality; and it is joined by the word and, but it is just a shorthand way of writing it.*1025

*I wrote this out as two related inequalities, solved each, and found the intersection of their solution sets.*1033

*The second example is a compound inequality joined by the word or.*1042

*I am going to go ahead and solve both of these and find the union of the solution sets.*1045

*Adding 7 to both sides gives me 3y ≤ 15; divide both sides by 3: y ≤ 5.*1058

*On this side, I am subtracting 10 from both sides to get 2y > 16; dividing both sides by 2 gives me y > 8.*1067

*OK, since this is or, it is the union of the solution sets; so any element of either set,*1082

*or that is in both sets, would be included in the solution set for this compound inequality.*1089

*So here, I have y ≤ 5, and here I have y > 8.*1104

*So, all of this is included, and all of this is part of the solution set.*1115

*It could also be written as y ≤ 5, or y > 8.*1122

*OK, the third example involves an absolute value inequality.*1132

*When I look at this, I just want to think about which form this is in; and this is in the general form |x| < n.*1138

*And recall, for those, that you could rewrite this and get rid of the absolute value bars by saying x < n, and x > -n.*1145

*That is the general form; so I am going to rewrite this as 2z - 6 ≤ 8 (that is this form), and 2z - 6 ≥ -8 (which is this form).*1158

*It is really important to memorize these or understand them well enough that you can apply them to more complex situations.*1174

*Now, I am going to solve each.*1181

*2z - 6 ≤ 8: I am adding 6 to both sides to get 14, then dividing both sides by 2 to get z ≤ 7.*1184

*And I need to solve this other one.*1200

*I am going to add 6 to both sides to get 2z ≥ -6, z ≥ -3.*1205

*And since I am dividing by a positive number, I don't have to worry about reversing the inequality symbol.*1218

*OK, I end up with...again, I had an absolute value inequality in this form; I solved both inequalities.*1226

*Actually, I made a little mistake here; let me go ahead and correct that.*1246

*2z - 6 ≥ -8, so 2z ≥...this is actually going to be -2; so adding 6 to both sides is going to give me -2.*1250

*Now, if I divide both sides by 2, that is going to give me z ≥ -1.*1263

*OK, you can either leave this as it is, or you can go ahead and graph it out.*1271

*And since this is "and," the solution set needs to meet both of these conditions.*1277

*So, -1, 0, 1, 2, 3, 4, 5, 6, 7...a little more...let's go to 8; OK, z is less than or equal to 7, so this is going to go on and continue on;*1284

*but it also has to meet the condition that z is greater than or equal to -1.*1301

*So, anything in this range is going to be the intersection of the solution set for these inequalities.*1305

*Also, writing it out using set notation: what we have is that z is greater than or equal to -1, and less than or equal to 7.*1313

*These are three different ways of writing out the solution set.*1331

*OK, the next example is also an absolute value inequality.*1335

*And this one is in the form |x| > n, in this general form, which can be rewritten as x > n or x < -n.*1340

*So, rewriting this and removing the absolute value bars as two different inequalities: 3w - 8, divided by 5, is greater than 4;*1354

*so that is this first form; or 3w - 8, divided by 5, is less than -4.*1364

*OK, the first thing to do is get rid of the fraction: multiply both sides of this inequality by 5.*1374

*Next, add 8 to both sides; and finally, divide both sides by 3.*1385

*For this inequality, again, get rid of the fraction; multiply both sides by 5; add 8 to both sides; and divide by 3.*1398

*This is joined by the word "or"; and 28/3...I am going to rewrite that as 9 and 1/3.*1415

*So, w is greater than 9 and 1/3, or w is less than -4.*1425

*To graph this, -5, -4, -3, -2, -1, 0, 1...all the way to 9 and 1/3.*1430

*OK, so this is saying that w is greater than 9.3, or it is less than -4.*1445

*Open circle, because it is a strict inequality...*1456

*Set notation would be w is greater than 28/3, or w is less than -4.*1460

*Recognizing that this is in the general form |x| > n, I rewrote this as 3w - 8, divided by 5, is greater than 4, or the same expression is less than -4.*1472

*Solve each one: and the solution set is the intersection of the solution sets of those two inequalities.*1487

*That concludes this lesson of Educator.com; and I will see you soon!*1496

*Welcome to Educator.com.*0000

*Today is the first in a series of lectures on conic sections.*0002

*And we are going to start out with a review of two formulas that you will be applying later on, when we work with circles and hyperbolas and other conic sections.*0005

*Starting out with the ***midpoint formula**: this formula gives you the midpoint of line segments with endpoints (x_{1},y_{1}) and (x_{2},y_{2}).0015

*So, let's take an example, just to illustrate this: let's say you were asked to find the midpoint of a line segment with endpoints at (2,3) and (5,1).*0028

*If these were given as the endpoints, and you were asked to find the midpoint, you could go ahead and apply this formula.*0044

*So, to visualize this, let's draw out this line segment: this is at (2,3); that is one endpoint, and (5,1) is the other endpoint.*0053

*I draw a line between these two; and I am looking for the midpoint, which is somewhere around here.*0065

*Looking at this formula, it is actually somewhat intuitive, because what you are doing is finding*0072

*the average of the x-values and the average of the y-values, which will give you the middle of each (the midpoint).*0078

*So, x*_{1} + x_{2} would give me 2 + 5 (the two x-values), divided by 2.0085

*That is going to give me the x-coordinate of the midpoint of this line segment.*0095

*For the y-coordinate, I am going to add the two y's and divide by 2; so I will average those two y-values, which are 3 and 1.*0099

*2 + 5 gives me 7/2; 3 + 1 is 4/2; I can simplify to (7/2,2); and I could rewrite this, even, as (3 1/2,2) to make it a little easier to visualize on the graph.*0110

*This is the midpoint; it is going to be at 3 1/2 (that is going to be my x-coordinate); and 2 is going to be my y-coordinate.*0128

*And that is the midpoint; this is a fairly straightforward formula; you will need to apply it in a little while, when we start working with circles.*0138

*Also, to review the distance formula: if you recall, the distance formula is based on the Pythagorean theorem.*0151

*And the distance formula tells us that we can find the distance between two sets of points*0159

*if we take the square root of x*_{2} - x_{1}, squared, plus y_{2} - y_{1}, squared.0163

*Now recall: if I am given two points; if I am asked to find the distance between a set of two points, (2,1) and (5,6),*0172

*I can assign either one to be...I could assign this as (x*_{1},y_{1}), (x_{2},y_{2});0192

*or I could do it the other way around: I could say this is (x*_{2},y_{2}); this is (x_{1},y_{1}).0198

*It doesn't matter, as long as you assign it and stick with that; you don't mix and match and say this is (x*_{1},y_{2}).0202

*Just assign either set; it doesn't matter which set you call which; just stay consistent with the order that you use the points in.*0209

*OK, so let's look at what this would graph out to.*0216

*(2,1) and (5,6) would be right about up there; so I have a line segment, and I am asked to find the distance.*0221

*So, the distance is going to be equal to the square root of...I am going to call this (x*_{1},y_{1}) and this (x_{2},y_{2}).0231

*So, x*_{2} is 5, minus x_{1}, which is 2; take that squared,0239

*and add it to y*_{2}, which is 6, minus y_{1}, which is 1, squared.0247

*Therefore, distance equals 3*^{2}, plus 6 minus 1 is 5^{2}.0253

*The distance equals the square root of 3*^{2}, which is 9, plus 25.0259

*Therefore, distance equals √36; distance equals 6.*0269

*The distance from this point to this point is equal to 6.*0277

*And again, this is review from an earlier lecture, when we discussed the Pythagorean theorem in Algebra I*0282

*and talked about how the distance formula comes from that.*0288

*So, you can always go back and review that information; but this is the application of the distance formula.*0291

*In the first example, we are asked to find the midpoint and distance of the segments with these endpoints (-9,-7) and (-3,-1).*0299

*So, recall that the midpoint formula just involves finding the average of the x-coordinates*0310

*of the two points, and the average of the y-coordinates of the two points.*0315

*Applying these values to this set of equations gives me -9 + -3, divided by 2, and -7 + -1, divided by 2.*0324

*Adding -9 and -3 gives me -12, divided by 2; and that is -8 divided by 2.*0349

*This becomes -6 (-12/2 is -6), and then -8/2 is -4.*0359

*So, that is the midpoint; that was the midpoint formula.*0367

*Recall that the distance formula is (x*_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}.0377

*I am going to assign this as (x*_{1},y_{1}) and this as (x_{2},y_{2}).0392

*Again, it doesn't matter; you can do it the other way around.*0404

*Distance equals...I have x*_{2} as -3, minus -9, all of this squared; plus y_{2}, which is -1, minus -7, and that whole thing squared.0406

*This gives me -3; a negative and a negative is a positive, so plus 9, squared, plus -1...and a negative and a negative gives me + 7...squared.*0433

*Therefore, distance equals the square root of...9 - 3 is 6, squared, and then 7 - 1 is also 6, squared.*0444

*Distance equals √(36 + 36), or radic;72.*0453

*But I can take this a step farther, and say, "OK, this is the square root of 36 times 2," and then simplify that,*0461

*because this is a perfect square, to 6√2.*0467

*So, I found the midpoint; it is (-6,-4); that is the midpoint of this segment with these endpoints.*0475

*And the distance between these endpoints is 6√2.*0481

*In Example 2, we are asked to find the midpoint and distance of the segment with these endpoints.*0488

*We go about it as we usually do; but this time we are working with radicals,*0493

*so we have to be really careful that we keep everything straight.*0496

*Recall that the midpoint formula is the average of the x-coordinates, and then the average of the y-coordinates.*0499

*Therefore, the midpoint equals x*_{1}...the square root of 2 + 3, plus x_{2}, 2√2 - 4√3, divided by 2;0512

*for the y-coordinate of the midpoint, we are going to get √3 - 5, plus 4√3, plus 2√5.*0526

*And then, we simplify this as much as we can.*0536

*Here, I have two like radicals: they have the same radicand, so I can combine those to make this 3√2.*0540

*I have a constant, and then this -4√3.*0549

*So, this is the x-coordinate of the midpoint; the y-coordinate is √3 here and 4√3; those can be combined into 5√3;*0555

*and then, let's see: this actually should have a radical over it, because that is a radical up there;*0567

*I have -√5 + 2√5; that is going to leave me with just + √5, divided by 2.*0577

*And the midpoint is given by this; it is a little bit messy-looking, but it is correct--you can't simplify, really, any farther.*0585

*So, we are just going to leave it as it is.*0592

*Now, we are also asked to find the distance; so I am going to go ahead and work that out here.*0594

*We found the midpoint; we are working with these same endpoints, but we are finding the distance.*0599

*Recall that the distance formula is the square root of (x*_{2} - x_{1})^{2} + (x_{2} - y_{1})^{2}.0603

*So, I am going to let this be (x*_{1},y_{1}), (x_{2},y_{2}).0617

*I could have done it the other way around; it doesn't matter, as long as you are consistent.*0625

*Therefore, the distance is going to be equal to y*_{2}, which is 2√2, minus 4√3;0630

*and I am going to take that, and from that I am going to subtract y*_{1}, which is √2 + 3.0645

*So, this covers my (x*_{2} - x_{1})^{2}.0657

*I am going to add that to (y*_{2} - y_{1}): I go over here, and I have y_{2}; that is 4√3 + 2√5.0664

*And I am going to subtract y*_{1} from that, which is over here; and that is √3 - √5; and this whole thing is also going to be squared.0678

*Let's apply these negative signs to everything inside the parentheses, so we can start doing some combining.*0692

*This is going to give me 2√2 - 4√3; negative...that is going to give me -√2;*0700

*apply the negative to that 3--it is going to give me -3; all of this is going to be squared.*0713

*Plus 4√3, plus 2√5...apply the negative to each term inside the parentheses to get -√3;*0718

*and this negative, and then the negative √5, gives me a positive √5; and this whole thing is squared.*0728

*All right, so let's see if I can do some combining to simplify a little bit before I start squaring everything,*0736

*because that is going to be the most difficult step.*0742

*All right, this gives me 2√2, and this is minus √2; so I can combine these two to get √2.*0747

*-4√3, minus 3; this whole thing squared, plus 4√3 - √3--that simplifies to 3√3.*0757

*2√5 + √5 gives me 3√5; squared.*0773

*Now, I need to just square everything, and then combine it.*0780

*There is no easy way around this first one; what I am going to do is multiply √2 times itself, times this, times this,*0784

*and then go on with the second, and finally the third, term.*0792

*So, this √2 times √2 is simply going to give me 2; √2 times -4√3 is going to give me -4... 2 times 3 is √6.*0795

*Then, √2 times -3 is going to give me -3√2.*0817

*OK, now I take -4√3 and multiply it by this first term, √2, to get -4; 3 times 2 is 6.*0828

*When I multiply this times itself, I am going to get -4 times -4; that is going to give me 16,*0840

*times √3 times √3 is going to give me 3.*0846

*Then, I multiply this times -3 to get -4 times -3 is 12√3.*0851

*Finally, multiplying -3 times √2 gives me -3√2; -3 times -4 is 12√3; -3 times -3 is 12√3.*0861

*And then, -3 times -3 is 9; all of this is the trinomial squared; now let's square the binomial.*0878

*3√3 times 3√3 is going to be...3 times 3 is 9; and then √3 times √3 is just going to be 3.*0887

*So, looking up here, just to make this a little clearer: 3√3 + 3√5, times itself (squared)...*0902

*I am going to do my first terms, and I get this; then I do the outer terms, plus the inner terms,*0919

*which would just be this times this times 2; so that is going to give me 2 times 3 times 3, which is 9,*0924

*times 3 times 5, with a radical over it; so that is this.*0941

*So, what I did is took...really, it is just outer plus inner, which is going to give me the outer, which is 9√15, plus the inner, which also 9√15.*0951

*And this is going to end up giving me 18√15, which is the same as what I have here.*0969

*And then, finally, the last terms are going to give me 3 times 3 is 9, and then √5 times √5 is just going to give me 5.*0976

*OK, now combining what we can in order to just make this a bit simpler: I have a constant here;*0989

*I am just going to cluster all of my constants together in the beginning; that is 2; 16 times 3 is 48, so that is 48;*1007

*what other constants do I have?--I have a 9, and then I have 27, and I have 45.*1016

*All right, so those are my constants.*1035

*Now, for terms with a √6 in them, I have 2 of those: -4√6 and -4√6 is going to give me -8√6.*1037

*So, I took care of the constants and the terms with a radicand of 6.*1048

*Next, let's look at terms with a radicand of 2: -3√2, and I have -3√2 here; and that is it.*1052

*I add those two together, and I am going to get -6√2.*1062

*So, these are all taken care of: √2, √6, constant: now I have √3.*1066

*12√3--do I have any other terms like that?--yes: 12√3, so I have 2 of those; and that is going to be...12 and 12 is 24√3.*1072

*Let's see what else I have left: I took care of those, those...that just leaves me with...2 times 9 is 18 times √15.*1083

*Therefore, the distance equals...putting this all together, if you added these up, you will get 131 - 8√6 - 6√2 + 24 √3 + 18 √15.*1094

*So, this is the distance; and this was really a lot of practice just working with multiplying and adding and subtracting radicals.*1115

*We found the midpoint in the previous slide; and here is the distance for the segment with these endpoints, applying the distance formula.*1123

*Example 3: Triangle XYZ has vertices X (4,9), Y (8,-9), and Z (-5,2).*1132

*Find the length of the median from X to line YZ.*1143

*Definitely, a sketch would help us to solve this; so let's sketch this triangle.*1146

*And it has...we will call this X at (4,9); at (8,-9), we are going to have Y; and then, Z is going to be over here at (-5,2).*1152

*The median is going to be a line going here from here right to this midpoint.*1173

*So, what we are asked to find is the length: we can use the distance formula to find the length.*1179

*But in order to use the distance formula, I need this endpoint and this endpoint.*1185

*But I don't have this endpoint; however, I can find it, because if you look at this, this is going to draw out right to here, right in the middle.*1190

*So, this is the midpoint; if I find the midpoint of YZ, then I have the endpoint of this median.*1197

*Using the midpoint formula: (x*_{1} + x_{2})/2, and then for the y-coordinate, (y_{1} + y_{2})/2.1208

*So, I am looking for the midpoint of YZ: that is going to give me...*1227

*for Y, the x-value is 8; for Z, it is -5; divided by 2; for Y, the y-value is -9; for Z, the y-value is 2; divided by 2.*1235

*This is going to give me a midpoint of 8 - 5; that is 3/2; -9 + 2 is -7/2.*1252

*This is the midpoint, which is (3/2, -7/2); now I can use my distance formula,*1262

*distance equals √((x*_{2} - x _{1})^{2} + (y_{2} - y_{1})^{2}).1271

*I can use that distance formula to find this.*1282

*And I am going to call (4,9)...I need the distance from x to this point m...I am going to call this*1283

*(x*_{1},y_{1}), and this (x_{2},y_{2}), applying my distance formula.1292

*So first, x*_{2}, which is 3/2, minus x_{1}, which is 4--I am going to rewrite that as 8/2 to make my subtraction easier.1299

*But this is just 4; squared; plus y*_{2}, which is -7/2, minus 9, which I am going to rewrite as 18/2.1311

*Again, I am just moving on to the next step of finding the common denominator.*1324

*But this is my (x*_{1},y_{1}), (x_{2},y_{2}); that is where these came from.1328

*Therefore, the distance equals...3/2 - 8/2 is -5/2; and we are going to square that;*1339

*plus -18 and -7 combines to -25/2, squared.*1347

*Therefore, the distance equals the square root of...this is 25 (5 squared), over 4, plus...25 squared is actually 625, divided by 4,*1357

*which equals the square root of 625 and 25 is 650, and they have the common denominator of 4.*1370

*And you could leave it like this, or take it a step farther and write this as 650 times 1/4, which equals 1/2.*1379

*And you could even look and see if there are any other perfect squares that you could factor out.*1389

*But this does give us the length of the median; in order to find the length of the median, we had to find this other endpoint,*1393

*which conveniently was the midpoint of YZ, so we found the midpoint of YZ and used that as the other endpoint.*1401

*Then I had X as an endpoint and the midpoint as the endpoint.*1408

*Plug that into the distance formula to get the length.*1412

*Working with triangles again: find the perimeter and area of the triangle with vertices at (1,-4), (-1,-2), and (6,1).*1418

*So, to help visualize this, we are going to draw it out on the coordinate plane.*1427

*(1,-4) is my first point; (-1,-2) is my second point; and then 6 is going to be about here; this is 6; 1 will be right here.*1433

*Let's see what this triangle looks like, and if the graph can help me.*1449

*Now, I am asked to find the perimeter and area: if I want to find the area of a right triangle,*1459

*I know that it is going to be 1/2 the base times height.*1462

*The problem is that I don't know if I am working with a right triangle; I may be; but this definitely does not look like a right angle.*1465

*This may be a right angle, but I am not sure.*1472

*In order to determine if it is a right angle (let's call this A, B, and C)--is this a right angle?--*1475

*well, if it is a right angle, that means that this AC and BC are going to be perpendicular.*1485

*And recall that perpendicular lines...the product of their slope equals -1.*1492

*So, what I am going to do is go ahead and find the slope of these two.*1510

*And I am going to determine what this slope is of AC; what BC is; find their product; and if they are a right angle, then I can proceed.*1515

*So, this point right here is (-1,-2); this point is (1,-4); and this point is (6,1).*1523

*Slope is just change in y, divided by change in x.*1534

*So, let's find the slope of AC: that is going to be the change in y (-2 - -4), divided by change in x (which is -1 - 1).*1538

*So, this is going to be -2 plus 4, divided by -2; this is going to give me 4 divided by -2, which is -2.*1561

*I just double-check this: -2 minus -4...actually, correction: 4 minus 2 is going to give me, of course, 2; and the slope, therefore, will be -1.*1579

*OK, now I also need to find, over here (to keep this separate), the slope of BC.*1596

*So, change in y: y is 1 - -4, over change in x: x is 6 - 1; this is going to be equal to 1 + 4, divided by 6 - 1 is 5, which equals 5/5.*1606

*So, the slope of BC equals 1.*1630

*Now, this means that, if I have the slope of BC, times the slope of AC, equals 1 times -1, or -1.*1634

*This is a right angle; I have a right triangle; I can use my formula, 1/2 the base times the height.*1649

*What I need to do next is find the length of these sides, using the distance formula.*1656

*So, let's start out by finding the length of side AC.*1661

*Recall your distance formula: for the distance formula, I am going to take x*_{2}, side AC,1668

*and I am going to call...it doesn't matter which one...I am going to call this, for this first one, (x*_{1},y_{1}), (x_{2},y_{2}).1678

*x*_{2} is -1, minus x_{1}, which is 1, squared, plus y_{2}, which is going to be -2, minus y_{1}, which is -4.1689

*So, the length of AC equals...-1 and -1 is -2, squared; plus -2...and a negative and a negative is a positive, so -2 + 4 is going to give me 2, squared.*1705

*Therefore, the length of AC is going to equal the square root of 4 plus 4, which equals the square root of 8.*1723

*So, that is the length of AC.*1733

*Now, I need to find the length of another side to find the area.*1735

*I found AC; let's go for BC next--the length of BC.*1739

*I am going to make this (x*_{1},y_{1}),(x_{2},y_{2}).1748

*So, I am starting out with my x*_{2}, which is 6, minus x_{1}, which is 1, squared,1753

*plus x*_{2}, which is 1, minus -4 (that is y_{2} - y_{1}), squared,1761

*equals the square root of...6 minus 1 is 5, squared; 1 minus -4...this becomes 1 + 4, so that is 5, squared;*1773

*So, this equals the square root of 25 + 25, or the square root of 50.*1783

*All right, so now I have two sides: I know that this side, AC, has a length of √8; and I know that BC is √50.*1789

*So, the area equals 1/2 the base times the height; so the area equals 1/2 (√8)(√50), which equals 1/2√400.*1804

*All right, and you could go on and then simplify this, because this would give you the perfect square of 20 times 20,*1828

*because 20*^{2} would give you 400; so then I could make this 1/2(20) is 10.1845

*All right, now we still need to find the perimeter.*1856

*In order to find the perimeter, I need this third side; and I can use the Pythagorean theorem, because this is the hypotenuse.*1859

*And I know that a*^{2} + b^{2} = c^{2}.1864

*So, a*^{2} is...one side is the square root of 8, squared, plus b^{2}, the square root of 50 squared, equals c^{2}.1869

*Well, the square root of 8 squared is 8, plus 50 (because the square root of 50 squared is going to be 50), equals c*^{2}.1879

*So, 58 = c*^{2}; therefore, c equals √58.1890

*All right, so I found the area right here; now, the perimeter.*1895

*The perimeter equals the sum of the three sides: so that is √8 + √50 + √58.*1907

*And you could go on and do a little simplification.*1920

*You could pull the perfect squares out of here.*1922

*So, I could go a little farther with this and say, "OK, 8 is equal to 4 minus 2; 4 is a perfect square, so this is 2√2."*1927

*50 is the square root of 25 times 2, so that is going to give me 5√2, plus √58.*1935

*But you can't combine any of these, because they are not like radicals.*1942

*So, in this example, we applied the distance formula, as well as the Pythagorean theorem and some geometry,*1947

*to find the area and the perimeter of this triangle.*1953

*Thanks for visiting Educator.com; and that concludes this lesson on the midpoint and distance formulas.*1957

*Welcome to Educator.com.*0000

*Today, we are going to talk about parabolas.*0002

*And in some earlier lectures in this series on quadratic equations, we talked about parabolas and did some graphing.*0004

*But now, we are going to go on and give a specific definition to parabolas, and learn about some other features of parabolas.*0011

*Although you have seen parabolas previously, when we graphed, we didn't form a specific definition of them.*0021

*So, the definition of a parabola is that it is the set of points in the plane whose distance from a given point,*0027

*called the focus, is equal to its distance from a given line, called the directrix.*0033

*Let's talk about that before we go on to talk about the axis of symmetry.*0041

*So, if you had a parabola (let's say right here; and we will do an upward-facing parabola), you would have some point,*0045

*which is known as the focus, and a line (I'm going to put that right about here) called the directrix.*0059

*By definition, every point on this parabola is equidistant from the focus and the directrix.*0080

*So, if I took a point right here, and I measured the distance from the focus, it would be equal to the distance from the directrix.*0085

*And this is just a very rough sketch; but these distances actually would be equal; they are theoretically equal.*0095

*Looking right here at the vertex, these distances would be equal; so that would be, say, y.*0108

*If I took some other point, say here, and I measured here to here, these two distances would be equal.*0117

*So, a couple things to note: the focus is inside the parabola; the directrix is outside.*0128

*And this is because the focus and the directrix are on the opposite sides of the vertex.*0150

*So, you could have a parabola facing downward, and then it would have a focus here and a directrix up here.*0155

*We are also going to talk, today, about parabolas that face to the left and right--horizontal parabolas.*0165

*But right now, we are going to stick with just (for this discussion) focusing on vertical ones,*0173

*the definition being that every point in the parabola equidistant between the focus and the directrix.*0180

*The axis of symmetry of the parabola passes through the focus; and it is perpendicular to the directrix.*0189

*In this case, the y-axis is the axis of symmetry; it is right here.*0195

*And you see that it passes through the focus, and it forms a right angle; it is perpendicular to the directrix.*0204

*Again, we talked about some of these concepts in earlier lectures.*0214

*But to review, ***vertex**: the vertex of a parabola is the point at which the axis of symmetry intersects the parabola.0216

*And it is a maximum or minimum point on the parabola, if the axis of symmetry is vertical.*0224

*If the axis of symmetry is horizontal (say we have a parabola like this, then the axis of symmetry would be horizontal),*0230

*we still have a vertex, but it is not a maximum or minimum.*0242

*And again, we are going to focus a little more on vertical parabolas right now, and then we will talk about horizontal parabolas.*0247

*So, if I have a downward-facing parabola, the vertex is here; the axis of symmetry is right here.*0254

*And this vertex is the maximum; this is as large as y gets--it is the largest value that the function attains.*0265

*If I am looking at a vertex that is upward-facing, then the axis of symmetry...we will put it right here; and the vertex is here.*0274

*In this case, the vertex is a minimum; this is the smallest value that the function will attain.*0286

*The standard form of a parabola with vertex at (h,k) is y = a(x - h)*^{2} + k.0299

*And this is for vertical parabolas; there is a slightly different form when we are talking about horizontal parabolas.*0308

*And you might recall this form of the equation that we covered earlier on, under the lecture on quadratic equations.*0314

*And we called this the vertex form of the equation; now we are going to refer to it as standard form.*0319

*And it is a very useful form, because it tells you a lot about the parabola.*0324

*The axis of symmetry is x = h: so I know a few things just from looking at this.*0330

*I know the vertex, because it is (h,k); I know the axis of symmetry--it is at x = h;*0335

*and if I look at a, I will know if the parabola is upward- or downward-facing.*0341

*If a is greater than 0, the parabola will open upward; and k gives you the minimum.*0348

*If a is a negative value--if it is less than 0--the parabola opens downward, and k is the maximum value of the function.*0357

*Let's look at an example: y = 2(x - 1)*^{2} + 4.0367

*So, this is in standard form: this means that I have h = 1, k = 4, and a = 2.*0374

*So, I know that my vertex is going to be at (1,4); the axis of symmetry is going to be at x = h, so at x = 1.*0384

*And since a is greater than 0, this opens upward.*0404

*So, I can sketch this out: I have a vertex at (1,4), right here, and it opens upward.*0411

*And the axis of symmetry is going to be right here at x = 1.*0421

*Here is my vertex at (1,4); and this vertex is a minimum, because this opens upward.*0428

*The minimum value is k, which is 4.*0433

*If I were to take a similar situation, but say y = -2(x - 1)*^{2} + 4,0441

*I would have, again, an h equal to 1 and a k equal to 4, but this time a would be -2, so this would open downward.*0453

*What I would end up with would be a parabola here, again, with the vertex at (1,4).*0465

*But it would open downward, and therefore, this would be a maximum.*0473

*Also, if the absolute value of a is greater than 1, you end up with a relatively narrow parabola.*0481

*If the absolute value of a is less than 1, you end up with a relatively wide parabola.*0490

*So, this form is very useful, because just by having the equation in this form, we can at least sketch the graph.*0500

*Let's talk a little bit more about graphing parabolas.*0508

*You can use symmetry and translations to graph a parabola: and by translations, we mean a shift.*0511

*Looking at the standard form: what this really is: if you took a graph of y = ax*^{2}, this is letting h equal 0 and k equal 0.0519

*And then, if you altered what h is, it is going to shift the graph horizontally by that number of units.*0530

*If you alter what k is, it is going to translate or shift that graph upward and downward by a certain number of units.*0538

*In order to graph a parabola, you often need to put it in standard form.*0547

*Let's start out by just talking about putting an equation or a parabola in standard form.*0552

*And then we will go on and look at some graphs, and how different values of h and k can affect the graph.*0556

*So, in order to put the equation into standard form...let's say you are given an equation such as this, y = x*^{2} + 6x - 8,0562

*and I want it in this standard form, y = a(x - h)*^{2} + k.0572

*The first thing to do (and this is, again, review from an earlier lesson--you can go back and look at the lesson*0580

*on completing the square as part of this lecture series, but we will review it again now): first, I am going*0586

*to isolate the x variable terms on the right side of the equation.*0592

*I am going to add 8 to both sides: now I am going to complete the square.*0596

*I am going to focus on this, and I need to add a term to it to make this a perfect square trinomial.*0602

*The term I am going to add is going to be b*^{2}/4.0608

*In this case, b is 6, so this is going to give me 6*^{2}/4, which is 36/4, which is equal to 9.0613

*So, that is what I need to add in here: y + 8, plus I need to add 9 to both sides.*0627

*It is easy to forget to add it to the other side, because you get so focused on completing the square.*0640

*But if you don't, the equation will no longer be balanced.*0645

*So, I am going to add 9 to both sides.*0648

*And I want this to end up in this form; so I am going to rewrite this.*0653

*First I will add these two together to simplify to get y + 17 =...well, this is a perfect square trinomial, so I just take (x + 3)*^{2}.0657

*And I look at what I have, and it is almost in this form, but not quite.*0668

*I want to isolate y on the left, so I am going to subtract 17 from both sides to get y = (x + 3)*^{2} - 17.0671

*And this is in this form: a happens to be equal to 1 in this case.*0679

*And so, if you are given an equation that is not in standard form, and you want to get it in standard form,*0683

*isolate the x variable values on the right (although if we are working with horizontal parabolas,*0690

*it is going to be the other way around, as we will see in a minute--we are actually going to end up*0697

*getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square*0701

*by adding the b*^{2}/4 term to both sides of the equation; and then simplify;0707

*shift things around as needed to get it in this form.*0716

*Remember, also, that if you have a leading coefficient that is something other than 1,*0718

*when you get to this step after isolating the x variable terms, you are going to need to factor out that term before completing the square.*0724

*All right, assuming that you have gotten your equation in standard form, and you are ready to graph the parabola, you are going to use symmetry.*0734

*The two halves of the parabola are symmetrical; if you graph half the points, you can use reflection across the axis of symmetry to graph the other points.*0741

*And translation is knowing how h and k, and changes in h and k, affect the graph, in order to graph.*0749

*All right, so let's just start out with something in this form--a very basic equation for a parabola.*0760

*Let's let f(x) equal x*^{2}, so it is in this form: y = ax^{2}.0767

*And so, here, what is happening is: if you think about what we have, we have a = 1, and then h is 0 and k is 0.*0775

*What this tells me is that the vertex is going to be at (0,0), and the axis of symmetry is going to be at x = 0.*0784

*And you can also very easily find some points to graph this.*0796

*right now, I am just going to sketch it out, and not worry about exact points, just so you get the idea.*0800

*So, since a = 1, this is going to open upward; this is going to be upward-opening, so the vertex is here at (0,0);*0805

*it is upward-opening; and it is going to look something like this.*0819

*So, this is my graph here of y, or f(x), = x*^{2}.0833

*Now, let's say I change this slightly: let's say I have another function, g(x) = x*^{2} + 2.0838

*So, looking at this form, h is still 0; but now I have k = 2.*0847

*And according to this, this is going to shift the graph up 2 units; so k is going to translate this graph up 2 units.*0855

*I have a similar graph, but it is going to be with the vertex right here at (0,2).*0867

*And remember that the axis of symmetry is at x = h, so the axis of symmetry is going to still be at x = 0; right here--this is the axis of symmetry.*0880

*This is shifted upward; it still opens upward, because a is positive.*0888

*So, now I am just going to have a similar idea, but shifted upward by 2.*0892

*So here, I have y = x*^{2} + 2.0901

*If this had been a -2, then it would have been shifted down by 2, and I would have had a graph right here.*0907

*So, let's see what happens when I change h.*0912

*Let's get a third function: we will call it h(x) = (x - 1)*^{2}.0917

*OK, now what I have here is h = 1; k, if I look here, is 0.*0935

*Therefore, the vertex (this is the vertex right here) equals (1,0), and the axis of symmetry is going to be at x = 1.*0947

*So, this is going to be shifted to the right; so I am going to have a graph something...let me move this out of the way...like this.*0962

*So, this one is y = x*^{2}, and this is y = (x - 1)^{2}.0979

*Important take-home points: a change in h will shift the graph horizontally, to the right or left.*0991

*A change in k will shift the basic graph either up or down, by k number of units.*1001

*Using symmetry: if I were to graph these out exactly, I would need to find points.*1009

*And I don't need to find all of the points: for example, if I had a parabola that was a downward-facing parabola*1013

*somewhere, then I could use the axis of symmetry, and I could just find the points over here and reflect across that axis in order to graph.*1021

*All right, this concept is another one adding on to our knowledge of parabolas from prior lessons.*1034

*And it is defining a segment called the latus rectum.*1040

*The ***latus rectum** is the segment passing through the focus and perpendicular to the axis of symmetry.1044

*Let's see what that means--let's visualize that.*1051

*Let's say I have a parabola like this, and let's say the focus is here.*1054

*So, this passes through the focus, and is perpendicular to the axis of symmetry.*1065

*This is the focus, and here we have the axis of symmetry.*1072

*That means the latus rectum is going to pass through here, and it is going to be perpendicular to the axis of symmetry.*1083

*So, that is this line; this is the latus rectum.*1090

*The equation for its length is the absolute value of 1/a; and if you have the equation of the parabola in standard form,*1095

*then this a is the same a as you will see in that formula.*1108

*So, this is something you might occasionally need to use.*1112

*For example, if I were given an equation of a parabola y = 2(x - 3)*^{2} + 5,1115

*and I was asked to find the length of the latus rectum of this parabola, then I would just say,*1122

*"OK, a equals 2; therefore, the length equals the absolute value of 1/2."*1127

*Horizontal parabolas: I mentioned that you can also have parabolas that open to the right or left, not just up and down,*1138

*although up to this point in the course, we have just talked about vertical parabolas, or parabolas that open upward or downward.*1143

*For parabolas whose axis of symmetry is horizontal, we end up with equations in this form: y = a(x - k)*^{2} + h.1150

*So, one thing to note: the positions of the x's and y's are reversed, but so are the h's and k's.*1160

*In the vertical formula, the h was in here, and the k was out here.*1168

*So, be careful when you are working with this formula to notice that the positions of h and k are reversed.*1171

*And there are translations of x = ay*^{2}, and then again, h and k shift this graph around horizontally and vertically.1177

*So, it would look something like this, for example: the axis of symmetry would be right here;*1189

*and it would be a horizontal axis of symmetry; or maybe I have one that opens to the left, and it has an axis of symmetry right here.*1197

*These do not represent functions; and you can see that they don't represent functions*1208

*by trying to pass a vertical line through them: they fail the vertical line test.*1212

*Remember: with a function, the vertical line test tells us that a vertical line drawn...you could try*1216

*any possible area of the curve, and the vertical line will only cross the curve once.*1224

*If the vertical line crosses the curve more than once, it is not a function.*1230

*So, this fails the vertical line test.*1233

*It is not a function; it is still an equation--you can still make a graph of it; but horizontal parabolas do not represent functions.*1241

*I am working on graphing some horizontal parabolas.*1250

*When you look at the equation in standard form, y = a...and remember, the k and h are in opposite positions;*1253

*they are reversed...looking at a, if a is positive (if a is greater than 0), then the parabola is going to open to the right.*1263

*If a is negative, then the parabola is going to open to the left.*1273

*So, let's look at a very simple horizontal parabola, x = y*^{2}.1278

*OK, the vertex is at (h,k); and I can see that h and k are both 0, so the vertex equals (0,0).*1284

*The axis of symmetry is at y = k, so that is going to be at y = 0.*1294

*And the a here is 1: a = 1, so this opens to the right.*1298

*So, you are going to have a parabola that looks something like this.*1308

*You could have another parabola, x = -y*^{2}.1321

*Here we would have the same vertex and the same axis of symmetry; here the x-axis is actually the axis of symmetry.*1325

*And I look at a now, and a equals -1, so this parabola is going to open to the left.*1333

*So, I am going to end up with a parabola like this.*1343

*Now again, change in h or change in k is going to shift this parabola a bit.*1350

*Let's change h and see what happens: let's let x equal y*^{2} + 2.1358

*Here I have h = 2, k = 0; so (2,0) is the vertex; a = 1, so it is positive, so this still opens to the right.*1366

*If I look at this, x = y*^{2}...here is my graph of x = y^{2}; over here is x = -y^{2}.1379

*Now, I am going to have h = 2, so that is going to shift horizontally by 2.*1386

*(2,0) will be the vertex; and it is going to open to the right.*1397

*So, this is x = y*^{2} over here; right here, this is actually x = y^{2} + 2 now.1402

*And k, as discussed before, shifts the graph of a parabola vertically.*1416

*The same idea here: if I were to change k, then I would shift this graph up or down by k units.*1423

*So, with horizontal parabolas, you need to be familiar with this equation.*1430

*You need to know that they open to the right if a is greater than 0; they open to the left if a is less than 0.*1435

*The vertex is at (h,k), and the axis of symmetry is y = k.*1441

*And you also need to keep in mind that these do not represent functions.*1446

*In the beginning of today's lesson, we talked about the focus and directrix.*1452

*And here are formulas to allow you to find those if you need to.*1455

*If you have a vertical parabola, the coordinates of the focus are h for the x-coordinate, and k plus 1/4a.*1460

*And the equation for the directrix is y = k - 1/4a; remember that the directrix is a line, so this is giving you the equation for that line.*1473

*And this would be for a vertical parabola; for a horizontal parabola, the focus is found at the coordinates h + 1/4a;*1485

*and then the y-coordinate is k, so the focus is a point, and this gives the coordinates of that point.*1493

*The directrix is a line, and the equation for this line for a horizontal parabola is x = h - 1/4a.*1498

*And you might need to occasionally use these when we are working problems.*1505

*And we will see that in one of the examples, actually, shortly.*1508

*Starting out with Example 1: Write in standard form and identify the key features: x = 3y*^{2} - 12y + 10.1513

*We have x equal to all of this; so this tells me, since I have x set equal to this y*^{2} term, that I am looking at a horizontal parabola.1524

*So, the standard form of this equation is going to be x = a(y - k)*^{2} + h.1536

*Remember, h and k are going to be in opposite positions.*1547

*In order to get this equation in standard form, we need to complete the square.*1550

*This time, since I am working with a horizontal parabola, I am going to isolate all of the y variable terms on the right.*1554

*And I am going to do that by subtracting 10 from both sides to get x - 10 = 3y*^{2} - 12y.1561

*This leading coefficient is not 1, so I have to factor it out.*1569

*And then, I have to be really careful when I am adding to both sides of the equation, because this is factored out.*1573

*So, factor out a 3 to get y*^{2} - 4y.1580

*I need to complete the square: that means I need to add something over here.*1585

*And the term that I need to add is going to be b*^{2}/4.1589

*b is actually 4; so this is going to be 4*^{2}/4, equals 16/4, equals 4.1594

*Here is where I need to be careful: on the right, I am adding 4 inside these parentheses, which is pretty straightforward.*1604

*But what I need to do on the left is realize that I am actually going to be adding 3 times 4, which is 12.*1613

*So, if I were just to add 4, this equation would not be balanced,*1628

*because in reality, what I am doing over here is adding 3 times 4.*1631

*So, on the right, I am going to add 12; and I got that from 3 times 4.*1635

*Simplifying the left: 12 - 10 is 2; on the right, inside here, I now have a perfect square.*1640

*And I want this to end up in this form, so I am going to write this as (y - 2) (and it is negative, because I end up with a negative sign in here) squared.*1649

*I am almost done; I just need to move this constant over to the right to have it in this form.*1661

*x = 3 times (y - 2)*^{2}, minus 2.1666

*So, now that I have this in standard form, I can identify key features.*1671

*Key features: 1: this is a horizontal parabola, as you can see from looking at this equation.*1677

*2: The vertex is at (h,k); h is 2, and k is also 2.*1686

*Actually, being careful with the signs, h is actually -2, because remember, standard form has a plus here.*1703

*I don't have a plus here; I could rewrite this so that I do, and that would give me + -2.*1710

*And it is good practice, actually, to write it exactly in this form, although this is correct--you could leave it like this.*1718

*By writing it in this form...and the same thing if I had ended up with a plus here--then I would need to rewrite that,*1725

*because here I need a negative to be in standard form; if I ended up with a plus here,*1736

*then I would have needed to rewrite that, as well, which would have been equal to minus -2.*1741

*Standard form, just like this, looking here, gives me a vertex at (-2,2).*1748

*And because a equals 3, that means that a is greater than 0; a is positive, so the parabola opens to the right.*1755

*OK, so key features: horizontal parabola; it has a vertex at (-2,2); a = 3, so this tells me that the parabola opens to the right.*1773

*We can also say that the axis of symmetry is at y = k, and therefore the axis of symmetry is at y = 2.*1784

*OK, in Example 2, we are asked to graph.*1797

*And you will notice that this is the same equation that we worked with in Example 1.*1802

*We already figured out standard form: and standard form is x = 3(y - 2)*^{2} - 2.1806

*And for clarity, we can actually write this as I did at the end, which is 3(y - 2)*^{2} + -2,1816

*so that we truly have it in standard form, with the plus here to make it easy to see what is going on.*1826

*To graph this, I want to know the vertex: the vertex is (h,k): h here is -2; k is 2.*1831

*The axis of symmetry is going to be at y = k; k is 2, so it is going to be at y = 2.*1840

*I know that this opens to the right, so I have a general sense of this graph.*1854

*But I can also just find a few points.*1859

*And we are used to working with a situation where x is the input and y is the output.*1868

*It is the opposite here, so we need to be really careful.*1873

*I also want to note that, since the vertex is here at (-2,2), and this opens to the right,*1876

*for this graph, we are not going to have values of x that are smaller than -2.*1881

*So, if I end up with something where an x is smaller than -2, then it is going to be off the graph.*1885

*Let's let y equal 1: if y is 1, 1 - 2 is -1, squared gives me 1; 1 times 3 is 3, minus 2 is 1; so, when y is 1, x is 1.*1891

*Let's let y equal 3: when y is 3, 3 minus 2 is 1, squared is 1; 1 times 3 is 3, minus 2 is 1.*1906

*And you can see, as I mentioned, that this is not a function; it failed the vertical line test (as horizontal parabolas do).*1915

*And you can see that there is an x-value, 1, that is assigned 2 values of y; so it does not meet the definition of a function.*1921

*So, just a couple of points...let's do one more: 0...0 minus 2 is -2; squared is 4; 4 times 3 is 12; 12 minus 2 is 10.*1929

*So, that is off this graph; but it gives us an idea of the shape.*1941

*So, I know that my axis of symmetry is going to be here; and I have a point at (1,1);*1945

*I have another point at (1,3); and then I have a point way out here at (10,0).*1954

*I know that this is going to be a fairly narrow graph, because a equals 3.*1960

*This is the graph of the horizontal parabola described by this equation; and here it is, written in standard form.*1971

*So, it opens to the right; it is fairly narrow, because a equals 3.*1979

*It has a vertex at (-2,2), and it has an axis of symmetry at y = 2.*1984

*Example 3: we are asked to graph; this is also going to be a horizontal parabola.*1994

*We are going to start out by putting it in the standard form, x = (y - k)*^{2} + h.1999

*We need to complete the square; start out by isolating the y variable terms on the right.*2009

*So, I am going to add 6 to both sides to get -2y*^{2} + 8y.2014

*Since the leading coefficient is not 1, I need to factor it out; so I am going to factor this -2 to get y*^{2}.2021

*Factoring a -2 from here would give me a -4.*2032

*And I need to add something to this to complete the square.*2035

*What I need to add is b*^{2}/4.2040

*b is 4, so I am going to be adding 4*^{2}/4; that is 16, divided by 4; that is 4.2044

*So, I am going to be adding 4 to the right; but to the left, I am actually adding -2 times 4, which is -8.*2061

*So, we subtract 8 from that side; to this side, since I am adding inside the parentheses, I am just adding 4.*2073

*But then, 4 times -2--that is how I got the -8 on the left.*2082

*This gives me x - 2 = -2; and I want it in this form, so I am going to rewrite this as (y - 2)*^{2}.2085

*The last thing I need to do is add 2 to both sides; and I have it in standard form.*2097

*Now that I have this in standard form, it is much easier to graph.*2106

*The vertex is going to be at (h,k); so h is here; k is here; the vertex is at (2,2).*2110

*There is going to be an axis of symmetry at y = k, and so that is going to be at y = 2; my axis of symmetry is going to be at y = 2.*2119

*Now, to finish out graphing this, I am going to find a few points.*2142

*I have the vertex at (2,2); I also know that a is less than 0 (a is negative), so I know this is going to open to the left.*2146

*So, I know it is going to look something like this; but I will find a couple of points.*2155

*And I know that x is (actually, (2,2) is right here)...I know that this opens to the left, and that x is not going to get any larger than that.*2158

*The graph is just going to go this way.*2172

*So, I can't use values that end up giving me an x that is greater than 2.*2174

*Let's try some simple values: I am going to try 1 for y, and looking in standard form, 1 - 2 gives me -1, squared is 1, times -2 is -2, plus 2 is 0.*2181

*And 3: 3 minus 2 is 1, squared is 1; 1 times -2 is -2, plus 2 is 0.*2195

*So, I have a couple of points here: this is at 0...when x is 0, y is 1; when x is 0, y is 3.*2204

*And this is going to give me a parabola shaped like this, opening to the left with a vertex at (2,2).*2214

*The axis of symmetry would be right through here; and I have a couple of points, just to make it a bit more precise.*2226

*So, the first step in graphing a parabola is always to get it into this form by completing the square.*2235

*And then, using the features you can see from here, sketch it out, and finding a few points, make the graph more accurate.*2240

*Find the equation of the parabola with a vertex of (2,3) and focus at (2,7); draw the graph.*2250

*This is a very challenging problem: we are not given an equation--we actually have to find the equation based on some key points that we are given.*2256

*Well, I am given that the vertex is at (2,3); so I know that the vertex is right here; that is the vertex.*2266

*This time, I am also given the focus; the focus is at (2,7), which is going to be up here somewhere...5, 6, 7...about here.*2278

*So, the vertex is (2,3); the focus is (2,7).*2290

*Remember, in the beginning of this lesson, I mentioned that the focus is always inside the parabola.*2296

*Since the focus is inside the parabola, I already know that this has to open upward.*2301

*So, I know something about the shape of the graph.*2306

*Let's find the equation: now, I know that this is a vertical parabola, because the focus is inside the parabola.*2311

*That told me that this has to open upward, so I know I am dealing with a vertical parabola.*2316

*And that helps me to find the equation, because the standard form is going to be y = a(x - h)*^{2} + k.2320

*I am given the vertex, so I am given h and k: I know that h = 2 and k = 3.*2330

*In order to write this equation, I need a, h, and k; all I am missing is a.*2340

*I am given the piece of information, though, that the focus is (2,7).*2347

*And that is going to allow me to find a.*2350

*You will recall that I mentioned the formulas for focus and directrix.*2353

*And for a vertical parabola, the focus is at h...the x-coordinate is h, which we see here; and the y-coordinate is k + 1/4a.*2359

*And I know the focus is at (2,7): so 2 = h, and 7 = k + 1/4a, according to this definition.*2377

*Well, since I know that k is 3, then I can solve this.*2392

*So, I know k; so I can solve for a.*2401

*Subtract 3 from both sides to get 1/4a; 1/a equals 16; multiply both sides by a, and then divide both sides by 16,*2408

*or just take the reciprocal of each side (essentially, that is what you are doing) to get a = 1/16.*2422

*Now, I have h and k given; I was able to figure out a, based on the definition of focus.*2427

*So, I end up with the equation y = 1/16(x - 2)*^{2} + 3.2433

*So, this is the equation.*2444

*And as you know, once we have the equation, the graphing is pretty easy.*2446

*I know that this opens upward; and since I know what a is, I know that this is going to be a pretty wide parabola; the a is a small value.*2451

*I am going to have a parabola that opens upward, with a vertex of (2,3), and fairly wide in shape.*2461

*That was a pretty challenging problem, because you had to go back*2470

*and think about how you could use a formula to find the focus; and knowing the focus allowed you to find a.*2473

*That concludes this lesson on parabolas at Educator.com; thanks for visiting!*2483

*Welcome to Educator.com.*0000

*Today we are going to talk about circles, beginning with the definition of a circle.*0002

*A ***circle** is defined as the set of points in the plane equidistant from a given point, called the **center**.0008

*For example, if you had a center of a circle here, and you measured any point's distance from the center, these would all be equal.*0018

*And the radius is the segment with endpoints at the center and at a point on the circle.*0031

*The equation for the circle is given as follows: if the center is at (h,k) and the radius is r,*0044

*then the equation is (x - h)*^{2} + (y - k)^{2} = r^{2}.0052

*And this is the standard form; and just as with parabolas, the standard form gives you a lot of useful information*0060

*and allows you to graph what you are trying to graph.*0067

*For example, if I were given (x - 4)*^{2} + (y - 5)^{2} = 9, then I would have a lot of information.0071

*I would know that my center is at (h,k), so it is at (4,5).*0084

*And the radius...r*^{2} = 9; therefore, r = √9, which is 3.0091

*So, based on this information, I could work on graphing out my circle.*0099

*Use symmetry to graph a circle, as well as what you discover from looking at the equation in standard form.*0108

*Looking at a different equation, (x - 1)*^{2} + (y - 3)^{2} = 4: this equation describes the circle0114

*with the center at (h,k), which is (1,3); r*^{2} = 4; therefore, r = 2.0124

*So, I have a circle with a radius of 2 and the center at (1,3).*0134

*So, if this is (1,3) up here, and I know that the radius is 2, I would have a point here; I would have a point up here.*0139

*Symmetry: I know that, if I divide a circle up, I could divide it into four symmetrical quarters, for example.*0156

*So, if I have this graphed, I could use symmetry to find the other three sections of this circle.*0169

*All right, if the center is not at the origin, then we need to use completing the square to get the equation in standard form.*0187

*Remember that standard form of a circle is (x - h)*^{2} + (y - k)^{2} = r^{2}.0198

*With parabolas, we put those in standard form by completing the square.*0210

*But at that time, we were just having to complete the square of either the x variable terms or the y variable terms.*0213

*Now, we are going to be working with both; and as always, we need to remember to add the same thing to both sides to keep the equation balanced.*0219

*If I was looking at something such as x*^{2} + y^{2} - 4x - 8y - 5 = 0,0226

*what I am going to do is keep all the x variable and y variable terms together, and then just move the constant over to the right.*0241

*The other thing I am going to do is group the x variable terms together: x*^{2} - 4x is grouped together, just like up here.0249

*And then, I am going to have y*^{2} - 8y grouped together, and add 5 to both sides.0258

*Now, I have to complete the square for both of these.*0266

*This is going to give me x*^{2} - 4x, and then here I need to have b^{2}/4.0269

*Since b is 4, that is going to give me 4*^{2}/4 = 16/4 = 4; so, I am going to put a 4 in here.0277

*For the y expression, I am going to have...let's do this up here...b*^{2}/4 = 8^{2}/4, which is going to come out to 16.0292

*So, I am going to add 16 here; and I need to make sure I do the same thing on the right.*0307

*So, I need to add 4, and I need to add 16; if I don't, this won't end up being balanced.*0312

*Now, I want this in this form; so let's change it to (x - 2)*^{2} +...here it is going to be (y - 4)^{2} =...0321

*4 and 16 is 20, plus 5; so that is 25.*0335

*This gives me the equation in standard form; and the center is at (h,k), (2,4).*0341

*And the radius...well, r*^{2} is 25; therefore, the radius equals 5.0350

*And as always, you need to be careful: let's say I ended up with something in this form, (x + 3)*^{2} + (y - 2) = 9.0355

*The temptation for the center might be just to put (3,2); but standard form says that this should be a negative.*0365

*So, I may even want to rewrite this as (x - (-3))*^{2} + (y - 2) = 9, just to make it clear that the center is actually at (-3,2).0372

*And then, the radius is going to be the square root of 9, which is 3.*0387

*So, be careful that you look at the signs; and if the signs aren't exactly the same as standard form,*0390

*you need to compensate for that, or even just write it out--because -(-3) would give me +3, so these two are interchangeable.*0396

*All right, in this example, we are asked to find the equation of the circle which has a diameter with the endpoints (-3,-7) and (9,-1).*0405

*So, let's see what we are working with.*0413

*Just sketch this out at (-3,-7), right about there; over here is (9,-1); there we have the diameter of the circle.*0418

*We have a circle like this, and we want to find the equation.*0433

*Recall that the formula for the equation of a circle is (x - h)*^{2} + (y - k)^{2} = r^{2}.0439

*Therefore, I need to find h; I need to find k; and I need to find the radius.*0449

*Recall that (h,k) gives you the center of the circle.*0453

*Since this is the diameter of the circle, the center of this line segment is going to be the center of the circle.*0460

*So, (-3,-7)...over here I have (9,-1); and here I have the center--the center is going to be equal to the midpoint of this segment.*0467

*Recall the midpoint formula equals (x*_{1} + x_{2})/2, and then (y_{1} + y_{2})/2.0477

*So, the center--the coordinates for that are going to be equal to (-3 + 9)/2, and then (-7 + -1)/2,*0491

*which is going to be equal to 6/2...-7 and -1 is -8/2; which is equal to (3,-4).*0507

*This means that h and k are 3 and -4; so I have h and k; I need to find the radius.*0520

*Well, half of the diameter...this is all the diameter; this is my midpoint; and I know that this is at (3,-4).*0526

*So, I just need to find this length--this is the radius.*0535

*And I now have endpoints; so I can use either one of these--I have a set of endpoints here and here, and here and here.*0539

*I am going to go ahead and use these two, and put them into the distance formula.*0549

*(-3,-7) and (3,-4)--I can use these in the distance formula: the distance here equals the radius,*0553

*which is the square root of...I am going to make this (x*_{1},y_{1}), and then this (x_{2},y_{2}).0562

*So, this is going to give me...x*_{2} is 3, minus -3, squared, plus...y_{2} is -4, minus -7, squared.0574

*So, the radius equals the square root of 3 + 3; a negative and a negative is a positive; all of this squared,*0589

*plus -4...and a negative and a negative is a positive, so -4 + 7, squared.*0599

*So, the radius equals the square root of...3 + 3 gives me 6, squared, plus...7 - 4 gives me 3, squared.*0608

*So, the radius equals √(36 + 9); 36 plus 9 is 45, so the radius equals √45.*0619

*But what I really want for this is r*^{2}, so r^{2} is going to equal (√45)^{2}, which is going to equal 45.0632

*Putting this all together, I can write my equation, because I now have h; I have k; and I have r.*0651

*So, writing the equation up here gives me (x - 3)*^{2} + (y...I can either write this as - -4,0658

*or I can rewrite this as (y + 4)*^{2} = r^{2}, which is 45.0672

*So, this, or a little more neatly, like this: (y + 4)*^{2} = 45--this is the equation for the circle.0679

*And I found that information based on simply knowing the diameter.*0689

*Knowing the diameter, I could use the midpoint formula to find the center, which gave me h and k.*0693

*And then, I could use the distance formula to find the distance from the center to the end of the diameter, which gave me the radius.*0698

*And then, I squared the radius and applied it to that formula.*0707

*Example 2: Find the center and radius of the circle with this equation.*0711

*In order to achieve that, we need to put this equation in standard form.*0716

*And recall that standard form of a circle is (x - h)*^{2} + (y - k)^{2} = r^{2}.0719

*So, we need to complete the square: group the x variable terms together on the left; also group the y variable terms on the left.*0730

*Add 8 to both sides to move the constant over.*0742

*I need x*^{2} - 8x + something to complete the square, and y^{2} + 10y + something to complete the square, equals 8.0746

*So, for the x variable terms, I want b*^{2}/4, and this is going to be 8^{2}/4, or 64/4, equals 16.0759

*So, I am going to add 16 here.*0773

*For the y variable terms, b*^{2}/4 is going to equal 10^{2}/4, which is 100/4, which is 25.0774

*I need to be careful that I add the same thing to both sides to keep this equation balanced, so I am also going to add 16 and 10 to the right side.*0785

*Correction: it is 16 and 25--there we go.*0805

*(x - 4)*^{2} equals this perfect square trinomial, and I am trying to get it in this form; that is what I want it to look like.0809

*Plus...(y + 5)*^{2} comes out to this perfect square trinomial.0820

*On the right, if I add 8 and 16 and 25, I am going to end up with 49.*0830

*8 and 16 is going to give me 24, plus 25 is going to give me 49.*0840

*Now, I have this in standard form: because the center of a circle is (h,k), I know I have h here,*0848

*and I know I have k here, this is going to give me...h is 4; k is -5.*0861

*Be careful with the sign here, because notice: this is (y + 5), but standard form is - 5, so this is equal to (y - -5)*^{2}--the same thing.0869

*It is just simpler to write it like this; but make sure you are careful with that.*0880

*The radius here: well, I have r*^{2}; the radius, r^{2}, I know, is equal to 49.0884

*Therefore, r = √49, so the radius equals 7.*0892

*Therefore, the center of this circle is at (4,-5), and the radius is equal to 7.*0898

*Example 3: Find the radius of the circle with the center at (-3,-4) and tangent to the y-axis.*0909

*This one takes more drawing and just thinking, versus calculating.*0919

*The center is at (-3,-4), right about here.*0924

*The other thing I know about this circle is that it is tangent to the y-axis.*0933

*That means that, if I drew the circle, it is going to extend around, and it is going to touch this y-axis.*0939

*Well, the radius is going to have one endpoint on the circle, and the other endpoint at the center.*0948

*Therefore, the radius has to extend from (-3,-4) over here.*0954

*And at this point, we are able to then find the length, because, since x is -3, and it has to go all the way to x = 0,*0965

*then this distance must be 3; therefore, the radius equals 3.*0976

*And again, that is because I know the center is here at -3, and I know*0984

*that the other endpoint of the radius is going to be out here, forming the circle,*0988

*and that, because it is tangent to the y-axis, x is going to be equal to 0 right here.*0996

*So, I know that x is equal to 0 here; and I know that x is equal to -3 over here; so it is just 1, 2, 3 over--this distance here is going to be 3.*1003

*And that is going to be the same as the radius, so the radius is equal to 3.*1013

*Example 4: Find the equation of the circle that is tangent to the x-axis, to x = 7, and to x = -5.*1018

*We are given a bunch of information about this circle and told to put it in standard form.*1026

*The first thing we are told is that it is tangent to the x-axis; so this circle is touching the x-axis; let's just draw a line here to emphasize that.*1040

*It is also tangent to x = 7; x = 7 is going to be right here--it is tangent to this.*1049

*It is also tangent to x = -5, out here.*1061

*I am going to end up with a circle that is touching, that is tangent to, these three things.*1068

*Let's think about what that tells me.*1074

*I need to find h and k (I need to find the center).*1077

*I also need to find the radius, so I can find r*^{2}.1081

*If this extends from -5 to 7, that gives me the diameter.*1086

*So, the diameter goes from 7 all the way to -5; so if I just add 7 and 5 (the distance from here to here,*1097

*plus the distance from here to here), I am going to get that the diameter equals 12.*1104

*The radius is 1/2 the diameter, so the radius equals 6.*1110

*I found that the radius equals 6.*1117

*The radius is going to extend from these endpoints to the center.*1122

*And I know that it is 6, so I know that the radius is going to go from 7 over here, 6 away from that.*1127

*7 - 6 is 1; it is going to go up to x = 1.*1136

*Again, that is because the length of the radius is 6, so the distance between the center and this endpoint has to be 6.*1141

*7 - 6 is 1; the radius is going to extend from there to there.*1152

*Therefore, the x-value of the center has to be 1.*1156

*Now, what is the y-value of the radius? The other thing I know is that this circle is tangent to this x-axis.*1164

*So, I know that it is going to have an endpoint on the x-axis.*1171

*And then, if it is going to extend from here to here, it is going to have to go up to 6; therefore, the center is at (1,6).*1175

*All right, so the radius equals 6, and the center is at (1,6); and that gives me an equation:*1193

*(x - 1)*^{2} + (y - 6)^{2} = the radius squared.1200

*If r = 6, then r*^{2} = 36.1208

*Again, that is based on knowing that this is tangent to x = -5, x = 7, and the x-axis.*1212

*So, I had the diameter, 12; I divided that by 2 to get the radius; I know that I have an endpoint here and an endpoint at the center.*1221

*So, that gives me the x-value of the center, which is 1.*1231

*I know I have an endpoint here, and also an endpoint at the center; so it has to be up at 6.*1233

*That gives me (1,6) for my value.*1238

*OK, and this is just drawn schematically, because the center would actually be higher up here.*1243

*This is just to give you...the center is actually going to be up here, now that I have my value: it is going to be at (1,6).*1247

*OK, that concludes this lesson of Educator.com on circles; thanks for visiting!*1258

*Welcome to Educator.com.*0000

*So far in conic sections, we have discussed parabolas and circles.*0002

*The next type of conic section we are going to cover is the ellipse.*0006

*First of all, what is an ellipse? Well, an ***ellipse** is formally defined as the set of points in a plane0012

*such that the sum of the distances from two fixed points is constant.*0019

*Well, what does that mean? First of all, this is the general shape of an ellipse.*0024

*And these two points here are the foci of the ellipse: we will call them f*_{1} and f_{2}.0029

*And looking back at this definition, it is the set of points in the plane...*0036

*and if you pick any of these points (say right here) and measure the distance from this point to one focus*0041

*(we will call that d*_{1}), and then we measure the distance from that same point to f_{2},0051

*to the other focus, we will call that distance d*_{2}.0059

*This definition states that, if I add up these two distances, d*_{1} + d_{2}, they will equal a constant.0063

*I could pick any point; I could then pick this point and say, "OK, here is another distance," calling that, say, d*_{3}, and this one d_{4}.0074

*And if I added up this distance and this distance, d*_{3} + d_{4}, I could get that same constant.0087

*And I could do that with any point on the ellipse.*0094

*Continuing on with some properties of the ellipse: an ellipse actually has two axes of symmetry.*0104

*One is called the major axis, and the other is the minor axis, and these intersect at the center of the ellipse.*0110

*Here, we are going to look at an ellipse that is centered at the origin (it has a center at (0,0)).*0117

*And again, I have foci f*_{1} and f_{2}.0122

*This is one of the vertices of the ellipse; here is a vertex; this is the second vertex of the ellipse.*0127

*And the major axis runs from one vertex to the other vertex.*0135

*And you can see that it passes through both of the foci; this is the major axis.*0140

*And what we are looking at here is an ellipse that has a horizontal major axis.*0149

*In a few minutes, we will look at ellipses that are oriented the other way--ellipses that are oriented with a vertical major axis.*0154

*So, that does exist; but right now we will just focus on this for the general discussion.*0163

*Here we have the major axis; and intersecting at the center is a second axis called the minor axis.*0167

*Looking more closely at the relationships between the major axis, the minor axis, the foci, and the distances relating them:*0182

*let's call the distance from one vertex to the center A: this distance is A.*0193

*Therefore, the length of the major axis is 2A.*0200

*And this is going to become important later on, when we are working with writing equations for ellipses,*0209

*or taking an equation and then trying to graph the ellipse.*0214

*That is my first distance: this is actually called the semimajor axis.*0218

*The length of the semimajor axis is A; the length of the major axis is 2A.*0226

*Now, looking at the minor axis: from this point to the center, this length is B.*0230

*Therefore, the length of the minor axis is equal to 2B.*0237

*Now, looking at the foci: the distance from one focus to the center is going to be C.*0246

*Therefore, the distance from one focus to another, or the distance between the foci, is equal to 2C.*0255

*There is also an equation relating these A, B, and C; and that is A*^{2} = B^{2} + C^{2}.0276

*So, keep this equation in mind: again, it becomes important, because you might be given A and B, but not C;*0287

*or you may be given a drawing, a sketch of the ellipse, and then asked to write an equation based on it.*0294

*And sometimes you need to find this third component in order to write that equation.*0300

*And you can do that by knowing that A*^{2} = B^{2} + C^{2}.0304

*Again, A is the length of the semimajor axis; B is the distance from here to the center of the minor axis*0309

*(2B is the length of the minor axis); and C is the distance between one focus and the center (2C gives the distance between the two foci).*0319

*Those are important relationships to understand when working with the ellipse.*0328

*Standard form: looking at what the standard form of the equation of an ellipse with a center at (0,0) and a horizontal major axis is:*0333

*it is x*^{2}/A^{2} + y^{2}/B^{2} = 1.0343

*This is the standard form; and again, we already discussed what A is and B is.*0353

*And if you figure out those from the graph, and are given those, then you can go ahead and write your equation.*0359

*Or, given the equation, you can graph the ellipse.*0365

*Let's take a look at an example: A*^{2}/9 + y^{2}/4 = 1.0368

*So again, this is just the equation for an ellipse with a horizontal major axis, so it is sketched out here that way.*0379

*We can make this more precise by saying, "OK, A*^{2} = 9; therefore, A equals √9, or 3."0388

*That means that the distance from here to here is going to be 3.*0405

*And since this is centered at the origin, this will actually be at the point (3,0).*0409

*So, let's write that as a coordinate pair, (0,3).*0415

*OK, that means that this length of A is 3; over here, the other vertex is going to be at (0,-3).*0420

*So, I have one vertex at (0,3) and the other at (0,-3); and I have f*_{1} here and f_{2} here.0431

*And the length of the major axis is actually 6: it is going to be equal to 2A.*0437

*2A equals 6, and that is the length of the major axis.*0443

*B*^{2} = 4: since B^{2} = 4, B = √4, which equals 2.0452

*Therefore...actually, this needs to be written the other way; this is actually (3,0), and (-3,0).*0461

*Now, up here, we have (0,2) and (0,-2).*0474

*And 2, then, is the length; that is B--that is the length of half of the minor axis.*0485

*2B = 4, and this is the length of the minor axis.*0492

*You can get a lot of information just by looking at the equation of the ellipse in standard form.*0498

*If I needed to, I could figure out the distance between the foci, and I could figure out what C is.*0505

*And remember, C is this length; because recall that A*^{2} = B^{2} + C^{2}.0512

*And I know that A*^{2} is 9; I know that B^{2} is 4; so I am looking for C^{2}.0518

*9 - 4 gives me 5, so C*^{2} = 5; therefore, C = √5, and that is approximately equal to 2.2.0524

*So, this distance here is roughly 2.2; and the distance between these foci would be about 4.4.*0535

*So, just by having this equation, I could graph the ellipse, and I could find this third component that was missing.*0544

*OK, so that was standard form for an ellipse that has a horizontal major axis.*0555

*For an ellipse that has a vertical major axis, you are going to see that the A*^{2} is associated with the y^{2} term.0560

*In the horizontal major axis, we had x*^{2}/A^{2}.0567

*Now, if you were given an equation, how would you know what you are dealing with--*0572

*if you were dealing with a vertical major axis or a horizontal major axis?*0578

*Well, A*^{2} is the larger number; so let's say I was given something like y^{2}/16 + x^{2}/9 = 1.0581

*When I look at this, I see that the larger number is associated with y*^{2}; so that tells me that I have a vertical major axis.0594

*If it had been x*^{2} associated with 16, then I would have said, "OK, that is a horizontal major axis."0602

*Again, I can graph this ellipse by having this equation written in standard form.*0609

*I know that A*^{2} equals 16; therefore, A equals the square root of 16; it equals 4.0614

*This time, I am going to go along here for the major axis; and that makes sense,*0622

*because I have a focus here and another focus here, and the major axis passes through the two foci.*0628

*OK, so (0,4) is going to give me one vertex; (0,-4) will give me the other vertex.*0638

*And remember that 2A = 8, and that is going to be the length of the major axis.*0647

*And again, right now, we are limiting our discussion to ellipses with a center at (0,0).*0658

*Later on, we will expand the discussion to talk about the graphs of ellipses with centers in other areas of the coordinate plane.*0664

*OK, so now I have B*^{2} = 9; therefore, √9 is going to equal B; this tells me that B equals 3.0672

*So, since B equals 3, the length of half of the minor axis is going to be 3.*0689

*So, right here at (3,0) is going to be one point, and I am going to have the other point right here at (-3,0).*0696

*And the length of the minor axis...2B = 6, and that is the length of the minor axis.*0704

*Again, I can use the relationship A*^{2} = B^{2} + C^{2}0719

*to figure out what C is, and to figure out where the foci are located.*0725

*I know that A*^{2} is 16; I know that B^{2} is 9; and I am trying to figure out C^{2}.0730

*So, I take 16 - 9; that gives me 7 = C*^{2}; therefore, C = √7.0738

*And the square root of 7 is approximately equal to 2.6, so this is going to be up here at about (0,2.6).*0746

*And f*_{2} is going to be down here at about (0,-2.6).0755

*So again, using standard form, you can graph the ellipse, and you can find where the foci are, based on the values of A*^{2} and B^{2}.0758

*We talked a little bit about graphing ellipses; but sometimes you are not given the equation in standard form.*0771

*If the equation is not in standard form, you have to put it that way.*0776

*And working with other conic sections, we have learned that you can put an equation into standard form for a conic section by completing the square.*0780

*You can also use symmetry, just as we did when graphing parabolas or circles.*0788

*So, let's say you are given an equation like this: 3x*^{2} + 4y^{2} - 18x - 16y = -19; and you are asked to graph it.0794

*You are going to put it in standard form; but it is nice, first of all, to know what kind of shape you are working with.*0807

*And you can tell that just by looking at this, even though it is not in standard form yet.*0811

*And the reason is: I see that I have an x*^{2} term and a y^{2} term0815

*on the same side of the equation, with the same sign, with different coefficients; that tells me that I am working with an ellipse.*0820

*And we are going to go into more detail in the lecture on conic sections.*0827

*We are going to review how to tell apart the equations for various conic sections.*0831

*But just briefly now: recall that a parabola would have either an x*^{2} term or a y^{2} term, not both.0835

*With a circle, the coefficients of x*^{2} and y^{2} are the same; that is for a circle.0845

*For an ellipse, the coefficients of the x*^{2} and y^{2} terms are different.0861

*Now, you see that these have the same sign; so with an ellipse,*0876

*it is important to note that the x*^{2} and y^{2} terms have the same sign.0881

*If they don't have the same sign, that is actually a different shape.*0884

*They have the same sign; but this coefficient (the x*^{2} coefficient) is 3, and the y^{2} coefficient is 4.0887

*That tells me I have an ellipse, not a circle.*0894

*My next step is to complete the square and write this in standard form,*0898

*so that, if I wanted to graph it, I would have all of the information that I need.*0903

*The first thing to do is to group the x variable terms and the y variable terms.*0906

*This gives me x*^{2} - 18x + 4y^{2} - 16y = -19.0911

*Now, when I am looking at this, I remember that, to complete the square, I want to end up with a leading coefficient of 1.*0931

*I am going to factor out this 3 to get 3(x*^{2} - 6x); and I am going to need to add0937

*something else over here to complete the square--a third term.*0944

*Here, factor out the 4 to give me (y*^{2} - 4y) = -19.0949

*Recall that, to complete the square, you are going to add b*^{2}/4 to each set of terms.0957

*So, for this x group, we have b*^{2}/4 = -(6)^{2}/4 = 36/4 = 9.0963

*So, I am going to add a 9 here; and very importantly, to the right side, I am going to add 9 times 3, which is 27.*0978

*Add that to the right, because if I don't, the equation won't be balanced anymore.*0995

*Now, the y variable terms: I need to add something here to complete the square.*1003

*And I will work over here for this: b*^{2}/4 equals (-4)^{2}/4 = 16/4 = 4.1013

*So, I am going to add 4 here; but to the right, what I need to add is 4 times 4; so I need to add 16.*1026

*Now that I have this, my next step is to write this in standard form: 3...and this is x - 3, the quantity squared;*1038

*if I squared this, I would get this back; plus 4y - 2, the quantity squared, equals...*1049

*-19 plus 16 gives me, let's see, -3; 27 minus 3 gives me 24 on the right.*1057

*Well, recall that for standard form, I want a 1 over here on the right; I want this to equal 1.*1069

*So, this gives me...I need to divide both sides by 24; this is going to give me 3(x - 3)*^{2}/24 + 4(y - 2)^{2}/24 = 24/24.1076

*This finally leaves me with (x - 3)*^{2}...this 3 will cancel, and I am going to get 8 in the denominator;1101

*this 4 will cancel, and I will get 6 in the denominator; equals 1.*1109

*Now, I have the bigger term associated with x; that tells me that I have a horizontal major axis.*1115

*So, the major axis is horizontal; and let me go ahead and write that up here.*1126

*x*^{2}/a^{2} + y^{2}/b^{2} = 1.1144

*OK, now you can see that this looks slightly different; and we have these extra terms here.*1154

*And what that tells us is something that we are going to discuss in a moment.*1160

*And what we are going to discuss is situations where the center of the ellipse is not at (0,0).*1164

*But we have the same information that we would have with the center being at (0,0).*1174

*And that is that I have A*^{2}, which equals 8; and I know I have a horizontal major axis; and I have B^{2} = 6.1180

*Looking, now, at equations for ellipses with centers at (h,k), at somewhere other than 0:*1198

*if you look at the situation where we have a horizontal major axis, versus a vertical major axis,*1206

*you can again see that it is very similar to when the center is at (0,0).*1212

*The x is associated with the A*^{2} term when the major axis is horizontal.1217

*The y is associated with the A*^{2} term when the major axis is vertical.1221

*The only difference is that we now have these terms telling us where the center is.*1227

*And if the center was going to be 0, all that would happen is that you would have x - 0 and y - 0,*1231

*which then gives you back the equation we worked with before, x*^{2}/A^{2} + y^{2}/B^{2},1237

*or y*^{2}/A^{2} + x^{2}/B^{2}, all equal to 1.1244

*So, let's talk about an example for this: (x - 4)*^{2} + (y + 6)^{2} = 1.1252

*This is all over 100, and this is all divided by 25.*1264

*What this tells me is that the center is at (h,k); so this is h (that is 4).*1269

*You need to be careful here, because what you have is a positive; but the standard form is a negative.*1278

*And it is perfectly fine to write this like this, but you need to keep in mind that this is really saying...*1284

*if you think about it, y + 6 is the same as y - -6.*1291

*So, when I look at it this way, I realize that, if it is in this form, k is actually going to be -6.*1296

*And if you are not careful about that, you can end up putting your center in the wrong place.*1302

*Let's let this be 2, 4, 6, -2, -4, -6; the center is at (4,-6), right here.*1309

*A*^{2} = 100; my larger term is my A^{2}term, and I have a horizontal major axis.1323

*Therefore, A = √100, which is 10.*1331

*So, since A equals 10 and the length of the major axis is horizontal, then I am going to need to go 10 over from 4.*1337

*It is going to be all the way out here at 14.*1351

*So, this is where one vertex would be--this is going to be at 4 + 10 gives me (14,-6).*1357

*That is one vertex: -2, 4, 6, 8, 10, 12, 14...so -14 is going to be about here.*1368

*And I am going to have the other vertex here at (-14,-6).*1376

*I have a minor axis: B*^{2} = 25; therefore B = 5.1384

*So, what that tells me is that I come here at -6; I add 5 to that; and that is going to bring me right here at -1; that is going to be right about here.*1389

*And then, -8, -10, -12...-(6 + 5) is going to be down here at -11; so this ellipse is going to roughly look like this.*1400

*OK, and standard form allowed me to determine that the center is at (4,-6);*1419

*that I have a vertex; if I add 10, that is going to give me this vertex right over here*1427

*(let's draw this more at the vertex); that I have another vertex here; that the major axis is going to pass through here;*1433

*and then, I am going to have a minor axis passing through here.*1441

*All right, in the first example, we are asked to find the equation of the ellipse that is shown.*1447

*And I am going to go ahead and label some of these points.*1453

*This is going to be f*_{1}; and let's say we are given f_{1} as being at (0,3).1457

*This is going to be 1, 2, 3: each mark is going to stand for one.*1463

*Down here, I have f*_{2}; that should actually be down here a little bit lower; so that is f_{2} at (0,-3).1467

*Let's say we are also given a vertex at (0,5), and the other vertex at (0,-5).*1480

*All right, so I am asked to find the equation of the ellipse shown.*1492

*And the one thing I see is that there is a vertical major axis.*1495

*Since there is a vertical major axis, it is going to be in the general form (y - k)*^{2}/A^{2} + (x - h)^{2}/B^{2} = 1.1503

*Now, I notice that the center is at (0,0); so that tells me that h and k are (0,0).*1520

*So, this is actually going to become the simpler form, y/A*^{2} + x/B^{2} = 1.1528

*The next piece of information I have is the length of A.*1537

*So, this line is going to give me A; and I know that, since the center is here at (0,0), the length of this is 5; therefore, A = 5.*1541

*Since A equals 5, A*^{2} is going to equal 5^{2}, or 25.1553

*I don't know B; this is going to be B, but I don't know what it is.*1561

*However, I do know C; if I look here, this is C--from -3 to the center is 3, so C = 3.*1567

*Therefore, C*^{2} = 3^{2}, or 9.1583

*The other piece of information I have is the equation we have been working with,*1587

*that states that A*^{2} = B^{2} + C^{2}.1591

*Therefore, I can find B*^{2}, which I need in order to write my equation.1602

*So, I know that A*^{2} is 25; I know that C^{2} is 9; and I am looking for B^{2}.1606

*25 - 9 is 16; therefore, B*^{2} = 16, and B = 4.1620

*Now, I can go ahead and write my equation; I have all the information I need, so let's put it all together right here.*1628

*y, divided by A*^{2}...I know that A^{2} is 25...(that is actually y^{2} right here--1633

*let's not forget the squares)...plus x*^{2} divided by B^{2}...B^{2} is 16...equals 1.1647

*So again, this had a center at (0,0); so remember that this is the general form of the equation.*1655

*When you have a center at (0,0), it becomes this form.*1660

*I had a vertical major axis, so I am using this form, where the A*^{2} is associated with the y^{2} term.1663

*And by having A*^{2} and C^{2}, I was able to determine what B^{2} is.1671

*Example 2: let's find the equation of the ellipse satisfying a major axis 10 units long and parallel to the y-axis; minor axis 4 units long; center at (4,2).*1678

*Let's start with the center--the center is at (4,2)--the center is right here at (4,2).*1690

*And it says that the major axis is parallel to the y-axis.*1699

*What that tells me is that I have a vertical major axis.*1703

*Since this is a vertical major axis and I know the center (I know (h,k)), I am going to be working with the general form*1711

*(y - k)*^{2}/A^{2} + (x - h)^{2}/B^{2} = 1.1719

*I have h; I have k; I need to find A*^{2} and B^{2}.1728

*The other piece of information I have is that the major axis is 10 units long; and I know that it is vertical.*1732

*Since it is 10 units long, that means that the major axis length is equal to 2A, and I know that that length is 10;*1738

*therefore, I just take 10 divided by 2, and I get that A = 5.*1751

*So, I am starting here; and if I take 5 + 2, I am going to get that I am going to have the vertex up here at 7.*1757

*5 + 2 is going to give me 7, right there; so I go to (4,7); that is where this is going to be.*1775

*And I got that by keeping the x where it was, and then adding 2 where the center was and adding 5 to that, which is the length of A.*1793

*Then, I am going to go down; again, it is going to be at 4.*1803

*But then, if I take 2 and subtract 5 from it, I am going to get -3.*1810

*So, right here at (4,-3) is the other vertex.*1815

*Now, I was not asked to graph this; but I am graphing it so that I have an understanding of what each of these means,*1823

*so that I can go ahead and write the equation.*1829

*What I have now is...I know what A is, which means I can figure out A*^{2}; and I know the center.1832

*The last thing I know is that the minor axis length equals 2B: I am given that that is 4 units long; therefore, B = 2.*1844

*So, if I started here at 4, and I added 2 to that, I am going to end up with (6,2).*1857

*And I am going to end up with...if I start at 4 and I subtract 2, I am going to end up with (2,2).*1867

*And this is B; this is A.*1881

*I have a minor axis that has a length of 2B, which tells me that B = 2.*1887

*From this information, I can go ahead and write this equation.*1891

*I know A equals 5, so A*^{2} equals 5^{2}, which is 25.1897

*B = 2; B*^{2}, therefore, equals 2^{2}, or 4.1904

*So, I have everything I need to write this: (y - k)...well, k is 2; the quantity squared, divided by A*^{2};1911

*A*^{2} is 25; plus (x - h)...h is 4...the quantity squared, divided by B^{2}, which is 4, equals 1.1920

*So, this standard form describes the ellipse with the major and minor axis here.*1936

*And you could finish that out by just connecting these points and drawing the ellipse if you wanted to finish graphing it.*1941

*Find the equation of the ellipse satisfying endpoints of the major axis at (11,3) and (-7,3) and foci at (7,3) and (-3,3).*1953

*All right, endpoints of the major axis: let's do 2, 4, 6, 8, 10, 12, and -2, 4, -6...OK.*1962

*The endpoint is at (11,3): 11 is right here, and then we will have 3 be right here.*1979

*The other endpoint is at -7, which is going to be here, 3.*1991

*And that tells me the major axis: since the major axis is horizontal, we are going to be working with the general form*1999

*(x - h)*^{2}/A^{2} + (y - k)^{2}/B^{2} = 1.2010

*So, that is my first piece of information.*2018

*I can also figure out the length of the major axis.*2020

*So, since the major axis goes from -7 to 11, if I take -7 minus 11, I am going to get -18.*2024

*And a length is going to be an absolute value, so I am just going to take 18; the length is going to be the absolute value of this difference.*2045

*All right, I know that the major axis length is 18; the other thing I know is that the major axis length equals 2A, as we have discussed.*2053

*So, 2A = 18; therefore, A = 9; so I know that the distance from the center to this endpoint is 9.*2061

*Therefore, I could just say, "OK, 9 minus 11 gives me 2; and I know that the y-coordinate will be 3; so that is (2,3)."*2076

*Another way to solve this, without using all this graphing, would have simply been to find the midpoint.*2087

*I am going to go ahead and do that, as well, because the midpoint of the major axis is the center of the ellipse.*2093

*Let's try that, as well--the center, using the midpoint formula.*2100

*Recall the midpoint formula: we are going to take x*_{1} + x_{2} (that is 11 + -7);2105

*and we are going to take the average of that (we will divide it by 2); that is going to give me the x-coordinate.*2112

*For the y-coordinate, I am going to take 3 + 3, and I am going to divide that by 2.*2117

*And this will give you a center at...11 - 7 gives you 4; 4 divided by 2 is 2.*2126

*3 + 3 is 6, divided by 2 gives you 3; and that is exactly what I came up with using the graphing method.*2138

*So, just to show you: you could have figured this out algebraically (where the center is); or you could have figured it out using graphing.*2144

*This gives me (h,k), so I have (h,k), and I have A, which is 9, so I can get A*^{2}.2154

*The next thing I need to do is figure out B, and they don't give me B; but what they do give me are the foci.*2162

*There are foci at (7,3) (7 is here--focus at (7,3)--we will call this one f*_{1} at (7,3)); and this is the center right here.2167

*There is another focus at (-3,3): f*_{2} is going to be at (-3,3).2187

*Recall that the distance from one focus to the other is 2C; the distance from one focus to the center is C.*2201

*Let's just work with this; this is C; 7 - 2 is 5; therefore, C = 5.*2210

*So, I have A = 9; I have C = 5 (again, that is the distance from the center to a focus);*2221

*or I could have figured out the distance from one focus to another--that is 2C--and then divided by 2;*2229

*So, I have A, and I have C, and I know that A*^{2} = B^{2} + C^{2}.2233

*So, this gives me 9*^{2} = B^{2} + 5^{2}: 9^{2} is 81, equals B^{2} + 25.2240

*If I take 81 - 25, I am going to get 56 = B*^{2}.2254

*And I don't even need to take the square root of that, because to put this into standard form, I actually need B*^{2}.2259

*So, I get (x - h); remember, the center is (h,k), so that is 2; the quantity squared, divided by A*^{2}...2265

*recall that A*^{2} is 9^{2}, so it is 81; plus (y - k)^{2}...k is 3, the quantity squared;2273

*divided by B*^{2}...I determined that B^{2} is 56; all of this equal to 1.2284

*Just by knowing the endpoints of the major axis and the location of the foci, I could figure out A*^{2}2289

*and B*^{2}, as well as the center, and then write this equation for the ellipse in standard form.2295

*Finally, we are asked to graph an ellipse that is not given to us in standard form.*2307

*We have some extra work to do: we are going to actually have to complete the square in order to even graph this.*2311

*So, let's go ahead and start by grouping the x terms together and y terms together.*2318

*Also note that, since this has an x*^{2} term and a y^{2} term on the same side of the equation,2329

*with the same sign (they are both positive), but different coefficients, I know I have an ellipse.*2335

*It is not a circle, because for a circle, these coefficients would be the same.*2341

*Grouping the terms together gives me 14x*^{2} - 56x + 6y^{2} - 24y = -38.2347

*Looking at this, I see that I have a common factor of 2.*2368

*If I divide both sides by 2, I can simplify this equation; so the numbers I will work with will be smaller.*2372

*So, let's divide both sides by 2 to get 7x*^{2} - 28x + 3y^{2} - 12y = -19.2378

*The next thing to do is to factor out the leading coefficient, since it is not 1.*2391

*I am going to factor out a 7 to get x*^{2} - 4x, plus...2397

*over here, I am going to factor out a 3; that gives me y*^{2} - 4y, all equals -19.2402

*I need to complete the square, so I need to get b*^{2}/4.2411

*In this case, that is going to give me 4*^{2}/4 = 16/4 = 4.2415

*So, over here, I am going to add a 4; very important--on the right, I have -19, and I am adding to the left 7 times 4; that is 28.*2423

*So, I need to add 28 to the right.*2440

*All right, over here, for the y terms: b*^{2}/4: again, we have a b term that is 4, so I am going to end up with the same thing, 4.2446

*Now, here I am actually adding 4 times 3 (is 12), so I need to add that to the right, as well, to keep the equation balanced.*2462

*This is the easiest step to mess up on: you are focused here on completing the square,*2470

*and then you sometimes don't remember that you have to add the same thing to both sides.*2474

*Now, working on writing this in standard form: this is going to give me (x - 2)*^{2} + 3(y - 2)^{2} = -19 + 12...2479

*that is going to leave me with 28; -19 + 12 is going to be -7; 28 minus 7 equals 21.*2493

*To get this into standard form, I need to have a 1 on the right, so I am going to divide both sides by 21.*2504

*This is going to give me 7(x - 2)*^{2}/21 + 3(y - 2)^{2}/21 = 21/21.2510

*This cancels to (x - 2)*^{2}/3; this becomes (y - 2)^{2}/7; and this just becomes 1.2528

*Now, I have standard form; I can do some graphing.*2547

*I have a center at, let's see, (2,2); that is right here; the center is at (2,2), so that is h and k.*2550

*And I notice, actually, that the larger term is under the y; it is associated with the y.*2577

*That tells me that this has a vertical major axis; and therefore, I am going to keep that in mind--that it is going to be oriented this way.*2587

*The ellipse is going to end up like this, instead of like this.*2600

*I have my center at (2,2); therefore, A*^{2} = 7; A = √7.2604

*The square root of 7 is about 2.6, so it gets messy, as always, when you are working with radicals.*2620

*But it is about 2.6; so what I have to do is say, "OK, my vertex up here is going to be at x = 2, and then y is going to equal 2 + 2.6, which is 4.6."*2626

*So, (2,4.6): and again, I got that by saying the length of A (the length from the vertex to the center) is 2.6.*2643

*So, 2 + 2.6 gives me 4.6.*2655

*Over here, I am going to take 2, and I am going to subtract 2.6 from it; so x is still going to be 2; now y is going to be about here, which is (2, -.6).*2658

*Again, 2 is here; the length of A is 2.6, so it is going to be 2 - 2.6; this gives me my major axis.*2674

*I know, from vertex to vertex, where this ellipse is going to land.*2685

*Now, the minor axis, B: I know that B*^{2} = 3; therefore, B = √3, which is approximately equal to 1.7.2690

*This is not to do the same thing, but going along the x direction, the horizontal direction.*2700

*So again, this is A; OK, now to get B, I am going to have 2, and I am going to add 1.7 to it.*2706

*That is going to give me 3.7; it is going to land about here.*2715

*In this direction, I am going to subtract; I am going to say that I have 2 - 1.7, so that is going to land here, at about .3.*2722

*And in the y direction (in the vertical direction) it is still going to be up at 2.*2739

*So again, to get this, I said 2 + 1.7; that brought me to (3.7,2)--that is this point.*2742

*This point is at 2 minus 1.7, so that is (.3,2).*2753

*So, this gives me the general shape of this ellipse, like this; so I can get a good sketch, based on this equation.*2761

*I took this equation, and I recognized that it was an ellipse.*2782

*I completed the square to get it in standard form, and saw that it had a vertical major axis and that it had a center at (2,2).*2787

*I then found A to determine where the vertices would be, and then B to determine the width of the ellipse here; and then I could get a good sketch.*2795

*That concludes this session of Educator.com on ellipses; thanks for visiting!*2806

*Welcome to Educator.com.*0000

*Today, we are going to be discussing the last type of conic section, which is hyperbolas.*0002

*So far, we have covered parabolas, circles, and ellipses.*0006

*As you can see, hyperbolas are a bit different in shape than the other conic sections we have worked with.*0012

*And one thing that makes them unique is that there are two sections referred to as branches; there are two branches in this hyperbola.*0018

*The formal definition is that a ***hyperbola** is a set of points in the plane,0025

*such that the absolute value of the differences of the distances from two fixed points is constant.*0029

*What does that mean? First, let's look at the foci.*0037

*These two fixed points are the ***foci**; and here is a focus, f_{1}, and here is the other, f_{2}.0041

*If I take a point on the hyperbola, and I measure the distance to f*_{1},0048

*and then the distance to f*_{2}, that is going to give me d_{1} and d_{2}.0056

*Recall that, with ellipses, we said that the distance from a point on the ellipse--if you measured the distance to one focus,*0067

*and then the other focus, and then added those, that the sum would be a constant.*0074

*Here, we are talking about the difference: the absolute value of this distance, d*_{1}, minus (the difference) d_{2}, equals a constant.0078

*That is the formal definition of a hyperbola.*0092

*Again, I could take some other point: I could take a point up here on this other branch.*0094

*And I could find a distance, say, d*_{3}, and then the distance to f_{1} could be something--say d_{4}.0100

*Again, the absolute value of those differences would be equal to that same constant.*0107

*All right, properties of hyperbolas: A hyperbola, like an ellipse, has two axes of symmetry.*0117

*But these have different names: here you have a transverse axis and a conjugate axis, and they intersect at the center.*0124

*We are looking here at a hyperbola with the center at (0,0).*0133

*One thing to note is that you can also have a hyperbola that is oriented as such.*0139

*But right now, we are looking at this one, with a more horizontal orientation.*0147

*But just to note: this does exist, and we will be covering it.*0153

*All right, first discussing the transverse axis: the transverse axis is going to go right through here--it is going to pass through the center.*0156

*And this is the vertex, and this is the vertex of the other branch, and this is the transverse axis.*0166

*The distance from one vertex to the center along this transverse axis is going to be A.*0179

*Again, a lot of this is going to be similar from when we worked with ellipses; but there are some important differences, as well.*0186

*So, the length of the transverse axis equals 2A; from here to here would be 2A.*0195

*The foci: if you look at foci, say f*_{1}, f_{2}...let's look at f_{2}...0212

*it would be the same over on f*_{1}: if I looked at the distance from one focus to the center, that is going to be C.0218

*The distance between the foci is therefore 2C; if I measured from here to here, that length is going to be 2C.*0229

*There is a second axis called the conjugate axis; and the two axes intersect here at the center.*0246

*The length of half...if you take half the length of this conjugate axis, it is going to be equal to B.*0260

*The transverse axis lies along here; the conjugate axis, in this case, is actually along the y-axis (and the transverse axis is along the x-axis).*0273

*The length of the conjugate axis is 2B.*0288

*As with the ellipse, there is an equation that relates A, B, and C; but it is a slightly different equation.*0297

*Here, A, B, and C are related by C*^{2} = A^{2} + B^{2}.0303

*This relationship will help us to look at the equation for a hyperbola and graph the hyperbola,*0311

*or look at the graph, and then go back and write the equation.*0317

*So again, there are two axes: transverse, which goes from vertex to vertex; and conjugate, which intersects the transverse axis*0321

*at the center of the hyperbola and has a length of 2B (the transverse axis has a length of 2A).*0331

*The distance from focus to focus (between the two foci) is 2C.*0338

*The standard form of the hyperbola is also going to look somewhat familiar, because it is similar to an ellipse, but with a very important difference.*0346

*Here we are talking about a difference instead of a sum.*0352

*So, if you have a hyperbola with a center at (0,0) and a horizontal transverse axis, the equation is x*^{2}/A^{2} - y^{2}/B^{2}.0355

* And here, we again have that the center is at the origin, (0,0).*0368

*Although the center certainly does not have to be at the origin, right now we are going to start out*0373

*working with hyperbolas with a center at the origin, just to keep things simple.*0377

*And again, by being given an equation in standard form, you can look at it and get a lot of information about what the hyperbola looks like.*0384

*Therefore, the vertex is going to be at (A,0); the other vertex will be at (-A,0).*0396

*You are going to have a point up here that is going to be (B,0); and this length, B, gives the length of half of the conjugate axis.*0408

*And then, you are going to have another point...actually, that is (0,B), because it is along the y-axis...another point, (0,-B).*0420

*This distance is B, from this point to the center; this distance is A.*0429

*And then here, I have f*_{1} and f_{2}; and the distance from one of those to the center is C.0436

*As I mentioned, you can have a hyperbola that is oriented vertically.*0447

*So, if the transverse axis is vertical, and the center is at (0,0), the standard form is such that y*^{2} is associated with the A^{2} term.0451

*And here, it is positive: so you are taking y*^{2}/A^{2} - x^{2}/B^{2} = 1.0463

*In this case, what you are going to have is a vertex right here at (0,A), the other vertex is here at (0,-A); here is the transverse axis.*0470

*And then, you are going to have the conjugate axis; the length of half of that is going to be B; the length of the entire thing is 2B.*0482

*So, this is going to be some point, (B,0); and B*^{2} is given here, so you could easily find B by taking the square root.0495

*And then, over here is (-B,0).*0502

*So again, there are two different standard forms, depending on if you are working with a hyperbola*0505

*that has a horizontal transverse axis or one that has a vertical transverse axis.*0508

*Something new that we didn't talk about with ellipses is asymptotes.*0516

*Recall that an asymptote is a line that a curve on a graph approaches, but it never actually reaches.*0520

*And asymptotes are very useful when you are trying to graph a hyperbola.*0528

*The equations are given here: let's go ahead and draw these first.*0534

*Now, recall that this vertex is at a point (A,0); this vertex is at (-A,0).*0536

*If I measure the length...this is the transverse axis, and it is horizontal...let's say that it turns out that B is right up here, (0,B).*0545

*And B is going to be the length from this point to the center; 2B will be the length of the conjugate axis.*0555

*Then, I am going to have another point down here, (0,-B).*0564

*What I can do is form a box, a rectangle; and the rectangle is going to have vertices...I am going to go straight up here and across here.*0568

*Therefore, this is going to be given by (-A,B); that is going to be one vertex.*0578

*I can go over here and do the same thing; I am going to go straight up from this vertex, and straight across from this point.*0586

*And that is going to give me the point (A,B).*0592

*I'll do the same thing here: I go down directly and draw a line across here; this point is going to be (-A,-B).*0597

*One final vertex is right here: and this is given by (A,-B).*0608

*OK, now you draw a box using these A and B points; and then you take that rectangle and draw the diagonals.*0618

*If you continue those diagonals out, you will have the two asymptotes for the hyperbola.*0630

*OK, so each of these lines is an asymptote.*0650

*And notice that the hyperbola is going to approach this, but it is not actually going to reach it.*0654

*So, it is going to continue on and approach, but not reach, it; it is going to approach like that.*0663

*All right, now this is one way to just graph out the asymptotes.*0671

*You can also find the equation; and for right here, we are working with a hyperbola with a horizontal transverse axis.*0676

*So, we are going to look at this equation.*0686

*If I was working with a vertical one, I would look at this equation.*0688

*Now, what does this mean? Well, y equals ±(B/A)x.*0691

*What this actually is: this B/A gives the slope of the asymptote.*0697

*Recall that y = mx + b; since the center is at (0,0), the y-intercept is 0; so here, b = 0, so I am going to have y = mx.*0703

*The slope, m, is B/A; B/A for this line is increasing to the right (m = B/A);*0714

*and the slope here equals -B/A, where the line is decreasing as we go towards the right.*0724

*So again, there are two ways to figure out these asymptotes.*0734

*You can just sketch it out by drawing this rectangle with vertices at (-A,B), (A,B), (-A,-B)...that is actually (A,B), positive (A,B)...or (-A,-B).*0737

*Draw that rectangle and extend the diagonals.*0754

*Or you can use the formula, which will give you the slope for these two asymptotes.*0758

*If you just started out knowing the A's and B's and drew these, then you could easily sketch the hyperbola,*0765

*because you know that it is going to approach these asymptotes.*0772

*OK, so we have talked a lot about graphing.*0778

*And just to bring it all together: you are going to begin by writing the equation in standard form.*0780

*And then, for hyperbolas, you are going to graph the two asymptotes, as I just showed.*0786

*So, let's start out with an example: let's make this x*^{2}/9 - y^{2}/4 = 1.0790

*Since I have this in the form x*^{2}/A^{2} (this x^{2} term is positive here),0801

*divided by y*^{2}/B^{2} = 1, what I have is a horizontal transverse axis.0809

*So, this tells me that there is a horizontal transverse axis.*0819

*So, that is how this is just roughly sketched out already, showing the transverse axis along here.*0830

*Since it is in this form, I know that A*^{2} = 9; therefore, A = 3.0841

*This has a center right here at (0,0).*0853

*And this point here is going to be A, which is 3, 0.*0860

*Right here, I am going to have -A, or -3, 0.*0867

*So, my goal is to make that rectangle extend out the diagonals.*0872

*And then, I would be able to graph this correctly.*0876

*OK, B*^{2} = 4; therefore, B = 2; so right up here at (I'll put that right there) (0,2)...that is going to be B.0882

*And then, right down here at (0,-2)...*0903

*Now, all I have to do is extend the line up here and here; and these are going to meet at (2,3).*0907

*Extend a line out here; I am going to have a vertex right here at (-3,2).*0916

*I am going to have another vertex here at (-3,-2), and then finally, one over here at (3,-2).*0924

*Now, this was already sketched on here for me; but assuming it was not there, I would have started out by drawing this box,*0936

*and then, drawing these lines extending out--the asymptotes.*0944

*And what is going to happen is that this hyperbola is actually going to approach, but it is never going to intersect with, the asymptote.*0960

*So again, write the equation in standard form, which might require completing the square.*0972

*I gave it to you in standard form already; use that to figure out this rectangle.*0977

*And you are going to need to know A and B to figure out this rectangle.*0982

*Draw the asymptotes, and draw then the hyperbola approaching (but not reaching) those asymptotes.*0985

*So far, we have been talking about hyperbolas with a center at the origin (0,0).*0993

*However, that is not going to always be the case.*1000

*If the center is at another point, (h,k), that is not (0,0), then standard form looks like this.*1002

*It is very similar to what we saw with the origin of the center, except instead of just x*^{2}/A^{2}, we now have an h and a k.1008

*For a horizontal transverse axis, you are going to have (x - h)*^{2}/A^{2} - (y - k)^{2}/B^{2}.1015

*For a vertical transverse axis, this term is going to be first; it will be positive.*1027

*And then, you are going to subtract (x - h)*^{2}/B^{2}.1033

*But the k stays associated with the y term.*1037

*For example, given this equation, (y - 3)*^{2} - (x - 2)^{2}...and we are going to divide that by 16,1041

*and divide this by 9, and set it all equal to 1: what this is telling me is that the center is at (2,3), because this is h;*1053

*that A*^{2} = 16, so A = 4; and that B^{2} = 9, so B = 3.1065

*From that, I can graph out this hyperbola.*1072

*And this has a vertical transverse axis; something else to be careful of--let's say I had something like this:*1076

*(y + 5)/10, the quantity squared, plus (x + 4), the quantity squared, divided by 12, equals 1.*1087

*The center is actually at (that actually should be a negative right here--this is a difference) (-4,-5).*1101

*And the reason for that is that this is the same as (y - -5)*^{2}, and then (x - -4)^{2}.1112

*A negative and a negative is a positive.*1125

*So, you need to be careful: even though it is acceptable to write it like this, it is good practice,*1128

*if you are trying to figure out what the center is, to maybe write it out like this,*1134

*so that you have a negative here, so that whatever is in here is already k, or already h.*1139

*You don't have to say, "Oh, I need to make that a negative; I need to change the sign."*1145

*So, that is just something to be careful of.*1148

*Here, I already had negative signs in here; they are completely in standard form--I have h and k here; h and k is (-4,-5).*1151

*All right, to get some practice, we are going to first find the equation of a hyperbola that I am going to give you some information on.*1160

*I will give you that one of the vertices is at (0,2); the other vertex is at (0,-2).*1168

*The other piece of information is that you have a focus at (0,4), and a focus called f*_{2} at (0,-4).1178

*So, looking at this, I can see that this is the transverse axis, and then the center is right there.*1188

*So, I have a horizontal transverse axis.*1196

*I can also see that the midpoint right here, the center, is at the origin; so the center equals (0,0).*1207

*So, this is actually vertical--correction--a vertical transverse axis, going up and down: a **vertical* transverse axis.1217

*Since this is actually a vertical transverse axis with a center at (0,0), I am working with this standard form:*1227

*y*^{2}/A^{2} - x^{2}/B^{2} = 1.1234

*So, the A*^{2} term is with the y^{2} term, since this is a vertical transverse axis.1240

*All right, in order to find the equation, I need to find A*^{2}.1247

*This distance, from 0 to the vertex, is 2, because this is at (0,2).*1251

*Therefore, A equals 2; since A = 2, A*^{2} = 2^{2}, or 4.1258

*I have A*^{2}; I need to find B^{2}; I am not given that.1267

*But what I am given is an additional piece of information, and that is that there is a focus here and a focus here.*1270

*This allows me to find C: the distance from the center to either focus (let's look at this one)--from the center, 0, down to -4--*1280

*the absolute value of that is 4; therefore, C = 4; the distance is 4.*1291

*C*^{2}, therefore, equals 4^{2}, or 16.1298

*Recall the relationship: C*^{2} = A^{2} + B^{2} for a hyperbola.1303

*So, I have C*^{2}, which is 16, equals A^{2}, which is 4, plus B^{2}.1309

*16 - 4 is 12; 12 = B*^{2}; therefore, B = √12, which is about 3.5.1315

*If you wanted to draw B, then you could, because that is right here at (3.5,0).*1330

*But what we are just asked to do is write the equation; and we have enough information to do that,*1338

*because I have that y*^{2} divided by A^{2}; I determined that A^{2} is 4;1342

*minus x*^{2}/B^{2}; I determined that that is 12; equals 1.1348

*So, this is the equation for this hyperbola, with a vertical transverse axis and a center at (0,0) in standard form.*1355

*The next example: Find the equation of the hyperbola satisfying vertices at (-5,0) and (5,0) and a conjugate axis that has a length of 12.*1368

*Just sketching this out to get a general idea of what we are looking at--just a rough sketch--vertices are at (-5,0) and (5,0).*1380

*That means that the center is going to be right here at (0,0).*1400

*So, the center is at the origin; since the vertices are here and here, then I have a horizontal transverse axis;*1406

*this is going to go through like this, and then like this.*1424

*So, my second piece of information is that I have a horizontal transverse axis.*1432

*Since I have a horizontal transverse axis, then I am going to have an equation in the form x*^{2}/A^{2} - y^{2}/B^{2} = 1.1440

*The center is at the origin; it has a horizontal transverse axis; this is a standard form that I am working with.*1451

*I need to find A: well, I know that the center is here, and that A is this length; so from this point to the center,*1457

*or from this point (the vertex) to the center, is 5: A = 5.*1466

*Since A = 5, A*^{2} = 5^{2}; it equals 25.1473

*The other information I have is that the conjugate axis has a length of 12.*1486

*So, the length of the conjugate axis, recall, is 2B; here they are telling me that that length is 12.*1490

*Therefore, 12/2 gives me B; B = 6; so, that would be up here and here: (0,6) and then (0,-6).*1503

*This would be the conjugate axis; so this is B = 6.*1517

*Since B equals 6, I want B*^{2} that equals 6^{2}, which equals 36.1523

*Now, I can write this equation: I have (this is my final one) x*^{2}/A^{2}, which is 25,1533

*minus y*^{2}/B^{2}, and I determined that that is 36, equals 1.1545

*So, this is a hyperbola with a center at the origin.*1551

*And A*^{2} is 25; B^{2} is 36; and it has a horizontal transverse axis.1554

*We are asked to graph this equation; and it is not in standard form.*1566

*But when I look at it, I see that I have a y*^{2} term and an x^{2} term, and they have opposite signs.1571

*So, I am working with the difference between a y*^{2} term1577

*and an x*^{2} term, which tells me that this is the equation for a hyperbola.1579

*If they were a sum, this would have been an ellipse, since they have different coefficients.*1585

*But it is a difference, so it is a graph of a hyperbola.*1589

*What I need to do is complete the square to get this in standard form.*1591

*OK, so first I am grouping y terms and x terms: y*^{2} + 12y - 6x^{2} + 12x - 36 = 0.1597

*What I am going to do is move this 36 to the other side and get that out of the way for a moment by adding 36 to both sides.*1617

*The next thing I need to do with completing the square is factor out the leading coefficient, since it is something other than 1.*1627

*So, from the y terms, I will factor out a 2; that is going to leave me with y*^{2} + 6y.1634

*You have to be careful here, because you are factoring out a -6, so I need to make sure that I worry about the signs.*1640

*And that is going to leave behind an x*^{2} here; here, it is going to leave behind, actually, -2x.1649

*So, checking that, -6 times x*^{2} is -6x^{2}--I got that back.1656

*-6 times -2x is + 12x; equals 36.*1661

*Now, to complete the square, I have to add b*^{2}/4 in here, which equals...b is 6; 6^{2}/4 is 36/4; that is 9.1669

*I need to be careful to keep this equation balanced.*1686

*Now, this is really 9 times 2 that I am adding; 9(2) = 18--I need to add that to the right.*1688

*Working with the x terms: b*^{2}/4 = 2^{2}/4, which is 4/4; that is 1, so I am going to add 1 here.1697

*-6 times 1 needs to be added to the other side; so I am going to actually subtract 6 from the right to keep it balanced.*1714

*Now, I am rewriting this as (y + 3)*^{2} - 6(x - 1)^{2} =...18 - 6 is 12; 36 + 12 gives me 48.1726

*The next step, because standard form would have a 1 on this side, is: I need to set all this equal to 1.*1747

*I need to divide both sides of the equation by 48.*1753

*This cancels, so it becomes y + 3*^{2}; the 2 is gone; this becomes a 24; minus...6 cancels out,1769

*and that leaves me with (x - 1)*^{2}; 6 goes into 48 eight times; and this is a 1.1780

*OK, so it is a lot of work just to get this to the point where it is in standard form.*1788

*But once it is in standard form, we can do the graph, because now I know the center; I know A*^{2} and B^{2}.1792

*We have this in standard form; so now we are going to go ahead and graph it.*1799

*I will rewrite the standard form that we came up with, (y + 3)*^{2}/24 - (x - 1)^{2}/8 = 1.1803

*Looking at this; since this is positive, I see that I have a vertical transverse axis.*1815

*The other thing to note is this plus here: recall that, if you have (y + 3)*^{2}, this is the same as (y - -3)^{2}.1824

*And when we look at standard form, we actually have a negative here.*1835

*So, you need to be careful to realize that the center is at (1,-3), not at (1,3).*1838

*Let's make this 2, 4, 6, 8, -2, -4, -6, -8; the center, then, is going to be at (1,-3).*1846

*The next piece of information: A*^{2} = 24; therefore, A = √24, which is approximately 4.9.1856

*B*^{2} = 8; therefore, B = √8; therefore, if you figure that out on your calculator, that is approximately 2.8.1871

*Since I have the center, and I have A and B, I can draw the rectangle that will allow me to extend diagonals out to form the asymptotes.*1883

*The goal is to write this in standard form, find A and B, find the center, make the rectangle, and make the asymptotes;*1892

*and then, you can finally draw both branches of the hyperbola.*1902

*All right, so if the center is here at (1,3), then I am going to have 2 vertices.*1906

*And what is going to happen, since this is a vertical transverse axis, is: one vertex is going to be up here; the other is going to be down here.*1912

*The center is at (1,3); that means I am going to have a vertex at 1, and then it is going to start at the center,*1926

*and then it is going to be 4.9 directly above that center; so that means this is going to be at -3 + 4.9.*1933

*The y-coordinate will be at -3 + 4.9, which equals (1,1.9).*1944

*Therefore, at (1,1.9) (that is right there)--that is where there is going to be one vertex.*1950

*And this is A--this is the length of A.*1958

*The second vertex is going to be at (1,-3); that is the center; and then I am going to go down 4.9--that is the length, again, of A.*1963

*That is -3 - 4.9 (or + -4.9; you can look at it that way) = (1,-7.9), down here.*1979

*OK, vertices are at (1,-7.9) and (1,1.9).*1999

*Now, I need to find where B is--where that endpoint over here is, horizontally--so that I can make this rectangle.*2014

*I know that B equals approximately 2.8; that means that I am going to have a point over here at 1 + 2.8...-3.*2024

*Well, 1 + 2.8 is (3.8,-3); so (3.8,-3) is right there.*2040

*I can reflect across; and I am going to have a point at 1 - 2.8, -3, which is going to give me...1 - 2.8 is (-1.8,-3).*2054

*That is going to be...this is 2...-2 is right here; so that is going to be right about there.*2068

*I now have these points; and recall that I can then extend out to make a box.*2076

*There is going to be a vertex here; I am going to extend across; there is going to be a vertex here.*2083

*Bring this directly down; there is a vertex here, and then another vertex right here.*2090

*Again, I got these points by knowing where the center is, knowing where the vertices of the hyperbola are, and then knowing the length of B.*2096

*This is B; then this length is A.*2109

*Once I have this rectangle, I can go ahead and draw the asymptotes by extending diagonals out.*2113

*Another way to approach this, recall, would have been to use the formula for the slope that we discussed, for the slope of the asymptotes.*2130

*Either method works.*2140

*I know that I am going to have a hyperbola branch up here; the vertex is right here, and it is going to approach, but never reach, this asymptote.*2143

*It is going to do the same thing with the other branch: a vertex is here; it is going to approach, but never reach, the asymptote.*2156

*OK, so this was a difficult problem; we were given an equation in this form.*2170

*We had to do a lot of work just to get it in standard form.*2175

*And then, once we did, we were able to find the center and form this rectangle, draw the asymptotes, and then (at last) graph the hyperbola.*2178

*Example 4: We don't need to do graphing on this one.*2189

*We are just finding the equation of a hyperbola with the center here, (0,0), and a horizontal transverse axis.*2192

*I am going to stop right there and think, "OK, I have a center at (0,0) and a horizontal transverse axis."*2200

*So, the standard form is going to be x*^{2}/A^{2} - y^{2}/B^{2} = 1.2206

*Since the center is at (0,0), I don't have to worry about h and k.*2215

*The horizontal transverse axis has a length of 12; well, the transverse axis length, recall, is equal to 2A.*2219

*I am given that that length is 12; if I take 12/2, that is going to give me A = 6.*2231

*The conjugate axis--recall that the length of the conjugate axis is equal to 2B, which is 6: B = 3.*2240

*Now, I need to find A*^{2}, which is 6^{2}, or 36, to put in here.2257

*B*^{2} is 3^{2}, which is 9.2264

*Now, x*^{2}/36 - y^{2}/B^{2} (which is 9) = 1.2267

*So, this is the equation for a hyperbola with the center at the origin, a horizontal transverse axis, and a conjugate axis with a length of 6.*2279

*That concludes this lesson on hyperbolas; thanks for visiting Educator.com!*2290

*Welcome to Educator.com.*0000

*In the past four lectures, we have discussed various conic sections: parabolas, circles, ellipses, and hyperbolas.*0003

*And this lecture is designed to bring that information together and to give you some context about this.*0010

*First of all, what are conic sections? We know we can name them; we know what they are; but where do they come from?*0016

*Well, they are literally sections of a cone: when you take a double cone (it is actually a double cone, as follows, with the points together),*0023

*and you section them (sectioning is slicing)--when you take slices of them, using a plane, you come up with these four types of curves.*0034

*So, as you can see, when you take a plane and section, or slice, the cone across, you are going to end up with a circle.*0043

*If you tip that plane at an angle, the result is an ellipse.*0054

*If you encompass the edge of the come, you end up with a parabola.*0065

*And if you slice through in such a way that you capture the edges of both cones, then you end up with a hyperbola; and there you can see the two branches.*0071

*So, this is where conic sections come from; and they have many applications in science.*0079

*We have talked about the standard form of each conic section (for example, the standard form of a circle, or the standard form of an ellipse).*0087

*This standard form that I am talking about now is a very general form.*0095

*It gives you a general equation, ax*^{2} + bxy + cy^{2} + dx + ey + f = 0.0098

*So, what we are going to do in a minute is talk about how you can look at this general form and determine*0109

*which type of conic section you are working with, so that you can put the equation*0115

*in the standard form particular to that type of conic section.*0119

*And as we have been going through, I have mentioned some ways that you can tell, if you just have an equation in the general form,*0124

*what type of conic section you are working with; and now I am going to bring that all together.*0131

*OK, if b = 0, we can analyze that standard form of the conic section to determine what type of conic section the equation represents.*0137

*Looking back at that general standard form again, ax*^{2} + bxy + cy^{2} + dx + ey + f = 0.0146

*Here, we are having the limitation that b = 0.*0168

*And throughout this course, when we work with conic sections, we have only worked with ones where b is 0.*0171

*When b is 0, you end up with this.*0178

*Once you have this standard form, then you can go ahead and analyze it in ways we are going to discuss in a minute*0192

*to determine which type of conic section you have (what the equation describes).*0197

*But let's talk for a minute about what the xy tells you.*0201

*So far, we have worked with shapes such as parabolas; and some were oriented vertically; some, horizontally.*0204

*We also worked with ellipses (some had a horizontal major axis; some had a vertical major axis); and the same with hyperbolas.*0215

*So, even though the center may have been shifted, these were all either strictly vertical or strictly horizontal.*0230

*What this bxy term does is rotates it so that instead of, say, having an ellipse that has*0237

*a completely vertical major axis or horizontal major axis, you could end up with an ellipse like this--*0243

*the major axis is at an angle--or a parabola that is like this.*0250

*And that is definitely more complicated to work with; and it doesn't allow us to complete the square, then,*0259

*to shift an equation from the general form to a specific standard form.*0265

*So, later on, if you continue on in math, you may end up working with these shapes.*0270

*But for this course, we are limiting it to conic sections that are either vertical or horizontal; but they are not tipped at any other type of angle.*0274

*In order to identify conic sections, you need to look at the coefficients of the x*^{2} and y^{2} terms.0286

*So, let's rewrite this; and again, the assumption is that b = 0.*0293

*So, I am just going to have ax*^{2} + cy^{2} + dx + ey + f = 0.0299

*Parabola: Recall that, with a parabola, you have an x*^{2} term or a y^{2} term, but not both.0310

*Therefore, either a is 0 (so this drops out) or c is 0 (so this drops out).*0325

*An example would be something like x = 3y*^{2} + 2y + 6.0331

*Or you might have y = 2x*^{2} - 4x + 8.0338

*So, neither of these has both an x*^{2} and a y^{2} term in the same equation.0344

*For a circle, recall that what you are going to end up with is an x*^{2} and a y^{2} term0352

*on the same side of the equation, with the same sign; and they are going to have the same coefficients.*0361

*Therefore, a is going to equal c.*0366

*An example would be x*^{2} + y^{2} + 3x - 5y - 10 = 0.0369

*Here, a equals 1, and c equals 1; those are the same coefficients; x*^{2} and y^{2}0378

*have the same sign and the same side of the equation; so it is a circle.*0386

*If we are working with an ellipse, this time the x*^{2} term and y^{2} terms are going to be0392

*on the same side of the equation, with the same sign (like with the circle), but a and c are different.*0399

*They are unequal; that tells me that I am working with an ellipse.*0405

*For example, 12x*^{2} + 9y^{2} + 25x + 28y + 40 = 0.0409

*Here, I have a = 12 and c = 9; so this is the equation describing an ellipse.*0423

*Finally, with a hyperbola, these are pretty straightforward to recognize, because you are going to have*0431

*an x*^{2} term and a y^{2} term, but they are going to have opposite signs.0436

*Their coefficients will have opposite signs.*0440

*For example, 4x*^{2} - 8y^{2} + 10x + 6y - 34 = 0.0443

*So, I have a = 4 and c = -8; since a and c have opposite signs, this is an equation describing a hyperbola.*0458

*You can use these rules to allow you to identify conic sections when you are given an equation in what we are going to call "general form."*0467

*It is standard form, but it is a very general standard form for any type of conic section.*0476

*OK, now we are going to work on identifying the various conic sections by looking at their equations.*0481

*First, write in standard form, and identify the conic section.*0488

*OK, so general standard form is what I am talking about right now: it is x*^{2} + 2y^{2}.0495

*I need to subtract 4x from both sides, subtract 12, and set everything equal to 0.*0503

*What this tells me is that I have a = 1 and c = 2.*0510

*Since a = 1 and c = 2, these have the same sign (the x*^{2} and the y^{2} terms); but they have different coefficients.0515

*And that means that what I am working with is an ellipse.*0525

*You could go on, then, and write this in the specific standard form for an ellipse.*0531

*Let's do that by completing the square: start out by grouping...let's rewrite it here.*0537

*And then, let's group the x and the y terms; so x*^{2} terms group together; y terms group together.0545

*And now, add 12 to both sides to move that over, to make completing the square a little bit easier.*0555

*To complete the square for x*^{2} - 4x, I need to add b^{2}/4.0563

*b*^{2}/4 is equal to 4^{2}/4, is 16/4; it is 4.0571

*So, I add x*^{2} - 4x + 4; and it is very important to remember to add the 4 to the right side, as well.0582

*There is no factor out here; I don't need to multiply--it is just 1; so 4 times 1 is 4; that gives me 12 + 4.*0592

*All right, that is x*^{2} - 4x + 4 + 2y^{2} = 16.0601

*This can be rewritten as (x - 2)*^{2} + 2y^{2} = 16.0609

*But recall, in standard form for an ellipse, you need to have a 1 on the right.*0616

*So, rewrite this up here, and then divide both sides by 16.*0622

*This is just (x - 2)*^{2}/16; this will cancel; this will become 8; and then 16/16 is 1.0633

*So, we started out with this equation, put it into the general standard form to identify that this is an ellipse,*0645

*and then went on to complete the square; and now I have it written in standard form specifically for an ellipse,*0652

*which is much more useful when you are working with that and trying to graph.*0657

*This time, without completing the square, all we are going to do is identify the conic section.*0664

*And this is already in standard form; therefore, a = 2; c = -3.*0668

*Since a and c have opposite signs, this is the equation for a hyperbola.*0678

*I have an x*^{2} term and a y^{2} term, both, so it is not a parabola.0687

*They have opposite signs; therefore, it must describe a hyperbola.*0691

*OK, write in standard form and identify the conic section.*0698

*Right now, this is not in any type of standard form; so I am going to work with the general standard form.*0702

*First, I am going to subtract 36x*^{2} from both sides.0708

*Then, I am going to subtract 128 from both sides.*0717

*This means that I have a = -36, and c = 16; since these two are opposite signs, this is an equation describing a hyperbola.*0727

*OK, now, let's go ahead and put this in standard form specific to a hyperbola.*0741

*And let's start out by moving this 128 back over to the right; this is actually 32.*0747

*Next, I do have a common factor of 4, so I am going to divide both sides by that, so that I am working with smaller numbers.*0763

*That is -9x*^{2} + 4y^{2} + 8y =...128/4 would give 32.0770

*All right, now to make this already move it more towards looking like a hyperbola, I am going to put the positive terms here in front:*0786

*4y*^{2} + 8y - 9x^{2}, because I am going to have a difference.0793

*To complete the square, I first need to factor out that 4; then I need to add b*^{2}/4 to this expression.0801

*This is going to equal 2*^{2}/4; that is 4/4, which is 1.0816

*Here is where I need to be careful, because I need to make sure I add 4 times 1 to the right, which is 4, to keep the equation balanced.*0825

*At this point, I am going to rewrite this as (y + 1)*^{2} - 9x^{2} = 36.0837

*The last step is: I want the right side to be 1, so I am going to divide both sides by 36.*0844

*4 goes into 36 nine times; 9 goes into 36 four times; and then this cancels out to 1.*0860

*OK, so I started out with an equation that wasn't in any kind of standard form.*0872

*I put it in general standard form, and then determined it was a hyperbola, completed the square, and ended up*0876

*with an equation in the standard form for a hyperbola, so that I can use that to graph the hyperbola, if needed.*0882

*Write in standard form and identify the conic section.*0891

*So, this is almost in the general standard form, but not quite.*0894

*I have 4x*^{2}; I need to move this -3y^{2} next, then -16x - 18y - 12 = 0.0897

*Now, I can easily see that a = 4 and c = -3; since these have opposite signs, that means that this is an equation describing a hyperbola.*0907

*OK, the next task is to complete the square.*0921

*I am going to first add 12 to both sides to remove the constant from the left side.*0926

*Then, I am going to group the x terms, which is 4x*^{2} - 16x.0937

*And then, I have a -3y*^{2} - 18y, and that all equals 12.0946

*I have a leading coefficient that is something other than 1, so I am going to factor out the 4, leaving behind x*^{2} - 4x.0955

*Here, I need to factor out a -3; that is going to leave behind y*^{2} + 6y.0965

*You need to be careful with the signs here; just double-checking: -3 times y*^{2} is -3y^{2}.0969

*-3 times positive 6y is -18y, when you factor out with that negative sign; equals 12.*0976

*Completing the square: b*^{2}/4, in this case, is 4^{2}/4, is 16/4; that is 4.0987

*So, I am going to add 4 here; I am also going to add 4 times 4, or 16, to the right, to keep the equation balanced.*0998

*For the y expression, I have y*^{2} + 6y; therefore, b^{2}/4 = 6^{2}/4, which is 36/4, which is 9.1012

*-3 times 9 is -27; so I am going to subtract 27 from the right side, again keeping the equation balanced.*1029

*I am rewriting this as (x - 2)*^{2} - 3(y + 3)^{2} = 16 + 12, is 28, minus 27; conveniently, I end up with a 1 on the right.1041

*Now, this is almost in standard form; generally, with standard form for a hyperbola, this term will be in the denominator.*1059

*So, it is possible to rewrite it like this; and it might be easier to look at it that way,*1071

*so that you can immediately know that this is a*^{2}, instead of having to think it out.1077

*Putting it in truly standard form is also a good idea, because recall that, if I have the numerator divided by 1/4, that is the same as this times 4.*1082

*And that tells me that I have a hyperbola with a center at (2,-3); you have to watch out for this positive sign.*1096

*And it has a horizontal transverse axis.*1103

*So, today we learned exactly what conic sections are, where they come from,*1109

*and how to look at an equation and determine what type of conic section it describes.*1114

*Thanks for visiting Educator.com; see you again soon!*1120

*Welcome to Educator.com.*0000

*Today, we are going to talk about solving quadratic systems of equations.*0002

*In earlier lectures, we discussed talking about linear systems of equations, and used various methods to solve those: for example, substitution and elimination.*0006

*And we are going to use some similar methods for quadratic systems, although these systems are more complex.*0017

*There are two types of quadratic systems that we are going to be working with today.*0023

*The first is a linear-quadratic system: ***linear-quadratic systems** are a set of two equations in x and y,0027

*in which one of the equations is linear, and the other is quadratic.*0035

*We will talk in a second about how to solve these; let's just stop at the definition right now.*0041

*x*^{2} + 3x + y = 9; x = 2y: these two equations, considered together, would be a linear-quadratic system.0046

*I have a linear equation here and a quadratic equation here.*0057

*And right here, this is a fairly simple linear equation to start out with.*0063

*And we can use that to apply the method discussed above.*0068

*So, given this type of system, these can be solved algebraically by isolating*0072

*one of the variables in the linear equation, and then substituting it into the quadratic equation.*0077

*OK, so conveniently, x is already isolated.*0083

*Often, you will be given a linear equation as part of a system where you have to do some manipulation to isolate either x or y.*0088

*But I already have x = 2y, so that is perfectly set up for me to substitute 2y in for the x variable in this quadratic equation.*0096

*So, I am rewriting the quadratic equation, and now substituting 2y in for x.*0105

*That gives me 2y*^{2} + 3(2y) + y = 9.0113

*What I am left is a quadratic equation, which I can solve by the usual methods that we have learned.*0122

*This is 2*^{2} is 4y^{2} + 6y + y = 9.0128

*That gives me 4y*^{2} + 7y = 9, or 4y^{2} + 7y - 9 = 0.0136

*And then, you could go on to solve this, using the quadratic formula.*0148

*Once you find y, you can go ahead and substitute that value(s) in here, and then determine the corresponding value of x.*0153

*In a linear quadratic system, you may have either 0, 1, or 2 solutions.*0170

*And the easiest way to understand this is to think about it in terms of a graph.*0176

*A linear equation is going to give you a line: a quadratic equation might give you a parabola or an ellipse, a hyperbola, a circle...*0180

*So, let's look at some possibilities: maybe I have a system that ends up graphing out like this.*0188

*It gives me that line and this parabola.*0196

*Well, the solutions are going to be where these two intersect; but in this system, they will never intersect.*0201

*So, this system is going to have 0 solutions.*0206

*I may have another situation (again, I will use a parabola as my example) where I have a line that goes through right here at one point.*0212

*And this is going to give me one solution.*0224

*Perhaps I have an ellipse, and I have a line going through it like this.*0228

*It intersects at two spots, so I would have two solutions.*0237

*This helps to illustrate why you may have no solutions, one, or two solutions.*0241

*More complicated systems involve quadratic-quadratic equations.*0251

*This would be a quadratic-quadratic system in which there is a set of two quadratic equations in x and y.*0255

*For example, x*^{2} + y^{2} = 5 and 2x^{2} - y = 4:0262

*you can see that you have two quadratic equations.*0272

*You could try to use substitution, but it could get a little bit messy.*0276

*So, often, elimination is the easiest way to use, so we solve them algebraically by elimination.*0281

*And you will recall, working with linear equations, that in elimination, what you try to do is get variables to have the same or opposite coefficients.*0286

*And then, either add or subtract the equations so that a variable drops out.*0294

*And here I want to get either x*^{2} or y^{2} to drop out.0298

*Well, I have an x*^{2} term in each; so that, I could end up having drop out if I multiplied this top equation by 2.0304

*So first, I am going to multiply the first equation by 2.*0314

*2 times x*^{2} + y^{2} = 5: and this is going to give me 2x^{2} + 2y^{2} = 10.0317

*So, I am going to then go ahead and write the other equation just below that: 2x*^{2} - y = 4.0333

*This is equation 1; this is equation 2; I multiplied equation 1 by 2, and then I ended up with this; and I have equation 2 right here.*0347

*Now, what I am going to do is subtract this second equation from the first.*0359

*Rewriting that: this gives me 2x*^{2} - 2x^{2}; these terms drop out.0366

*The x*^{2} terms drop out; a negative and a negative is a positive, so this is actually going to give me, let's see, 3...0375

*actually, there are slightly different coefficients, so I can't add those; this is 2y*^{2} - a negative0394

*(that becomes a positive) y, and then 10 - 4 is equal to 6.*0402

*OK, so 2x*^{2} - 2x^{2}--those drop out; I have a 2y^{2} down here.0407

*I have a negative and a negative, gives me a positive y; and then this becomes a -4; so 10 - 4 is 6.*0414

*So, I end up with 2y*^{2} + y = 6 by using elimination.0422

*Therefore, I have gotten rid of one of the variables.*0427

*From there, I have a regular quadratic equation that I can solve: that gives me 2y*^{2} + y - 6 = 0.0431

*I am going to try to solve that by factoring; and this is going to give me (2y + something) (y - something).*0440

*So, let's look at 6: if we take factors of 6, those are 1 and 6, and 2 and 3.*0448

*And I want them to be close together, so that I just end up with a coefficient of 1 here; so let's try 3 - 2 = 1.*0455

*Therefore, I am going to take + 3, - 2.*0466

*But I have to take this 2y into account; so let's go ahead and see if this works.*0478

*This becomes 2y*^{2}, and then this is minus 4y, plus 3y; so that gives me a -y; so that won't work--let's try it the other way.0482

*Let's actually make this a negative and see what happens.*0496

*This gives me 2y*^{2}, and then this is 4y, minus 3y; that gives me a y; and then, -3 times 2 is -6.0500

*So, this actually factored out, and it worked.*0510

*Using the zero product property, let's go back up here and get (2y - 3) (y + 2) = 0.*0514

*2y - 3 = 0, or y + 2 = 0; if either of these equals 0, the whole quantity equals 0.*0522

*Solving for y gives me 2y = 3 or y = 3/2; solving for y gives me y = -2.*0530

*Now, what I need to do is go back and find the corresponding x-value, so I have coordinate pairs as my answers.*0541

*All right, let's first work with y = -2.*0552

*And going back to this equation (because this one is simple to work with), I am going to substitute in that value for y*^{2} and see what I get.0558

*x*^{2} + (-2)^{2} = 5: that is x^{2} + 4 = 5, or x^{2} = 1.0568

*Therefore, x = 1 and -1; now, that is when y is -2.*0580

*So, my solutions are (1,-2) and (-1,-2)--I have two solutions there.*0588

*Now, I need to repeat that process for y = 3/2; let y equal 3/2, and then substitute in here.*0603

*This is going to give me x*^{2} + (3/2)^{2} = 5, so x^{2} + 9/4 = 5.0610

*So, x*^{2} = 5 - 9/4; that gives me x^{2} =...a common denominator of 4; it would be 20/4 - 94, so x^{2} = 11/4.0620

*Therefore, x = ±√11/4; now, looking over here, when y is 3/2, x could be √11/4, or it could be -√11/4.*0637

*So, I have another two members to my solutions: I have that x could be (√11/4,3/2), and (-√11/4,3/2)*0656

*Here, we have a situation where we actually have four solutions.*0673

*And I solved this by elimination; and then I had to go back and take each of the solutions I ended up with,*0679

*-2 and 3/2 for y, and find corresponding solutions for x.*0686

*And each of those yielded two values for x, so I ended up with 4 solutions altogether.*0691

*And I can also illustrate that point using graphing--of why you can end up with 0, 1, 2, 3, or 4 solutions.*0700

*Quadratic-quadratic systems: think about all of the different curves that you could come up with, and the possibilities.*0708

*You could maybe have a circle and a parabola as a system that never intersect: this is 0 solutions.*0714

*Perhaps you have an ellipse, and then another ellipse, like this: that gives you 1, 2, 3, 4 solutions.*0722

*You may have an ellipse, and then a circle, right here; and this gives you two solutions.*0735

*You could have, say, a parabola and an ellipse here that intersect at just one point.*0747

*So, you can see how that, with various combinations, you could get 0, 1, 2, 3, or 4 solutions.*0754

*And we just saw that demonstrated algebraically in the previous example--that we ended up with 4 solutions.*0760

*Systems of quadratic inequalities: these are systems of inequalities that have two inequalities, and at least one of these is a quadratic inequality.*0769

*So, for example, x*^{2}/9 + y^{2}/4 <1; and then that is considered along with (x - 1)^{2} + (y - 3)^{2} < 9.0781

*If you look at this, you have two quadratic equations; and you would recognize this*0807

*as an ellipse in standard form; and this gives the equation for a circle in standard form.*0811

*We have worked with some systems of inequalities before, and talked about how these can be solved by graphing.*0818

*And we are going to do the same thing here.*0823

*We are going to first graph the corresponding equation; and that will give us the boundary for the solution set.*0825

*Then, we will use a test point to determine on which side of the boundary the solution set lies.*0832

*We will do the same thing for the second equation that corresponds with the inequality:*0840

*find the boundary; use a test point; find where the solution set is.*0845

*And then, the overlap between those two solution sets gives you the solution for the system.*0849

*Illustrating this with this example: I am going to start out be graphing the boundary, and then finding the test point, for this first inequality.*0855

*So, the corresponding equation is going to be x*^{2}/9 + y^{2}/4 = 1.0866

*Since this is an ellipse, looking at it in this form, I know that it has a center at (0,0).*0883

*I know that A*^{2} = 9, so a = 3; and since the larger term is associated0892

*with the x*^{2} term, I also know that this has a horizontal major axis.0901

*The major axis is going to be oriented this way.*0907

*B*^{2}, right here, is equal to 4; therefore, B = 2.0911

*So, that allows me to at least sketch this out: since A = 3, then I am going to have a vertex here, and the other vertex here, at (-3,0).*0917

*B is 2, so I am going to go up 2 and down 2; and this allows me to just sketch out the ellipse.*0928

*All right, the next step is to use a test point, because now I have a boundary.*0941

*And I actually need to be careful; I need to determine if the boundary should be dotted, or if it should be solid.*0948

*And looking here, I actually have a strict inequality.*0956

*What that tells me, recall, is that the boundary is not part of the solution set.*0960

*And the way we make that known is by using a dotted or dashed line.*0966

*Clarifying that, so we have the correct type of boundary...these breaks in the boundary indicate that this boundary is not part of the solution set.*0976

*OK, so I graph the boundary; I checked, and I had a strict inequality.*0993

*Now, I am going to take a test point; this is a convenient test point,*0999

*because the boundary has divided this into two regions: the region outside the ellipse, and the region inside it.*1004

*And I need to determine where the solution set is.*1010

*So, let's take the test point (0,0): I am going to put these values back into that original inequality.*1013

*0 is less than 1; this is true; therefore, this test point is part of the solution set.*1033

*So, the solution set must be inside this boundary.*1040

*So, I graphed my first boundary, and I determined where the solution set for this inequality is.*1050

*Looking at the second inequality: the corresponding equation will give me the boundary line for that.*1055

*Looking at this, this is written in standard form for a circle.*1064

*So, this is a circle with the center at (1,3), and r*^{2} = 9, so the radius equals 3.1067

*And this is a circle; and the center is at (1,3), so the center is right here.*1078

*And this is also a strict inequality, so I am going to use the dotted line when I draw this boundary line.*1086

*And the radius is 3: since the radius is 3, then that would be 4, 5, 6: that would go up to here.*1099

*And then, this is at 1, so then I would have the end right here, here, here.*1106

*OK, this is enough to get a rough sketch of the circle.*1114

*So, first I will just draw it as solid, and then go back and make it dashed, since this is a strict inequality.*1133

*All right, so this is a dashed line; it is a circle; it has a center at (1,3), and a radius of 3.*1149

*So again, this boundary line is not part of the solution set.*1161

*I am going to use another test point and insert it into that inequality to figure out where my solution set is.*1165

*So, for this, let's go ahead and use (1,1) as a test point, right there...the test point is going to equal (1,1).*1173

*So, that is going to give me 1*^{2}/9 +...oops, I actually need to go back into that inequality...there is my equation1185

*for the circle, (x - 1)*^{2} + (y - 3)^{2} < 9; my test point is (1,1).1207

*So, that is going to give me (1 - 1)*^{2} + (1 - 3)^{2} < 9.1214

*1 - 1 is 0, plus 1 - 3...that is -2, squared is less than 9; so 0...and this is -2 times -2 is 4...so 4 is less than 9.*1222

*Yes, this is true; therefore, this point is part of the solution set.*1235

*So, the solution set lies inside the circle.*1240

*The solution set for the system of inequalities is going to be this area here that is the overlap*1251

*between the solution set for the circle and the solution set for the inequality involving the ellipse.*1260

*And you can see, right here, the area where there is both red and black; it is that area of overlap.*1266

*So again, this is similar to methods we have used before, involving solving systems of inequalities,*1270

*where we graph the boundary line for one inequality; we graph the boundary line for the other inequality;*1277

*we use test points to find the solution sets for each inequality; and then, we determine the area of overlap*1282

*between those two solution sets, and that is the solution set for the system of inequalities.*1289

*And here, we are doing that with quadratic inequalities involving a circle and an ellipse.*1295

*Example 1: this is a linear-quadratic system: I have a linear equation here, and a quadratic equation here.*1303

*Recall that the easiest way to solve for these is by substitution.*1310

*Therefore, I am going to isolate x; I am going to rewrite this as x = y + 2.*1314

*Then, I go back to this first equation (this is from equation 2); I go back, and I make sure I substitute this into the other equation.*1320

*So, wherever there is an x, I am going to then put y + 2 instead; and I need to square that in this case.*1331

*This equals 36; I am going to write this out as y*^{2} + 4x + 4 + y^{2} = 36.1339

*OK, y*^{2} + y^{2} gives me 2y^{2} + 4y (we are working with y here) + 4 = 36.1360

*Now, it is a regular quadratic equation that I can just solve as I usually would.*1372

*And I see here that I have a common factor of 2; let's first go ahead and subtract 4 from both sides.*1377

*And this is going to give me 32; I am going to divide both sides by 2 to make this simpler, which is going to give me y*^{2} + 2y = 16.1391

*And now, I am going to solve it as I would any other quadratic equation: y*^{2} + 2y - 16 = 0.1403

*You could try this out, but it actually doesn't really factor out.*1411

*This is a situation where we have to go back to the quadratic formula, y = -b ±√(b*^{2} - 4ac), divided by 2a.1415

*It is a little more time-consuming, but it will get us the answer when factoring doesn't work.*1431

*So, let's rewrite this up here: y*^{2} + 2y - 16 = 0.1436

*y =...well, b is 2, so that is -2 ±√((-2)*^{2} - 4 times a (is 1), and then c is negative 16), all divided by 2 times a, which is 1.1443

*Simplifying: y = -2 ±√...-2*^{2} is 4; -4 times 1 is -4; -4 times -16 is + 641461

*(those negatives, times each other, become positive) all over 2.*1478

*Therefore, y = -2±√...that is 64; -4 times 1 times -16 is 64, plus 4 is 68, divided by 2.*1485

*Now, you actually could determine that 68 is equal to 4 times 17; therefore, √68 equals the perfect square of 4, times 17.*1507

*So, I can then pull this 2 out by taking the square root of 4, which is 2; and it becomes 2√17.*1523

*So, I am going to do that over here, as well.*1531

*All right, you can factor out a 2, and those will cancel; I am just going to do that over here.*1540

*y = 2(-1) ± 1√17/2; that will cancel.*1547

*What you are left with -1 ± 1√17, divided by 2.*1555

*OK, so this is not an easy problem, because you ended up having to use the quadratic formula.*1567

*So, let's look at what we actually have: we have y = -1 + 1√17, and we also have y = -1 - √17.*1576

*So, we don't really need this 1 here.*1591

*The next step is to go back and substitute in: fortunately, we have an easy equation here*1595

*that we can substitute: x - y = 2, which is the same as x = y + 2.*1601

*I am going to work up here; and I am going to say, "OK, when y is -1 + √17, then x = this (that is y) + 2."*1607

*Therefore, x = 2 - 1 (is 1)...1 + √17.*1624

*That gives us an ordered pair: (1 + √17, -1 + √17).*1632

*That is an ordered pair; then I am going to take this second possibility, where y equals -1 - √17.*1643

*And I am going to substitute that into this, as well: x = -1 - √17 + 2.*1656

*This is going to give me x = -1 + 2...that is going to give me positive 1...so 1 - √17.*1669

*So, my second ordered pair is going to be (let's write it over here) x = 1 - √17, and then the y-value is -1 - √ 17.*1677

*So, it gets a little confusing with all of those signs; you want to make sure that you are careful*1692

*to check your work, and that you don't have any of the signs mixed up.*1697

*But what it came down to is using the quadratic equation to determine that y = -1 + √17, and y = -1 - √17.*1701

*Then, you take each possibility, starting with the first one; substitute for y in this equation,*1711

*which is the same as this (just rearranged); and substitute this to determine the corresponding value of x*1719

*(because remember, the solution is going to be an ordered pair, an x and a corresponding y value).*1724

*So, I did that for this first one; then I went and took the second one, repeated that process,*1730

*and got that, when y is -1 - √17, x is 1 - √17.*1734

*These two are an ordered pair, as well--that is where this came from; and these are the two solutions for this linear quadratic system of equations.*1743

*Example 2: we are going to solve...this is a quadratic-quadratic system, because I have a second-degree equation here and here.*1755

*I have two second-degree equations.*1761

*Recall that the best way to solve these is by elimination.*1763

*Looking at this, the first thing I can do to make this a little easier is actually to divide this first equation (equation 1 and equation 2) by 2,*1768

*because right now, it is kind of messy; it is bigger numbers than I need to be working with.*1781

*So, 2 goes into 8 four times; 2 goes into 2 once; and 2 goes into 40 twenty times.*1786

*All right, now I am going to go ahead and take equation 2, 4x*^{2} + y^{2} = 100, and rewrite it down here.1797

*And I see that, if I subtract this second equation, the y*^{2} terms will drop out, because they already have the same coefficient.1805

*I am going to rewrite this as 4x*^{2} + y^{2} = 100; plus...I am going to make this a plus,1819

*and then apply that as -4x; minus y*^{2}; and then I have a -20 here; I am just adding the opposite of each term.1828

*So here, I end up with 4x*^{2} - 4x (these are different, so I can't just combine them);1843

*y*^{2} - y^{2}...the y^{2}'s drop out; 100 - 20 is 80.1857

*All right, now I am going to divide both sides by a factor of 4 to simplify this.*1863

*This gives me x*^{2} - 4 = 20; now it is just a quadratic equation that I need to solve.1867

*x*^{2} - 4x - 20 = 0: let's rewrite that up here.1875

*x*^{2} - 4x - 20 = 0: and let's hope that we can solve this by factoring,1882

*instead of having to go back to the messy quadratic formula.*1888

*x...the factors of 20 are 1 and 20, 2 and 10...actually, this is just x, because we got rid of that when we divided by 4...*1893

*this is just x; we divided by 4 to give us x*^{2}, x, and 20, which will make this even easier to factor.1905

*OK, so since this is -1 for a coefficient, I want some factors that are close together, like 4 and 5.*1912

*And if I take 4 - 5, I am going to get -1; so I am going to use this combination.*1920

*This is going to give me (x - 5) (x + 4) = 0; just checking that, x times x is x*^{2}, and then outer terms is 4x,1926

*minus the inner terms--that is 4x + -5x gives me -x; -5 times 4 is -20.*1939

*The zero product property tells me that, if x - 5 equals 0, or x + 4 = 0, this whole thing will equal 0.*1948

*So, solving for x will give me these two solutions: x = 5 and x = -4.*1956

*OK, I have the x-values: the next thing is to find the y-values.*1962

*So, I need to go back and substitute into one of these equations.*1968

*I am just going to select the top one; and what I need to do is determine what y will be when x is 5, and what y will be when x is -4.*1973

*So, starting out, I am letting x = -4, and then using this 8x + 2y*^{2} = 40.1986

*So, this is 8 times -4, plus 2y*^{2}, equals 40; that is -32 + 2y^{2} = 40.1997

*Adding 32 to both sides gives me 2y*^{2} = 72; dividing both sides by 2 gives me y^{2} = 36 (72/2 is 36).2008

*Therefore, if I take the square root of 36, I get that y = ±6; so y = 6, and y = -6, when x = -4.*2023

*OK, let's write some ordered pairs up here as solutions.*2035

*When x = =4, you could equal 6; when x equals -4, another solution could be that y = -6.*2039

*I am going to repeat this process with x = 5, substituting into this equation.*2047

*This is going to give me 8(5) + 2y*^{2} = 40.2053

*That is 40 + 2y*^{2} = 40; 40 - 40 is 0; 2y^{2} = 0; divide both sides by 2; I get y^{2} = 0.2058

*The square root of 0 is 0, so I only get one solution for y here.*2072

*So, when x is 5, y is 0; I ended up with 3 solutions for this system of quadratic equations,*2076

*because it turned out that I had two values for x; one of these values of x, substituting in, yielded 2 values for y.*2085

*The second value for x yielded only one result for y; so I have three solutions here.*2094

*Another system: this is another quadratic-quadratic system, so I am going to use the approach of elimination.*2104

*Before I do anything, though, I can simplify these, because there are common factors.*2110

*I am going to take the first equation; and first, I will just rewrite it so that it is in more of a standard form.*2115

*So, I am going to subtract 4x*^{2} from both sides; so it is -4x^{2} + 4y^{2} = -28.2124

*OK, this is still equation 1; I am going to divide both sides by the common factor of 4.*2134

*That is going to give me -x*^{2} + y^{2} = -7.2141

*For the second equation, I have a common factor of 5; so I will divide both sides by 5*2147

*to get x*^{2} + y^{2}... 125 divided by 5 is 25.2152

*Now, all I have to do is add these two together, because I have a -1 for a coefficient here, and a 1 here; these cancel out.*2159

*y*^{2} + y^{2} is 2y^{2}; 25 - 7 is 18.2166

*Just solve for y*^{2}: y^{2} = 18/2, so I divided both sides by 2; y^{2} is 9.2175

*Therefore, y = ±3, by taking the square root of 9.*2183

*That means that y = 3, and y could also equal -3.*2188

*Now, I need to go back and substitute into one of these equations when y = 3, and figure out what x is.*2196

*Then, I need to see, when y is -3, what x is going to be.*2204

*Let's see, the easiest one to work with would be this: and I could go back in and use the top one,*2211

*but since I divided both sides by the same thing, I didn't really change this equation.*2221

*And it is a lot easier to work with this without these larger coefficients.*2228

*So, what I am going to do is say, "Let's let y equal 3."*2232

*And then, I am going to look at this: x*^{2} + y^{2} =...actually, this first one; the first one is a smaller number, over here.2237

*I am going to say -x*^{2} + y^{2} = -7.2247

*And let's rearrange this a bit, because we are looking for x.*2251

*So, let's move this y*^{2} to the other side; and now I am stuck with a bunch of negatives.2254

*And what I can do is just multiply both sides of the equation by -1, and that gives me x*^{2} = y^{2} + 7.2261

*Now, this is the one I am going to substitute back into: again, you could have taken either of these forms;*2270

*but I just took this and made it easier to work with, and solved for x*^{2}.2275

*I am going to substitute 3 in wherever this is a y: so x*^{2} = 3^{2} + 7; therefore, x^{2} = 9 + 7.2281

*So, x*^{2} = 16, which means that x = ±4; so x could equal 4, and x could equal -4, when y is 3.2291

*So, let's start our solutions up here.*2307

*When x is 4, y is 3; when x is -4, y is 3; that is two solutions so far.*2311

*That is when y is 3; but recall, y can also equal -3.*2322

*So, when y is -3, I am going to go back into that equation and substitute -3 and see what I get for x.*2326

*(-3)*^{2} is 9; 9 + 7 is 16; so again, I get x = ±4; so x can equal 4, and x can equal -4.2340

*But this time, y is -3; so this is two different solutions from what I had up here.*2352

*So, when x is 4, y is -3; that point is a solution for this system of equations.*2358

*When x is -4, y is -3; so there are four solutions.*2366

*If you graphed this out, you would find that these intersected at four points.*2372

*This is pretty complicated: the initial part actually wasn't that bad, but keeping track of all of the different solutions was a little bit challenging.*2377

*We started out with these two equations that I simplified by dividing the first by its common factor of 4, and the second by its common factor of 5.*2385

*Then, you added them together; the x*^{2} terms dropped out, which allowed you to just solve for y.2393

*I got two solutions for y: 3 and -3; I took each of those solutions, y = 3 and y = -3, and plugged them into this equation,*2399

*right here, to find corresponding values for x; that yielded these two solutions; and then y = -3,*2411

*when I found the x-values that corresponded to that...I got two more solutions, for a total of four solutions.*2422

*This time, we have a system of quadratic inequalities: I have x*^{2} ≤ y, and then 4x^{2} + 4y^{2} < 36.2429

*So, remember, we are going to solve these by graphing the corresponding equation to find the boundary line.*2440

*And then, we are going to use test points to find the solution sets for each, and then the solution set for the system.*2446

*Starting out with x*^{2} ≤ y: this is x^{2} = y--that is the corresponding equation.2452

*Finding a few points: x and y--recall that y = x*^{2}, so when x is 0, 0^{2} gives you 0, so y is 0.2465

*When x is 1, then we get 1*^{2}; y is 1; when x is 2, we get...2 times 2 is 4; y is 4.2482

*When x is -1, -1 times -1 is 1; when x is -2, -2 times -2 is 4; all right.*2493

*You recognize this as a parabola that opens upward; and it has its vertex, which is a minimum, right here at the center: (0,0).*2503

*I have a point here at (1,1), and a point at (-1,1); I have another point at (2,4), and then a point at (-2,4).*2513

*I could have also just graphed half, and used reflection symmetry to graph the other part of the parabola.*2524

*OK, now, this is less than or equal to; so I am actually using a solid line for my boundary,*2531

*because the boundary line is included as part of the solution set.*2540

*Now, I have the graph of the boundary; and I need to use a test point to determine where my solution set lies.*2544

*Does it lie inside the parabola, or outside?*2552

*And I am going to use the test point right here, (0,2); that would be convenient to work with.*2555

*x*^{2} ≤ y; and I have a test point at (0,2); so when x is 0, that would give me 0^{2} ≤ 2.2560

*Is 0 less than or equal to 2? Yes, this is true; therefore, the test point is part of the solution set.*2574

*So, this solution set for this first inequality lies inside the parabola.*2581

*That is shaded in; and it is going to include the boundary.*2594

*This second equation that I have is 4x*^{2} + 4y^{2} < 36.2600

*So, let's rewrite this down here, and find the corresponding equation, 4x*^{2} + 4y^{2} = 36.2608

*And as you can see, there is a common factor of 4; so I am dividing both sides by that.*2621

*Looking at this, you can see that this describes a circle; and the circle has a center at (0,0).*2630

*r*^{2} = 4; therefore...actually, that should be 9, because I divided both sides by 4; so 36/4 is 9: r^{2} = 9.2643

*Therefore, r = 3; so this equation--all I did is divide both sides by 4, and I can see now that I have*2658

*an x*^{2} and y^{2} term here on the same side of the equation, and they are equal to 9.2666

*Therefore, it is a circle at center (0,0); they have the same coefficient; and r*^{2} = 9, so the radius of the circle is 3.2674

*So, the center of the circle is here; the radius is 3.*2683

*There would be a point there...the edge of the circle there...there...and there.*2686

*Then, you can fill in; now, this is a strict inequality, so I am just using a dashed, or dotted, line, as the border.*2692

*I am not going to make this a solid line, because this border is not part of the solution set.*2706

*The next step is to use a test point to determine, just for this inequality (this is the second inequality), where the solution set is.*2711

*And I am going to use the test point for the circle, right, test point (0,0), at the origin; that is easy to work with.*2721

*So, I go back up here to the original, 4x*^{2} + 4y^{2} < 36.2730

*So, 4 times 0*^{2} + 4 times 0^{2} < 36.2736

*This just gives me 0: 4 times 0 is 0; and this is 0*^{2}...4 times 0 is also 0.2742

*Is 0 less than 36? Yes, so again, I have a solution set that is on the inside of this boundary.*2752

*OK, so this is the solution set for the second inequality; this is the solution set for the first inequality.*2766

*And the solution set for the system is going to be right in here, where the blue and the black overlap.*2771

*This boundary is included in part of the solution set: the boundary of this circle is not.*2780

*So, we solved this by graphing the boundary for the first inequality, then the boundary for the second inequality,*2786

*and using test points to find the solution sets for the individual inequalities.*2794

*And then, the overlap between the two is the solution set for the entire system.*2799

*So today, we worked on systems involving quadratic equations--systems of equations where one was linear,*2806

*one was quadratic, or both were quadratic, as well as some systems of quadratic inequalities.*2814

*And that concludes today's lesson; thanks for visiting Educator.com.*2822

*Welcome to Educator.com.*0000

*Today, we are going to start a series of lectures on sequences and series, starting with arithmetic sequences.*0002

*A ***sequence** is a list of numbers given in a certain number; and each of these numbers is called a **term**.0010

*So, the general form that a sequence is written in is this: a*_{1},0016

*or just a, is the first term; a*_{2} is the second term, and on and on.0021

*Each term is given by a; and then n is the term number.*0026

*For example, a typical sequence would be something like 5, 10, 15, 20; and this type of sequence is called a ***finite sequence**.0032

*Another type of sequence is one like this: 3, 9, 27, 81...and then it has the three dots, the ellipses, at the end,*0052

*indicating that it goes on forever: this is called an ***infinite sequence**.0064

*Looking back at this second function, looking a little bit more deeply, we could actually rewrite this second function as a*_{n} = 3^{n}.0074

*I will go into more detail about this in a few minutes.*0089

*But just in general, sequences in general (not just limiting ourselves to arithmetic sequences)...*0092

*The idea, then, would be: if I was looking for the first term, a*_{1} (here n = 1--it is the first term),0099

*I would then substitute in 1 where the n was; so that gives me 3.*0106

*So, the first term is 3: a*_{2} is 3 to the second power, 9; a_{3} is 3 to the third power, which is 27; and so on.0112

*So, you can develop an equation for this sequence that will tell you what a particular term is--what the value of that term is.*0126

*If I wanted to know the 17*^{th} term, then it would simply be 3^{17}, whatever that comes out to.0134

*So, this is just sequences in general; but today we are focusing on arithmetic sequences.*0142

*In an arithmetic sequence, each term after the first one is obtained by adding a constant--*0149

*not multiplying by a number or anything--it is just by adding; that is limiting this to arithmetic sequences.*0154

*The constant that you add to obtain the subsequent term is called the common difference.*0162

*In the previous slide, I showed you a typical sequence: 5, 10, 15, 12 and this is an arithmetic sequence.*0171

*The common difference, which is often just called d, is 5; so this is the common difference.*0181

*Therefore, I can start with the first term; and to get the second term, I am going to add 5 to that.*0191

*So, a*_{2} = 10; a_{3}, 10 + 5, equals 15; and so on.0199

*One thing to be aware of is that the common difference can be negative; it also can be a fraction.*0211

*So, the common difference can be a positive whole number; it could be a fraction; or it could be an integer that is negative.*0218

*And we are going to see examples of all of those in a few minutes.*0227

*If you want to find the value of a particular term in an arithmetic sequence, you can use the formula for the n*^{th} term.0233

*Here, this is called the formula for the general term sometimes (either for the n*^{th} term or the general term).0242

*a*_{n} = the first term, plus (n - 1) times the common difference.0249

*So, if I know the first term and the common difference, and I am looking for a particular term (let's say the 20*^{th} term),0257

*then I would know n, so I can find the value of a*_{20}.0266

*Looking at a little bit different sequence here: 5, 3, 1, -1, -3, and continuing on--looking at this,*0273

*I can see that the numbers are getting smaller.*0286

*Since the numbers are getting smaller, I know that the common difference is negative.*0289

*And I can easily find that common difference by taking one of the terms and subtracting the previous term.*0292

*So, I will go ahead and take 3; I will subtract the previous term; and that is going to be equal to -2, so the common difference is -2.*0299

*All right, so if I have this common difference, then if I wanted to look for a particular term, I could find it.*0312

*For example, I may want to find a*_{10}: then I could use this formula.0321

*a*_{10} would equal the first term, 5, plus 10 minus 1, times the common difference, which is -2.0336

*So, a*_{10} = 5 + (10 - 1)...is 9...times -2 (that is *times* -2, actually, making that clear).0349

*a*_{10} = 5 + -18, or a_{10} = -13.0365

*Therefore, by having this formula, I can find any term in here, without going through the cumbersome of just subtracting 2, subtracting 2, and so on.*0376

*I knew the first term; I knew n; and I knew the common difference; the rest is just calculating.*0386

*OK, the equation for the n*^{th} term: you can use the formula that I just gave in order to find the equation for the n^{th} term.0397

*And I gave an example of applying that general equation.*0404

*Let's talk about using that formula to find a specific equation again.*0409

*So, looking at another example: 9, 12, 15, and so on--I have been given some terms in the sequence.*0416

*And I think back to my formula: a*_{n} = a_{1} + (n - 1)d.0431

*I want to find an equation for the n*^{th} term specific to this sequence.0441

*So, I am going to look at what I have: I have a*_{1}...I have the first term, often just called a; that equals 9.0446

*I need to find the common difference: I can find the common difference by saying 12 - 9 = 3.*0453

*Therefore, I can write a general equation: the general term, or the n*^{th} term, a_{n}, equals 90459

*(9 is the first term), plus (n - 1), times the common difference, which is 3.*0472

*So, a*_{n} = 9 + 3n - 3; a_{n} = 6 + 3n.0479

*Now, I have a formula that I can use; so if I wanted to find any term, such as a*_{21}, I could use this.0493

*a*_{21} = 6 + 3(21); so, a_{21} would then equal 63 + 6, which equals 69.0504

*OK, so again, I have my general equation here; and then I said, "All right, I have my general formula; I can write a specific equation for this situation."*0532

*The first term is 9; and the common difference is 3; that comes out to 9 + 3n - 3; 9 - 3 is 6, so it is 6 + 3n.*0544

*Now, if I want to find the 21*^{st} term, then I will say, "OK, I am going to go ahead and use my specific equation0558

*for this scenario, for this sequence," which is a*_{21} = 6 + 3(21).0570

*That gives me 63 + 6; a*_{21} is 69.0582

*Arithmetic means: arithmetic means does not mean finding the mean of the sequence or anything; this is totally different.*0588

**Arithmetic means** are actually the terms between two non-successive terms of an arithmetic sequence.0595

*And you can actually use the formula that we learned for the n*^{th} term to find the common difference, d.0603

*Once you know the common difference, d, you can use d to find the arithmetic means between terms.*0610

*For example, if I were given a sequence, -3...and then there were some missing terms: one term is missing,*0616

*two terms are missing, three, four, and then they give me the last one; I have my first term, a*_{1};0625

*and I have 1, 2, 3, 4, 5, 6 terms in this; so a*_{1} = -3, a_{6} = 17.0634

*Then, I look back to that formula for the general term, a*_{n} = the first term, plus (n - 1) times the common difference.0645

*In this case, I am going to be using the formula, not to find a*_{n} yet; first I need to find the common difference.0654

*I then use the common difference to find these terms.*0661

*Let's go ahead and find the common difference.*0664

*If a*_{6} is 17, and I put that here; -3 is the first term, and I am working with a_{6};0666

*so in this case, there are 6 terms, a*_{6}...6 - 1 times the common difference, which is what I am looking for.0675

*Now, it is just a matter of solving for that: this gives me 17 = -3 + 5d; add 3 to both sides; that is going to give me 20 = 5d.*0685

*And then, I am going to divide 20 by 5 to give me 4 = the common difference.*0699

*So now that I have a common difference of 4, all I have to do is find a*_{2}; that is going to equal -3 + 4, so that is going to equal 1.0706

*To find a*_{3}, it is going to be 1 + 4 = 5; a_{4} = 5 + 4; that is 9.0715

*And a*_{5} = 9 + 4, which is 13.0726

*OK, we have -3; and now, our missing terms: 1, 5, 9, and 13; and then I have 17, which was already given.*0732

*So, I found the arithmetic means, or the missing terms, by using the formula for the n*^{th} term0746

*to find the common difference, and then taking that common difference and adding it to each term to find the next term.*0750

*Example 1: Write an equation for the n*^{th} term of this sequence.0758

*Recall that the equation for the n*^{th} term, just the general formula, is a_{n}, the general term,0764

*equals the first term, plus (n - 1) times the common difference.*0774

*Therefore, we need to find the common difference; and you can find that by taking any term and subtracting the one just before it.*0779

*Therefore, 26 - 19 = 7; so the common difference is 7.*0786

*a*_{n} equals the first term; well, I also have the first term--that is -2.0797

*So, that is -2 + (n - 1) times 7, so a*_{n} = -2 + 7n - 7; a_{n} =...-2 + -7 is -9, so 7n - 9.0801

*So, this is the equation to find any term of this sequence.*0822

*Example 2: Find the arithmetic means: this time, we need to find the missing terms.*0829

*And we can do that because we, again, have that equation for the n*^{th} term,0834

*which is a*_{n} = the first term + (n - 1) times the common difference.0840

*In order to do this, though, I need to find the common difference; and I can do that because I have the first term,*0846

*which is -7; I also have 1, 2, 3, 4, 5, 6, 7...I also have the seventh term, which is equal to 11.*0854

*And in this case, n = 7; so I just go ahead and use these values to find d.*0863

*Therefore, I am going to end up with 11 = the first term, which is -7, plus n, which is 7, minus 1, times d.*0872

*Therefore, 11 = -7 + 6d; I am going to add 7 to both sides to simplify--that gives me 18 = 6d.*0887

*Divide both sides by 6; I end up with d = 3.*0896

*Now that I have this, all I need to do is say, "OK, for the second term" (I have the first term),*0900

*"I am going to take the first term, -7, and just add 3 to that to get -4."*0908

*For a*_{3}, I am going to take -4 and add 3 to that to get -1.0914

*a*_{4} = -1 + 3, so that is going to give me 2.0925

*a*_{5} = 2 + 3, which is actually equal to 5; and then, a_{6}, which is also missing, is going to be 5 + 3, or 8.0933

*Therefore, the term I was given was -7; and then, I found the missing terms, -4, -1, 2, 5, and 8; these were the missing terms, the arithmetic means.*0943

*And I was given the last term, 11.*0960

*And the key thing is to find the common difference, using this general equation, and then to take that common difference and use it to fill in the missing terms.*0962

*Which term is 763 in the arithmetic sequence (and the arithmetic sequence is given)?*0974

*So, you have a term, a*_{n} = 763; and what this is really asking is what place that is.0983

*Is it the fifth term? Is it the seventeenth term? What number term is this?*0991

*I know that its value is 763; and let's look at what else I know.*0997

*Well, I know that the first term is -7; and I can easily find the common difference.*1002

*I can just take 15 - 4, for example, which is 11; so knowing this value, the first term, and the common difference,*1009

*I can go back to my equation for the n*^{th} term, a_{n}1017

*equals the first term, plus n - 1 times the common difference.*1022

*What I am looking for is n, the term number: for a*_{n} = 763, what is n?1027

*I can put in 763 = -7 +...n is my unknown, so I have (n - 1) times this common difference of 11.*1037

*So, 763 = -7 + 11n - 11; adding 7 to both sides is going to give me 770 = 11n - 1.*1047

*Now, adding 11 to both sides gives me 781 = 11n; and the last step is just to divide both sides by 11.*1065

*And if you figure that out, it comes out to n = 71.*1074

*Therefore, the term number equals 71, or a*_{71} = 763.1078

*So, this time, I was given a term, and was asked to figure out which term it is in the sequence.*1090

*Where does it land in this sequence?--I could do that because I had the first term; I had the common difference; and I had the value of that term.*1097

*All right, find an equation for the n*^{th} term of the arithmetic sequence with a_{101} = 100 and a common difference of 7.1106

*Well, to figure out this equation for the n*^{th} term, I need the first term.1118

*So, I have the common difference; but the thing that is missing here is the first term.*1125

*So, how do I figure that out? Well, I do know another term; I know an a*_{n}.1138

*Since I know a*_{n}, or a_{101}, and I know that n is 101 in that case, and I know the common difference, I can solve.1143

*So, I am going to first solve for the first term.*1151

*Then I will go back and use that first term to develop an equation for the n*^{th} term for this sequence.1156

*So, I have a*_{1} = 100, and I don't know my first term; I know my n for this term is 101,1163

*and I am going to say minus 1, times the common difference, which is 7: this gives me 100 = a*_{1} + 100 times 7.1176

*100 = a*_{1} + 700; subtract 700 from both sides; that is going to give me...actually, let's see:1187

*yes, it is going to give me -600 equals this first term.*1202

*So, I am just rewriting it in a more standard form: the first term equals -600.*1207

*Now that I know the first term, I go back again and look at that general equation for the n*^{th} term.1212

*And recall that I am asked to find a specific equation for the n*^{th} term for this sequence.1220

*And I can do that, because I know that the first term is -600, and the common difference, d, is equal to 7.*1226

*So, a*_{n} = -600 + 7n - 7; this equals a_{n} = 7n - 607, just simplifying.1236

*I had to take an extra step here, because I wasn't given the first term.*1251

*But since I was given another term, I could find the first term.*1254

*And then, I went ahead, and I used that to find the equation for the n*^{th} term for this sequence, which is a_{n} = 7n - 607.1258

*This concludes the lesson on arithmetic sequences on Educator.com; thanks for visiting!*1269

*Welcome to Educator.com.*0000

*In a previous lesson, we talked about arithmetic sequences; and in this lesson, we will continue on with the discussion, to discuss arithmetic s0ries.*0002

*First of all, what are arithmetic series? An ***arithmetic series** is the sum of the terms of an arithmetic sequence.0012

*So, I will briefly review arithmetic sequences; but if you need a complete review of that,*0019

*go and check out the previous lesson, and then move on to arithmetic series.*0024

*Recall that a series is a list of numbers in a particular order; and each term in the series is related to the previous one by a constant called the ***common difference**.0028

*A typical arithmetic sequence would be something like 20, 40, 60, 80, 100.*0043

*Taking 40 - 20 or 60 - 40, I get a common difference of 20, which means that, to go from one term to the next, I add 20.*0051

*So, this is the sequence: it is a list of numbers; the series is actually a sum of numbers,*0063

*so I am going to add a + between each number: 20 + 40 + 60 + 80 + 100.*0071

*So, an arithmetic series is in the general form: first term, a*_{1} + a_{2} + a_{3}, and on and on until the last term, a_{n}.0082

*This is a finite arithmetic series, because there is a specific endpoint; there is a finite number of terms.*0097

*As with arithmetic sequences (those could be finite or infinite), you could have finite or infinite arithmetic series.*0106

*So, looking at a different sequence: 100, 200, 300, 400, and then the ellipses to tell me that this is an infinite sequence:*0115

*I could have a corresponding series, 100 + 200 + 300 + 400 + ...on and on; this is an infinite arithmetic series.*0127

*There is a formula (or actually, two formulas) here to allow you to find the sum of an arithmetic series.*0148

*Sometimes you will be asked to find the sum of the first n terms of an arithmetic series--maybe the first 17 terms or the first 10 terms.*0154

*And that would be s*_{17} or s_{10}.0162

*There are two formulas: the first one involves taking the number of terms, dividing it by 2, and then multiplying it*0166

*by 2 times the first term, plus the quantity (n - 1) times the common difference.*0173

*The second term is the sum of the first n terms: again, it equals the number of terms divided by 2.*0179

*But this time, you are going to multiply that by the sum of the first and last terms.*0185

*So, you can see the difference: you use the first formula if you know the common difference, and the second formula if you know the last term.*0190

*So, you have to choose a formula, depending on what you know.*0199

*For example, let's say that a series has, as its first term, a 5: 5 is the first term.*0202

*And it has a common difference of 4, and you are asked to find the sum of the first 10 terms, s*_{10}.0211

*I am looking, and I see that I have the common difference; so I am going to go ahead and use the first formula.*0220

*And let's go ahead and do that: s*_{10} = 10, because, since I am finding the sum of the first 10 terms,0229

*n = 10, divided by 2, and that is times 2 times 5 (the first term), plus n (which is 10) minus 1, times the common difference (which is 4).*0241

*So, s*_{10} = 5(10) +...10 - 1 is, of course, 9, times 4.0253

*Therefore, s*_{10} = 5 times...9 times 4 is 36; 36 + 10 is 46.0262

*And if you multiply that out, or use your calculator, you will find that the sum of the first 10 terms*0272

*in a series with this first term and this common difference is actually 230.*0276

*So, that is one example; in another example, perhaps you are asked the sum of the 16 terms in a series,*0283

*if the first term is 20, and the last term of the ones we are trying to find, a*_{16}, is equal to -10.0292

*Well, we are going to use the second formula, because we have the first and last terms, but we don't have the common difference.*0305

*So, s*_{16} = n; in this case, I am looking for the sum of 16 terms, so n is 160311

*(I am going to put that right here--n is 16); that is going to give me 16/2, times the first term, which is 20, plus that last term, which is -10.*0323

*So, this is s*_{16} = 8, times 20 - 10, which is 10; that gives me 80 as the sum of the first 16 terms.0336

*You need to learn both formulas and simply know when to use a particular one.*0348

*Sigma notation: a series can be written in a concise form, using what is called sigma notation.*0353

*And what sigma refers to is the Greek letter Σ; and in this case, sigma means "sum."*0359

*So, what this symbol means, in the context in which we are using it, is "sum."*0366

*And you will see it written something like this: you will see a letter here, called the index (that variable is called the index).*0371

*And we have been working with n, but often i is used; it could be n; it could be i; it could be k; it could be something else.*0378

*i = 1; I am just giving an example; and let's say, up here, the upper index is going to be 10.*0385

*And then, there will be the formula for the general term, which we talked about earlier on,*0393

*when we talked about arithmetic sequences and the formula for a*_{n} that is particular to a sequence.0399

*So, if you need to go ahead and review that, it is in the previous lecture.*0406

*This is the formula for the general term of the sequence.*0409

*And this is read as "the sum of a*_{n} as i goes from 1 to 10."0412

*So, it is the sum of the terms that are found, using this particular formula, when you plug in 1, then 2, then 3, and up through 10.*0420

*That is in general; let's talk about a specific example with a specific formula.*0429

*You could have another arithmetic series, written in sigma notation, where n goes from 5 to 12.*0435

*So, n goes from 5 to 12; so I am going to start with 5, and I am going to end with 12.*0448

*That means that there are actually 8 terms; there are 8 terms in the series, because it is 5 through 12, inclusive.*0454

*And the formula, we are going to say, is n + 3: so the formula to find a particular term, a*_{n}, is n + 3.0460

*I could find the terms, then, by saying that for the first term, a*_{1}, I am going to use 5 as my n, so it equals 5 + 3; therefore, it is 8.0472

*For a*_{2}, I am then going to go to 6: so that is going to be 6 + 3; that is 9.0485

*a*_{3} =...then I am going to go to the next value, which is 7: 7 + 3 = 10; and on up.0497

*So, if I wanted to find the last term, a*_{8} (because there are 8 terms), then I would put in 12; and 12 + 3 is 15.0512

*There would be terms in between here, of course.*0524

*This is just a concise way of writing a series; and we have already talked about how to work with these series, and what the different terms mean, and formulas.*0527

*But now, this is just a different way of writing them.*0539

*We need to find the first term of the arithmetic series with a common difference of 3.5 and equal to 20,*0541

*and the sum of the terms, s*_{20}, equaling 1005.0547

*Since we know the common difference, we can use this formula: s*_{n} = n/2, times 2 times the first term, plus (n - 1)d.0553

*Except, in this case, I am not looking for the sum: I have the sum; I am looking for the first term.*0566

*Therefore, 1005 = 20/2, times 2 times a*_{1}, plus n (n is given as 20), minus 1, times the common difference of 3.5.0571

*This gives me 1005 = 10(2a*_{1}) + 19(3.5).0589

*19 times 3.5 is actually 66.5; therefore, 1005 equals 10 times 2a*_{1}, which is 20a_{1}, plus 10(66.5), which is 665.0601

*1005; subtract 665 from both sides to get 340 = 20 times that first term.*0622

*Divide both terms by 20, and you will get that the first term is equal to 17.*0631

*So, we use this formula for the sum of the series; but in this case, we were looking for the first term.*0636

*We had the sum; we were looking for the first term.*0646

*And the solution is that the first term is 17.*0647

*In the second problem, we are asked to find the first three terms of the arithmetic series with n = 17, a*_{n} = 103, s_{n} = 1102.0653

*In order to find the first three terms, I need to find the common difference.*0666

*And I also need to find the first term; I need the first term, and then I need the common difference, to find the second and third terms.*0670

*I can use the formula s*_{n} = n/2 times the first term plus the last term, because I don't have the common difference--I am looking for it.0680

*But what I do have is a*_{n}, so my first step is going to be to find the first term.0691

*The sum is 1102; n is 19; I don't know the first term; and I know that a*_{n} is 103.0698

*I am going to multiply both sides by 2 to get 2204 = 19 times this, a*_{1} + 103.0710

*I am then going to divide both sides by 19, and that comes out to 116 = the first term, plus 103.*0721

*And then, I just subtract 103 from both sides; and now I have my first term.*0729

*So, I am asked to find the first three terms: the first term is 13.*0734

*To find the next two terms, I find the common difference.*0737

*What I am going to do is switch to the other formula: that other formula, s*_{n},0740

*equals n/2 times 2a*_{1}, the first term, plus n - 1 times the common difference.0744

*I found the first term; now that I have that, I can find the common difference, because I can put the first term in here.*0754

*So again, s*_{n} is 1102; n is 19; and this gives me 2 times 13, plus I have an n of 19 - 1, and I am looking for the common difference.0761

*I am going to multiply, again, 1102 times 2 to get 2204 = 19...2 times 13 is 26, plus 19 - 1...that gives me 18d.*0775

*I am going to divide both sides by 19 to get 116 = 26 + 18d.*0790

*Subtracting 26 from both sides gives me 90 = 18d; the final step is to divide both sides by 18 to get a common difference of 5.*0798

*I have a*_{1} is 13; I am going to take 13 + the common difference of 5 to get the second term (that is 18).0810

*So, a*_{2} is 18; the third term--I am going to take 18, and I am going to add 5 to that to yield 23.0823

*So, I was asked to find the first three terms, and I did that by first using this formula to find the first term,*0836

*then going to the other formula and finding the common difference to get 13, 18, and 23 as my solutions.*0843

*The third example: I am asked to find the sum of the series 6 + 11 + 16 + 21, and on and on, with the last term of 126.*0854

*That means that what I have is the first term and the last term.*0867

*I am going to find the sum using this formula, because I can figure out my common difference.*0874

*So, I am going to use the formula n/2, times 2a*_{1}, plus a - 1, times d.0884

*I could easily find the common difference, because I know I will just take 11 - 6, so I have a common difference of 5.*0896

*Looking at this formula, the only issue is going to be that I don't know n.*0904

*But I can figure out n; and that is because I have another formula.*0910

*Recall the formula for the general term: we discussed this in the lecture on arithmetic sequences.*0917

*a*_{n} equals the first term, plus n - 1, times the common difference.0926

*a*_{n} is 126; so that shouldn't be 16--that is 126: a_{n} is 126, and I have the first term equal to 6; and I am solving for n.0937

*I know that the common difference is 5.*0956

*126...and then I am subtracting 6 from both sides: that gives me 120 = (n - 1)5, so 120 = 5n - 5.*0959

*Adding 5 to both sides gives me 125 = 5n; 125/5 is 25, so I have n = 25.*0970

*Now, again, I am asked to find the sum of the series; and I can use this formula, because I now have n; I have the first term; and I have the common difference.*0982

*So, let's go ahead and use that: it is actually s*_{25} = n, which is 25, divided by 2, times 2 times that first term0990

*(which is 6), plus (n - 1) (n is 25, minus 1), times the common difference of 5.*1000

*Now, it is just a matter of simplifying: this is going to give me 25/2 times 12, plus 24 times 5.*1009

*So, the sum equals 25/2, times 12; and then if you multiply out 24 times 5, you will get 120.*1022

*Therefore, the sum equals 25/2; 120 + 12 is going to give you 132.*1033

*Now that I have gotten it down to this point, I can simplify,*1050

*because this is going to give me 132/2, times 25; so that is s*_{25} = 25 times 66.1054

*You can multiply it out; or it is a good time to use your calculator to find that the sum equals 1650.*1065

*So again, in order to use this formula, I had my common difference.*1074

*I didn't have n; I solved for n; n equals 25.*1077

*I went back in, substituted those values in, and then came out with s*_{25}; the sum of this series is 1650.1081

*Example 4: we are working with sigma notation, so you need to know how to read this notation.*1093

*And I am going to start with 4 and end with 14; and I can find the first term, because I am also given the formula for a general term in this series.*1098

*So, to find the sum of the series, let's just start out by finding the first term, because we know we are going to need that.*1116

*a*_{1}...we are going to begin with 4, so for the first term, n is going to equal 4.1123

*It is 2 times 4, minus 3; a*_{1} = 8 - 3, so the first term is going to be equal to 5.1131

*Now, recall that we have two formulas that we can use to find the sum.*1143

*We have one formula that involves knowing the common difference.*1146

*We have another formula that requires us to know the first term and the last term.*1150

*I found the first term; since I know that, for the last term, n = 14, I can find that, as well.*1156

*And what that means is that I can use this formula: that the sum of the series is going to be equal to n/2, times the first term, plus the last term.*1162

*Therefore, let me find the last term: a*_{n} = 2(14) - 3; this is going to give me a_{n} = 28 - 3, so the last term equals 25.1172

*Now, what is n? Well, this is telling me that the number of terms...I would have to take each number from 4 through 14, inclusive.*1196

*And if you figure that out, that is actually 11 terms, because you are including 14.*1206

*So, starting with 4 and going up through 14, there are actually 11 terms, so n = 11.*1209

*It is really a*_{11} = 15 I am asked to find the sum of these 11 terms.1215

*And I can do that now, because I know that I have n = 11, divided by 2, and then I am going to get the first term*1222

*(that is 5), plus the last term, which is 25; so the sum is 11/2, times 5 + 25 (is 30).*1228

*Therefore, this cancels; I am going to get 11 times 15, and that is simply 165.*1239

*OK, so in sigma notation, this gives me a lot of information, because I saw that I knew the formula to find a particular term.*1251

*And I knew the n for the first term and the n for the last term.*1260

*So, I knew I could use this formula, because I could find the first and last terms.*1264

*So, I made n equal to 4 to find the first term, which is 5; I made n equal to 14, which is to help me find the last term, which is 25.*1268

*And then, I knew that, since it was going from 4 to 14, that n is equal to 11.*1277

*Once I had first term, last term, and n, it was just a matter of calculating the sum, which was 165.*1282

*That finishes up today's lesson on arithmetic series; thanks for visiting Educator.com!*1290

*Welcome to Educator.com.*0000

*In the previous lessons, we talked about arithmetic sequences and series.*0002

*So, we are going to go on to discuss geometric sequences.*0007

*What are geometric sequences? Recall that a sequence in general is a list of numbers in a certain order.*0012

*So, it is in the general form first term, second term, and on...and it could end at a particular term, a*_{n}, or it may go on indefinitely.0020

*In previous lessons, we talked about arithmetic sequences: for example, 5, 10, 15, 20.*0034

*And each term was related to the previous one by a common difference, d (here d = 5).*0043

*What we did is added whatever the common difference was to a term to get the next, to get the next, and so on.*0051

*So, if you need to review arithmetic sequences or sequences in general, it might be a good idea to go back and start with that lecture.*0057

*And now, we are going to continue on and learn about geometric sequences.*0064

*So again, a geometric sequence is a list of numbers; so a ***geometric sequence** is a sequence0067

*in which each term after the first is found by multiply the previous term by a non-zero constant r.*0073

*When we talked about arithmetic sequences, we added the common difference.*0081

*Here we are going to multiply a term by the common ratio r to get the next term.*0087

*Therefore, a*_{n} equals the previous term, which is a_{n - 1}, times the common ratio.0092

*So, if I were to look for a*_{4} in a particular series, I would find it by saying,0103

*"OK, the previous term, a*_{4 - 1}, times the common ratio..." so the term a_{4} would equal r times the third term.0109

*Looking at an example of this, this, again, was an arithmetic sequence.*0124

*Now, we are going on and talking about an example that gives you a geometric sequence: 3, 12, 48, 192.*0150

*Working with these, it is important to find a common ratio: the common ratio is given by taking a term*0166

*(any term--I am going to take 12) and dividing it by the previous term.*0173

*So, this is a*_{2}; the first term is a_{1}; therefore, if I take a_{2}/a_{2 - 1}, that is just a_{2}/a_{1}.0178

*So, take a term; divide it by the previous term; and this is going to give me the common ratio of 4.*0191

*So, you can see where these two equations come from.*0206

*To find the next term, you multiply the term previous to it by the common ratio.*0209

*To find the common ratio, you take a term, and you divide it by the term that came just before.*0215

*Now, as we talked about arithmetic sequences, I said that the common difference could be a negative number; it could also be a fraction.*0223

*And that is true, as well, with the geometric series.*0231

*For example, I could have 2, -6, 18, -54: this is another geometric sequence, and I want to find the common ratio.*0237

*So, I will take a term; I am going to go for -6, and I am going to divide it by the previous term, which is 2; and it is going to give me -3.*0250

*The thing that you will notice is: when you have a negative common ratio, the terms are going to alternate their signs.*0257

*So, I am going to have a positive, a negative, a positive, a negative.*0265

*If it is a positive common ratio, then the terms will just all be positive.*0269

*You can also have a common ratio that is a fraction; and we will work with those examples later on in the lesson.*0274

*For a geometric sequence, there is a formula for the n*^{th} term.0282

*So here, the formula for the n*^{th} term is that the general term, a_{n},0286

*equals the first term, times the common ratio, taken to the power n - 1.*0290

*Looking at a geometric sequence: 2, 6, 18, and we will make this an infinite sequence...*0296

*it is going to go on and on...perhaps I was asked to find the sixth term, a*_{6}.0302

*What is that? Well, I can use this formula.*0309

*a*_{6} equals the first term (I have the first term--it is 2), and I also need to find the common ratio.0312

*The common ratio is going to be any of the terms (I will take 18), divided by the previous term (6), which is 3.*0320

*So, I have the common ratio; I have the first term; and I have n; here, since I am looking for the sixth term, n will equal 6.*0329

*So, the sixth term is equal to the first term (which is 2), time the common ratio (which is 3), raised to the power of 6 - 1.*0339

*The sixth term is equal to 2, times 3*^{5}.0354

*So, recalling powers of 3: 3 times 3 is 9, times 3 is 27; so 3*^{3} is 27; 3^{4} is 81; and 3^{5} is 243, 81 times 3.0359

*So, a*_{6} = 2 times 243, or simply 486.0381

*So again, this is the formula for the n*^{th} term of a geometric sequence.0389

*Given the sequence, I could find the sixth term, because I know the first term, 2; I know the common ratio;*0394

*I was able to figure out that it is 3; and I know that n is equal to 6; so I got that a*_{6} is 486.0404

*Geometric means: again, thinking back to arithmetic sequences, we said that arithmetic means are missing terms in an arithmetic series.*0412

*And that is analogous to this situation: geometric means are missing terms between two non-successive terms of a geometric sequence.*0421

*2048 is the first term; then I have three missing terms--those are the geometric means; and then, I have my last term, 8.*0429

*Use the common ratio to find the geometric means.*0442

*If I find the common ratio, then all I need to do is take a term and multiply it by the common ratio to find the next term;*0444

*multiply that by the common ratio to find the next term; and so on.*0450

*If I want to find the missing terms here, I am going to use my formula for the n*^{th} term,0454

*a*_{n} = the first term, times r^{n - 1}.0458

*Let's look at what I am given: the first term is 2048: 1, 2, 3, 4, 5...the fifth term is 8; OK.*0463

*Using this formula: a...I need to find the common ratio, and I can do that because I have the first and last terms, and I have n.*0476

*1, 2, 3, 4, 5; n = 5, right here; so a*_{n} is 8; that is equal to the first term, times the common ratio, raised to the power 5 - 1.0485

*8 = 2048 times r*^{4}; so if I take 8/2048, equals r^{4}, this simplifies to 1/256 = r^{4}.0501

*If you think about your roots and your powers, the fourth power is actually plus or minus...*0524

*I am going to take the fourth power of 1/256; it is actually ± 1/4.*0535

*And if you multiply this out, you will find that 4*^{2} is 16, times 2 is 32, times 2 is 64...0540

*Excuse me, 1/4 times 1/4 is 1/16, and continuing on, you will find that the fourth power of 1/4 is 1/256.*0551

*OK, the important thing to note, though, is that it is not just that the fourth power is 1/256; it is actually plus or minus,*0568

*because I could take -1/4 times -1/4 times -1/4 times -1/4, and I would also get 1/256.*0579

*Therefore, with geometric means, you may end up with two sets of answers.*0591

*All right, so I found my common ratios, which could be ± 1/4; and I have my first term, 2048.*0599

*So, to find my second term, I just take 2048 times 1/4; that is 5/12.*0606

*To find my third term, I just take 512, times 1/4; and I am going to get 128.*0614

*To find the fourth term, I am going to take 128 times 1/4, and I am going to get 32.*0626

*So, that is one set of answers; I actually have two sets of answers.*0635

*If r = 1/4, then the missing terms (the geometric means) are 512, 128, and 32.*0639

*If r is actually equal to -1/4, then the signs will alternate, so what I am going to get is 2048, then -512, then positive 128, then -32.*0665

*So, I have two possible sets of geometric means.*0682

*Again, to find the geometric means, I am just going to find the common ratio.*0686

*And I was able to do that using this formula; I got two answers--the common ratio is either plus or minus 1/4.*0691

*I took the first term; I multiplied it by 1/4, and then multiplied that by 1/4, and on to get this set of geometric means.*0698

*I took my other possible solution, r = -1/4, and multiplied 2048 by that to get -512, times -1/4 is 128, times -1/4 is -32.*0705

*So, that is something to keep in mind: that, with geometric means, you can get two sets of solutions.*0718

*Example 1: Find the ninth term of the geometric sequence with a fifth term of 80 and a common ratio of 2.*0726

*The formula for the n*^{th} term is first term, times r^{n - 1}, the common ratio to the n - 1 power.0734

*Here, n = 9; and I am given a*_{5}, and I am already given r.0744

*So, let's go ahead and work on this.*0755

*I have my common ratio of 2, and n is 9, so that is 9 - 1; so a*_{9}(which is what I am looking for) equals0760

*the first term, times 2*^{8}; I am stuck.0776

*I can't go any farther; I need the first term--I need a*_{1}.0780

*But there is something else I haven't used yet, and that is the fact that I know a*_{5}.0786

*In order to go over here and find the first term, I can do that by saying, "OK, let's look at this formula again."*0791

*I know what a*_{5} is; so let's look at this formula again and use it to solve for the first term.0800

*a*_{5} equals the first term, times r; in this case, n will be 5; and that is to the 5 - 1.0809

*So, I can substitute in; I know that 80 equals the first term, times r*^{4}; and I already do know r, so let's put that in, as well.0818

*The first term is 2 to the fourth; 2 times 2 is 4, times 2 is 8, times 2 is 16.*0827

*So, 2*^{4} is actually 16; if I divide both sides by 16, I am going to get 80/16, and the first term is 5.0838

*Now, I can go back and finish my problem.*0849

*I know that a*_{1} is 5; since a_{1} = 5, let's finish this out.0852

*a*_{9} = 5 times 2^{8}; when you continue on with powers of 2, we know that 2^{4} is 16.0861

*If you continue on up, you are going to find that 2*^{8} is 256.0870

*So, a*_{9} = 5(256); 5 times 256 is 1280, and that is what we were looking for.0875

*All right, again, it looked straightforward, but we had to take a detour, because when we started out*0886

*using that formula to find the ninth term, we discovered that we got stuck at this step, because we didn't have the first term.*0893

*But, since they gave us another term and the common ratio, I was able to go back,*0902

*substitute in 80 here, put in my common ratio of 2, and solve for the first term.*0906

*Then, I finish out the problem to find that the ninth term is equal to 1280.*0914

*Find the geometric means: so we need to find the three missing terms.*0919

*I always look at what I am given first.*0923

*Well, I am given the first term, and I am given 1, 2, 3, 4, 5...the fifth term.*0925

*As always, I am going to use my formula here, that the n*^{th} term is equal to the first term, times the common ratio raised to the n - 1 power.0934

*To find the geometric means, I need the common ratio--what is r?*0944

*If I have r, I multiply it by 4 to get the second term, then the second term by the common ratio to get the third, and so on.*0948

*But I don't know r: what I do know are these two things, so I can find r.*0957

*I have that a*_{5} is 324; and my first term is 4; in this case, n is going to be 5; now I can find the common ratio.0961

*This gives me 324 = 4r*^{4}; divide both sides by 4--that gives me 81 = r^{4}.0978

*The fourth root of 81 is 3; 3 to the fourth power is 81.*0988

*But there is something I have to remember: the other fourth root of 81 is -3.*0995

*If I take -3 times -3 times -3 times -3, that is 9, -27, times -3 is also 81.*1002

*So, I have two possibilities here: r can equal plus or minus 3.*1013

*I am going to have two sets of results here.*1023

*Let's let r equal 3; if r equals 3, then I am going to end up with 4 as my first term; I add 3 to that--I am going to get 7.*1027

*I am going to go ahead and (let's see) add 3 to that...*1039

*Actually, a correction: I was thinking of arithmetic series; this is a geometric series--I am going to multiply.*1051

*I need to multiply each term, so 4 times 3 is going to give me 12, times 3 is going to give me 36, times 3 will give me 108, times 3 is 324.*1057

*So, make sure, when you are working with geometric series, that you are multiplying, not adding.*1076

*So again, if r = 3, I am going to get geometric means of 12, 36, and 108.*1080

*If r equals -3, I am going to get 4 times -3 is -12; -12 times -3 is going to give me positive 36, times -3 is -108.*1087

*They are alternating signs; so there are two possible solutions for the geometric means: 12, 36, 108; or -12, 36, -108.*1100

*Write an equation for the n*^{th} term of the geometric sequence -2, 1/2, and -1/8.1112

*The formula for the general term is the first term, times the common ratio raised to the n - 1 power.*1123

*So, if I am looking for an equation for the n*^{th} term here, I am going to need the common ratio.1133

*To find the common ratio, I will just take a term and divide it by the previous term.*1140

*The common ratio...I could take 1/2, and I am going to divide that by -2.*1145

*Recall that I could just rewrite this, to make it a little clearer, as 1/2 divided by -2--just write it out.*1148

*And that is the same as multiplying 1/2 by the inverse of -2, and the inverse of -2 is -1/2.*1156

*So, that is -1/4; the common ratio is -1/4.*1171

*Now, I can go ahead and write my equation: a*_{n} = the first term, which is -2, times (-1/4)^{n - 1}.1178

*And I only have three terms here; but just writing it in a more general form...n - 1, because it is just asking me for the n*^{th} term.1194

*-2 times -1/4...a negative times a negative is going to be a positive, so that is just going to be 2/4, or 1/2.*1210

*Oh, actually, I cannot simplify that any further--correction.*1225

*I can't simplify that any further, because it is (-1/4)*^{n - 1}; we are actually done at this step.1228

*We are done right here with the general formula, because I don't have n.*1234

*All right, so the equation for the n*^{th} term is simply going to be a_{n} = -2(-1/4)^{n - 1}.1238

*So, I could find any term from this geometric sequence, using this equation.*1252

*All right, write the next three terms of the geometric sequence: -1/3, 1/2, -3/4.*1260

*In order to find a term, I need to have the common ratio.*1268

*So, let's find that common ratio by taking 1/2 and dividing it by the previous term, which is -1/3.*1272

*This is the same as 1/2 divided by -1/3; and remember, I can always rewrite that as 1/2 times the inverse, which is -3, or -3/2.*1279

*Therefore, r = -3/2; now that I have the common ratio, I can find the next three terms.*1290

*So, we stopped with the third term--I am looking for the fourth term, the fifth, and the sixth.*1297

*So, the fourth term is going to be equal to -3/4, times -3/2; this is just going to be 9/8.*1306

*The fifth term is going to be equal to 9/8, times that common ratio of -3/2.*1317

*-3/2 times 9/8 is going to give me -27/16: a*_{6} (the sixth term) is going to be -27/16 times -3/2.1326

*A negative and a negative is going to give me a positive, and 27 times 3 is actually 81.*1343

*16 times 2 is 32; and you could leave these as fractions, or you could rewrite them as mixed numbers.*1349

*I am just going to leave them as they are; so the next three terms are 9/8, -27/16, and 81/32.*1357

*And I could have looked here and just predicted that the common ratio is negative, because I have these alternating signs: negative, positive, negative.*1366

*That concludes this lesson on Educator.com, covering geometric sequences; thanks for visiting!*1376

*Welcome to Educator.com.*0000

*Today we are going to talk about geometric series.*0002

*In the previous lesson, I introduced the concept of geometric sequences; so this continues on with that knowledge.*0004

*So, what are geometric series? A ***geometric series** is the sum of the terms in a geometric sequence.0012

*Again, make sure that you have geometric sequences learned, that you understand that well, before going on to geometric series.*0017

*But just briefly, recall that a geometric sequence is a list of numbers.*0025

*And what is unique about this list is that you find one term by multiplying the previous term by a number r, which is the common ratio.*0030

*For example, here the common ratio is 2: so I multiply 8 times 2 to get 16, times 2 is 32, times 2 is 64.*0042

*You can always find that common ratio by taking a term and dividing it by the previous term.*0053

*This is a geometric sequence: today we are going to move on to talk about geometric series.*0062

*And a geometric series is the sum of the terms; so from this, we could get a geometric series, 8 + 16 + 32 + 64.*0072

*This is the geometric series: first term + second term + third term, and on and on, until we get to that last term.*0083

*Now, this is a finite series, but you could also have an infinite series, where it just continues on indefinitely.*0100

*So, what we want to find, often, is the sum of a particular number of terms in the series.*0112

*Now, I could look up here and say, "OK, I want to find the first three terms: that is 8 + 16 + 32."*0121

*And then, I could just add that up and figure out what it is.*0128

*But that is going to get really cumbersome to add manually; so we have a formula for the sum of a geometric series.*0132

*The sum of the first n terms of a geometric series is given by this formula.*0139

*And we have the limitation that r does not equal 1, because if r equaled 1, if I had, say, 3,*0143

*and I just multiplied it by the common ratio of 1, I would just get 3 again and again and again, and it wouldn't really change.*0151

*So, the limitation is that the common ratio cannot equal 1.*0158

*Looking at an example, 6 + 18 + 54 + 162, let's say I wanted to find the sum of this entire series--the sum of all the four terms.*0162

*I could use this formula: my first term here is 6--what is my common ratio?*0174

*Well, I can say 18/6 is 3; so the common ratio is 3.*0179

*Therefore, the sum of these four terms would be 6, times (1 - 3*^{n}) (n = 4 in this case),0186

*divided by (1 - 3); this is going to give me 6(1 - 3*^{4}).0197

*Well, recall that 3 times 3 is 9, times 3 is 27, times 3 is 81; so this is 81, divided by 1 - 3.*0208

*So, the sum is 6 times...1 - 81 would give me -80, divided by...1 - 3 is -2.*0219

*Let's go up here to continue on: 6 times -80 is going to give me -480, because 6 times 8 is 48; add a 0;*0231

*divided by -2; the negatives cancel out; 480/2 is 240.*0241

*So, I was able to find this sum by using a formula, rather than just adding each number.*0248

*And the formula requires that I know the common ratio, the first term, and n.*0255

*Sigma notation: as with arithmetic series, we can also use sigma notation as a concise way to express a geometric series.*0261

*Again, the Greek letter Σ means sum; and the variable that we are going to use is called the index.*0270

*So here, I have a lower index; and then the upper index tells me how high to go.*0278

*So, here I have n going from 1 to 5; and then I am going to have the formula for the general term*0286

*written right here, so I know how to find each term in this geometric series.*0292

*Looking at an example that is specific: as I said, they often use the letter i for the index in sigma notation, so I am going to use i.*0297

*i going from 1 to 6...and the formula to find each term is going to be 3 raised to the i power.*0308

*Therefore, I can find this series; it is going to be 3 raised to the first power, plus 3...*0316

*I started out with 1, and I inserted that here; next I am going to go to 2.*0328

*Then, I am going to go to 3, then 4, 5, and 6.*0333

*And then, you could, of course, figure this out; this is 3 + 9 + 27 + 81 + 729 + 2187.*0343

*So, this notation means this; and again, this is just a different way of writing a geometric series.*0355

*But the concepts that we have discussed remain the same.*0365

*All right, we learned one sum formula: and there is a second formula.*0368

*This second formula is very valuable when you know the first and last terms, and you know the common ratio, but you don't know the number of terms.*0374

*If n is not known, use this formula.*0382

*For example, maybe I have a series, and I know that the first term is 128; that the last term is 4; and that the common ratio is 1/2.*0388

*But I don't know n--n is not known--and I am asked to find the sum.*0400

*I can do that using this formula: the sum is going to be the first term, minus this last term, times r, divided by (1 - r).*0410

*which is going to be equal to 128 minus...4 times 1/2 is 2...divided by...1 minus 1/2 is just 1/2.*0422

*This is going to be equal to 126 divided by 1/2, and that is the same as 126 multiplied by the reciprocal of 1/2, which is 2.*0432

*And that is 252; so the sum is going to be 252, and I was able to find that because I knew the first term; I knew the last term; and I knew the common ratio.*0442

*And I had a formula that did not require me to know n.*0457

*All right, the formula for s*_{n} can be used to find a specific term in the series.0462

*And a very important term is the first term: so we often use s*_{n} to find the first term in a series.0468

*So, looking at one of the sum formulas that we just discussed, that would be the first term, times (1 - r*^{n}), divided by (1 - r).0477

*Let's say that you are given that the sum is 62.*0489

*And you are also given that the common ratio is 2, and that the number of terms is actually 5.*0496

*And you want to find the first term; you can do that with this formula.*0504

*I know that 62 equals the first term, times 1 - 2, raised to the n power (here, n is 5), times 1 - 2.*0509

*So, 62 = a*_{1}...2 to the fifth...2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32; so that is 32.0521

*1 - 32, divided by -1...therefore, this equals...rewriting this as 62 =...the first term...1 - 32 would give me -31,*0533

*divided by -1; if you just pull the negative out front, then those are going to end up canceling; -31a/-1 is just going to give you a positive.*0551

*So, I am going to rewrite this as 62 = 31 times the first term, because this is just a -1 down here, and I can rewrite this much more simply.*0570

*OK, dividing both sides by 31 gives me that the first term equals 2.*0585

*So, I was able to find the first term by being given the sum and the common ratio and the number of terms.*0590

*Frequently, you will use one of your sum formulas to find the first term in a geometric series.*0597

*All right, on to Example 1: Find the sum, s*_{n}, for the geometric series with a first term of 162,0603

*with a common ratio of 1/3, and with a number of terms equal to 6.*0613

*All right, the formula for the sum: a*_{1}(1 - r^{n})...and since I know n, I can use this formula,0619

*rather than the other one...divided by (1 - r); and I want to find s*_{n}.0632

*s*_{n} equals the first term, which is 162, times (1 - 1/3^{6}), all of that divided by (1 - 1/3).0639

*Therefore, the sum equals...162 times 1, minus 162 times 1/3 to the sixth power, divided by...1 - 1/3 is going to leave me with 2/3 in the denominator.*0655

*Now, this looks like it might be complicated to work with; but there is a shortcut.*0672

*If you think of...I have my 162; I can think of 162 as 2 times 81, and 81 is a multiple of 3, so you might be able to see where I am going with this.*0680

*2 times 81...as I said, that is a multiple of 3; 3 times 3 is 9, times 3 is 27, times 3 is 81.*0698

*Therefore, I could rewrite this as 162 - 2...let's move this out of the way a bit...times 3 to the fourth power, times 1/3 to the sixth power.*0708

*I can use rules governing how I work with powers, and say, "OK, if this gives me 3*^{4}/3^{6}, 162 -2 times...0725

*if I look at this as 3*^{4}/3^{6}, they have the same base; if I take 6 - 4, this is going to give me 1/3^{2}."0742

*So, this is 2 times (1/3)*^{2}, divided by 2/3--it is much easier to work with now.0762

*This is simply going to be 162 minus 2 times 1/9, or 2/9; all divided by 2/3.*0772

*Coming up here to finish this out: the sum, therefore, equals 162 - 2/9; that is going to give me 161 and 7/9, all divided by 2/3.*0785

*Recall that, then, dividing by 2/3 is the same as multiplying by 3/2.*0799

*That is not a very pretty answer, but you can simplify this, calculate it out, use a calculator...but this does give the sum.*0810

*So again, I was given the first term, the common ratio, and the number of terms.*0818

*I can use this formula, since I have the number of terms.*0822

*This looked like it was going to be very messy to work with: 162 times (1/3)*^{6}.0825

*But by recognizing that I could break 162 into 2 times a multiple of 3, I was able to get this into a base with 3, 3*^{4}.0830

*Dividing 3*^{4} by 3^{6} canceled out, and I got (1/3)^{2};0842

*and that simplified things a lot to give me this answer that I have in the upper left.*0848

*Find the sum of the first 8 terms of the geometric series.*0855

*So, here we are asked to find the sum of the first 8 terms of this series, and that would be s*_{8}.0859

*Thinking about which formula I am going to use: I know n; since I am looking for the first 8 terms, I know n.*0867

*I can also easily find r, because I can just take 1 divided by 1/4; 1 divided by 1/4 is the same as 1 times 4, or 4.*0877

*So, I have r; I have n; the other thing is the first term, and I have that.*0888

*With these three pieces of information, I know that I can use the formula*0895

*that the sum equals the first term, times (1 - r*^{n}), divided by (1 - r).0898

*So, the first 8 terms...that is going to be 1/16, times 1 minus...r is 4...raised to the n power, where n is 8, divided by (1 - 8).*0904

*OK, let's look at what I have: I have 1/16 times 1, minus 1/16 times 4*^{8}.0932

*Actually, this is not 1 - n; it is 1 - r; so let's go ahead and change that to a 4.*0955

*All right, and then I have 1/16 times 4*^{8}; as with the previous problem,0960

*you might recognize that there is something to make this a lot easier than taking 4 to the eighth power and dividing it by 16.*0965

*You are going to recognize that you can rewrite this 1/16 as (1/4)*^{2}.0971

*So, 1/4 times 1/4 would give me 1/6; that times 4*^{8} allows me to do some canceling to make things simpler.0985

*1 - 4 just gives me -3; let's go up here and work this part out.*0998

*This is (1/4)*^{2} times 4^{8}; this is the same as taking 4^{8} and dividing it by 4^{2}.1003

*Using my rules governing exponents, 4*^{8}/4^{2}...if I want to divide, and I have like bases,1015

*I just subtract the exponents; so this is going to give me 4 raised to the sixth power.*1028

*So, knowing your rules of exponents is important to solve when you are working with this type of problem.*1032

*Therefore, I am going to get 1/16 - (1/4)*^{2} times 4^{8}1038

*(that is the same as 4*^{6}, so I am going to go ahead and rewrite it that way) divided by -3.1044

*At this point, you are going to have to do a lot of multiplying, or else use your calculator to determine that 4*^{6} is actually 4096.1052

*And we are dividing by -3; so the sum is 1/16 - 4096; that comes out to -4095 and 15/16, all divided by -3.*1059

*And a negative and a negative will give me a positive, so that would give me 4095 and 15/16, divided by 3.*1077

*You can work this out with your calculator and see that this sum is approximately equal to 1365.3.*1086

*This does give us our answer; but we can estimate to put it in a neater decimal form.*1094

*Finding the sum: this time, we are given the first term; we are given the last term; and we are given the common ratio.*1104

*But we don't know the number of terms--we don't know what n is.*1113

*But that is OK, because we have that other formula that allows us to find the sum of a geometric series when we know the first term;*1117

*we know the last term; and we know the common ratio (that second formula that we worked with).*1125

*All right, looking for the sum: I have the first term; this is 4 - a*_{n} (which is 8748), times the common ratio of 3, all of that divided by 1 - 3.1140

*That is going to give me that the sum equals 4 minus...multiply this out, or use your calculator, to get 26244, divided by 1 - 3 (gives me -2).*1159

*Coming up here to the second column: the sum is going to be equal to 4 - 26244, or -26240, divided by -2.*1174

*A negative divided by a negative gives me a positive; and 26240 is even--I divide it by 2, and I get a nice whole number, 13120.*1192

*So, the sum of this geometric series is 13120; and I found that using the formula*1201

*that only requires that I know the first term and the last term, and the common ratio.*1207

*I didn't have to know the number of terms in the series, and I could still find the sum of the terms.*1211

*Example 4: We are asked to find the first term.*1217

*As I mentioned, it is important to have the first term frequently when you are working with these series.*1219

*And you can use the sum formulas to find that first term.*1224

*I am going to look at the information I have: I have the sum; I have the common ratio; and I have the number; but I don't have the last term.*1227

*So, I use the formula that involves the number of terms, not the last term.*1233

*And what I am looking for is the first term; therefore, the sum is 1020, equals the first term,*1244

*times 1 - r (is 2), and it is 2 raised to the eighth power, divided by 1 - 2.*1254

*And it is 2 raised to the eighth power, divided by (1 - 2); this is going to give me 1020 equals the first term, times 1 minus...*1258

*if you go through your powers of 2, you will find that 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32.*1270

*So, 2*^{5} - 32; then, we are going to get 2^{6} is 64; 2^{7} is 128; and then, 2^{8} is 256.1283

*That is 1 - 256, divided by 1 - 2 (is -1); therefore, 1020 = a*_{1}...1 - 256 is -255; divided by -1.1297

*Therefore, 1020 = -a*_{1} times 255, divided by -1.1316

*Well, a negative and a negative gives me a positive, so that is 1020 = a*_{1} times 255.1325

*I just take 1020 and divide by 255; so I have divided both sides by 255.*1335

*And I am going to get that the first term equals 4.*1340

*I was asked to find the first term, and I found that using the sum formula requiring the first term, the common ratio, and the number of terms.*1345

*And I determined that the first term in this geometric series is 4.*1353

*That finishes up this lesson on geometric series on Educator.com; thank you for visiting!*1358

*Welcome to Educator.com.*0000

*Today, we continue on our discussion of sequences and series with infinite geometric series.*0002

*What are ***infinite geometric series**? Well, this is a type of series in which there is an infinite number of terms.0010

*Earlier on, I mentioned that, for either arithmetic or geometric series, you may have a limited number of terms,*0018

*which is a finite series; or it may go on indefinitely, which is an infinite series.*0024

*For example, the geometric series 1 + 1/4 + 1/16 + 1/64 is a geometric series with 4 terms; and it has a common ratio r = 1/4.*0030

*This is a finite series: it has a limited number of terms.*0047

*Consider another geometric series: 3 + 6 + 12 + 24, and then, when you see the ellipses (the three dots),*0052

*it indicates that it goes on indefinitely; so this is an infinite series.*0063

*The sums, s*_{n}, are called partial sums of the infinite series.0070

*For example, I may decide that I want to find the first 7 terms of this series, or the first 5 terms of an infinite geometric series.*0074

*That would just be a partial sum of this series.*0084

*And we will talk, in a minute, about special cases, when you actually can find the sum overall of an infinite geometric series.*0087

*So, recall from the previous lesson the formula for the sum of a geometric series,*0097

*s*_{n} = the first term, times (1 - r^{n}), divided by (1 - r).0103

*So, if I were to try to find the first seven terms of this series, s*_{7}, I could use this formula.0114

*I have the first term; I need to find r; recall that I can find r by taking a term and dividing it by the one that goes just before.*0120

*So here, r = 2; so this is 3(1 - 2*^{n}), so that is 7; and then, divided by 1 - 2.0130

*So here, I found the partial sum for this infinite geometric series.*0149

*You can actually find the sum of an infinite geometric series (not just the partial sum) in some cases.*0160

*And I know that this sounds counterintuitive--how can you find the sum of something that goes on forever?*0166

*But if you look, you can see why.*0172

*All right, first of all, it is very important to know that this is limited to cases in which the series is convergent.*0178

*And the word "convergent" indicates that the sum converges toward a particular number.*0184

*A series is convergent if and only if the absolute value of r is less than 1.*0191

*So, if you are working with a geometric series in which the value of the common ratio is either greater than -1 or less than 1,*0196

*such as 3/4, for example--if r is 3/4 (the absolute value of that is 3/4), or if r is -1/2 (I take the absolute value of that--it would be 1/2),*0205

*both of these are convergent; I could actually find the sum of those.*0219

*Consider the series 1/3 + 1/9 + 1/27, going on indefinitely.*0222

*Remember, to find r (let's go up here and find r), we are going to take 1/9, divided by 1/3.*0234

*This is the same as 1/9 times 3, which equals 3/9, which equals 1/3; therefore, r = 1/3.*0241

*If I am asked to find the sum of this, I go ahead and use this formula, x equals the first term, which is 1/3, divided by 1 - 1/3, equals (1/3)/(2/3).*0258

*1/3 divided by 2/3 is the same as 1/3 times 3/2, or 3/6.*0273

*The sum of this infinite geometric series is 1/2.*0283

*Let's look at this another way: just go ahead and add up some of the terms and see what happens.*0288

*s*_{1} for this term is 1/3; that is all you have: 1/3.0297

*So, s*_{2} would be adding 1/3 + 1/9; 1/3 + 1/9 equals...giving this a common denominator,0304

*I multiply both the numerator and the denominator by 3 to get 3/9 + 1/9 is 4/9; that is s*_{2}.0313

*s*_{3} = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.0322

*So again, I need to get a common denominator; and if you work that out, you will find that s*_{3} is 13/27.0333

*s*_{4}...I would continue on: 1/3 + 1/9 + 1/27 + 1/81...and if you figure that out, it becomes 40/81.0343

*I won't work out the rest of these right here; but I will just tell you that s*_{5} is 121/243.0360

*s*_{6} is 364/729; and s_{7} is 1093/2187.0370

*Let's look at the pattern here: it started out as 1/3; then it became 4/9, 13/27, 40/81, 121/243, 364/729, 1093/2187.*0383

*What is happening is: this sum is converging upon 1/2.*0400

*Another way that we say this is that the limit is 1/2; and that is terminology you will hear later on in higher math courses.*0414

*But for right now, just be aware that you can only find the sum of an infinite geometric series if it is convergent,*0422

*meaning that, as you take more and more terms and add them to the series, add them, and get their sum,*0428

*you will see that the sum of the series is converging upon a particular number.*0435

*And so, we just use this formula as a great shortcut to find the sum.*0440

*Sigma notation is something we discussed earlier on.*0452

*Go back and look at the lectures on geometric sequences and geometric series, if any of these concepts are new.*0456

*But sigma notation--you will recall that the Greek letter sigma means sum, and we used it with other geometric series.*0462

*For an infinite series, you are going to have something like this.*0470

*We have our lower index, i = 1; and this is the series as i goes from 1 to infinity.*0476

*The difference here is that, instead of stopping at a specific value up here, it goes on through infinity.*0484

*And again, we have the formula for the sequence right here.*0490

*Another example, or a specific example, would be a series, again, where i goes from 1 to infinity,*0494

*but we have a formula over here, 1/4 times 1/2, raised to the n - 1 power.*0501

*So, what you could do, then, is put 1 in here and find your first term, a*_{1}.0507

*Then, put 2 here; find your second term; and go on infinitely, because of the type of series that this is.*0513

*We can use the concepts that we just learned to actually write a repeating decimal as a fraction.*0523

*The sum formula I just described can be used in this way.*0529

*A repeating decimal would be something like this: .44444...and it just goes on and on that way.*0533

*What we do is rewrite this as a geometric series.*0542

*1: First step--how do you do that?*0547

*Well, look at what this really means: it really means 0.4 + 0.04 + 0.004 + 0.0004, and so on.*0556

*So, I rewrote this, but just as a series; and it is an infinite geometric series.*0573

*Then, find the sum of the series: recall that you can only find the sum of an infinite geometric series*0580

*if the absolute value of r is less than 1; this is required, or you can't find the sum.*0590

*Well, let's show here that we are OK, because the absolute value of r is actually less than 1.*0604

*So, in order to find the common ratio, r, I am going to take .04, and I am going to divide it by the term before it, which is .4.*0609

*So, I move the decimal over one place; that is going to give me .4/4, which is .1; so the common ratio, r, equals .1.*0618

*.1 is less than 1, so I am fine: I can use the sum formula, the first term divided by (1 - r).*0626

*So, the sum equals the first term, divided by (1 - .1), equals .4/.9.*0635

*Dividing that, move the decimal over; you can get rid of that decimal; that gives me 4/9.*0645

*So, I found that the sum of this...*0652

*I started out; I have this decimal that is a repeating decimal (it goes on forever); but I saw that I could rewrite this as a series.*0657

*So, this, therefore, is equivalent to the sum of the series.*0665

*My next step was to figure out what the sum of this series is, and it is actually 4/9.*0670

*Therefore, .444 repeating can be rewritten as 4/9; those are equivalent.*0674

*Recall that you can have a repeating decimal that doesn't just have one number repeat and repeat; it could be multiple numbers.*0683

*It could be .383838 repeating, and you could do the same thing.*0690

*You could rewrite this as .38 + .0038 + .000038, and so on; and then repeat as above, by finding the sum of the series.*0698

*So, this could also be used for repeating decimals where there is a longer repeat.*0717

*Also recall that we can write these as .4 with a bar over it, or .38 with a bar over it--that is just a different notation.*0722

*All right, let's find the sum of this infinite geometric series.*0736

*First, though, I am going to verify that I actually can find the sum of this by figuring out what r is.*0741

*So, I am going to take 24, the common ratio, divided by 32; one term divided by the previous one gives me the common ratio.*0746

*This simplifies out to 3/4: 3/4 is less than 1, so yes, I can find the sum, using this formula.*0753

*The formula: I need to use the first term: 32/(1 - 3/4) = 32/(1/4).*0769

*We can rewrite this as 32 times 4; and 32 times 4...4 times 2 is 8; 4 times 3 is 120; so that gives me 128.*0778

*The sum of this infinite geometric series is 128, and I was able to find that using this formula, because I had an absolute value of r that is less than 1.*0791

*Write as a fraction: recall that this notation means the same thing as .36363636, and so on.*0806

*The first step is to write this as a geometric series, so I am going to rewrite this as .36 + .0036 + .000036, and so on.*0821

*And I am going to use my formula for the sum of an infinite geometric series, which is the first term, divided by (1 - r).*0844

*What is r? Well, as usual, I can find the common ratio, r, by taking a term, .0036, and dividing by the previous term, .36.*0853

*Move the decimal over two places to get .36/36; and that is going to give me .01.*0864

*.01 is less than 1, so I can find the sum.*0871

*Take the first term; divide it by (1 - .01); this gives me .36/.99.*0875

*I can move the decimal over two places, to give me 36/99; simplify that, because there is a common factor of 3; this is actually 12/33.*0885

*I can see, again, that I have another common factor of 3; so this is going to give me 4/11.*0896

*Therefore, this 0.36 repeating decimal can be rewritten as 4/11; so I wrote this repeating decimal as a fraction.*0901

*Find the sum for this infinite geometric series: before I proceed, I check that the absolute value of the common ratio is actually less than 1.*0918

*10/12...to find the common ratio, divide that by the term that came just before, which is 5/4.*0927

*And this is going to give me 10/12 times 4/5 equals 40/60; so this is going to give me 2/3.*0933

*And since 2/3 is less than 1, I can find the sum of this series using my formula for the sum of an infinite geometric series.*0949

*OK, the first term is 5/4; I am dividing that by 1 - 2/3 to get 5/4 divided by 1/3.*0961

*This is the same as 5/4; and then I just take the inverse of 1/3, and 5/4 times the inverse of 1/3; that is 3, so this is going to give me 15/4.*0977

*And you can keep this as an improper fraction, or write it as a mixed number.*0989

*The sum of this infinite geometric series is 15/4.*0993

*Again, I was able to find that because the common ratio had an absolute value that was less than 1.*0999

*Write as a fraction: we have another decimal that is repeating--goes on infinitely: 0.99999, and so on.*1009

*Start out by rewriting this as a geometric series.*1021

*I have 0.9 + 0.09 + 0.009, and so on; it goes on infinitely.*1027

*Find the common ratio, r: r =...I am going to take .09, and divide that by .9.*1041

*Move the decimal over one place to get .9/9; therefore, the common ratio is .1.*1048

*Next, I use my formula for the sum of an infinite geometric series, which is the first term, divided by 1 - r.*1055

*And this is going to give me the first term, .9, divided by 1 - .1; that is .9/.9; this is simply 1.*1063

*So, this asked me to write it as a fraction; and .999 = 1, or you could say 1/1.*1081

*This is still a fraction, because you could just write it as 1/1.*1090

*And it is counterintuitive, if you think ".999 repeating is actually 1"; it doesn't seem like it is, but it is actually correct.*1093

*That concludes this lesson of Educator.com on infinite geometric series; thanks for visiting!*1105

*Welcome to Educator.com.*0000

*Today, we are going to be covering recursion and special sequences.*0002

*The Fibonacci sequence is a particular sequence that states that one term, a*_{n}, is equal to the previous two terms,0006

*a*_{n - 1} + a_{n - 2}--the term that came just before it and the one before that.0016

*Fibonacci is an Italian mathematician who lived in the 13*^{th} century.0023

*And he actually brought the knowledge of this type of sequence to Europe; and that is why the sequence is called the Fibonacci sequence.*0029

*This type of formula that you see right here is called a recursive formula.*0037

**Recursive formula** means that the value of a term depends on previous terms.0044

*In order to use this type of formula, you would have to start out with a couple of values.*0052

*Looking at the Fibonacci sequence, you need to start out with your first term, 1, and your second term,*0058

*which actually is also 1, because if you don't have those two terms, you can't find the next term.*0065

*So, a*_{3} is going to be equal to the term that came just before it, a_{n - 1} (and 3 - 10070

*is going to give me a*_{2}), and then the term just before that, which (in this case) is actually the first term.0086

*a*_{3} = 1 + 1, or 2; a_{4}...now I am going to add the previous two terms here, 2 and 1, to get 3.0092

*The previous two terms, 3 and 2, get me 5.*0102

*I can use this to form a sequence; so I take these terms, and I write them as a sequence:*0106

*1, 1, 2, 3, 5, and you could go on very easily: 5 + 3 is 8; and so on.*0111

*This gives you one recursive formula, which will provide a special sequence called the Fibonacci sequence.*0128

*But just to give you an example of a recursive formula, more generally, remember that a recursive formula*0135

*is one in which the value of a term depends upon previous terms.*0143

*I might say, "OK, a*_{n} = 4 times the previous term, plus 2 times the term before that."0148

*So, if I wanted to find a particular term in this series, and I was told the first term is equal to 1 and the second term is equal to 2,*0159

*and I was asked to find the third term--"What is a*_{3}?"--well, a_{3} = 4 times the term0169

*that came just before, which is 2, plus 2 times the term before that, which is equal to 1.*0178

*So here, a*_{3} = 8 + 2, which is actually going to be 10.0188

*So, in this case, I was given a formula, a*_{n} equals 4 times the previous term, 4 times 2,0203

*plus 2 times the term before that, giving me that the third term is going to be equal to 10.*0212

*Iteration is the process of composing a function with itself again and again; what does that mean?*0229

*Iteration is something that can be used to generate a sequence recursively.*0238

*So, if we are composing a function with itself again and again, what the next iteration is depends on the previous one.*0242

*And that goes back to what we talked about with recursion.*0249

*Again, you need a starting point, though; so here, we need to start out with a function and a first value that we are going to use with the function.*0252

*And that is x*_{0}, or "x naught."0264

*The easiest way to understand this is through an example.*0268

*f(x) = 3x - 1; x*_{0} = 1; so I have been given my function, and I am given this x_{0}.0272

*So, I could be asked to find the first iterate; I am going to find f(x*_{0}), which here equals f(1).0283

*So, as usual, to find the value of a function when you are given what you want x to be, you just substitute 1 for x.*0299

*3 times 1, minus 1, is 3 minus 1, which equals 2.*0309

*So, I found the first iterate; to find the next iterate, I am actually going to take f(f(x*_{0})), which is f(f(1).0318

*And we determined that f(1) is 2, so I just find f(2).*0335

*This gives me 3 times 2, minus 1 equals 6, minus 1, which equals 5.*0340

*If I want to find another iterate, then I am going to take f(f(f(x*_{0}))).0348

*Here, I know that this in here is 5; so I am taking f(5); this is 3(5) - 1 = 15 - 1 = 14.*0359

*And you can continue on like that: if I wanted to find the next term, it would just be f(f(f(f(x*_{0})))).0379

*And you go on that way; so again, this is the process of composing a function with itself again and again and again.*0388

*I could use these numbers to form a sequence; so I have generated a sequence recursively.*0396

*My first term, a*_{1}, is going to be right up here; it equals 1.0403

*My second term is going to be 2; my third term is 5; and then, my fourth term is going to be 14.*0410

*And I could write this out as a sequence: 1, 2, 5, 14.*0421

*In the first example, I am asked to find the first 5 terms; and I am given that the first term is 3.*0429

*And when I look at this, this is a recursive formula, because the term that I am looking for depends on the term before it.*0435

*So, if I am looking for a*_{n + 1}, it is going to be equal to 2 times the previous term, minus 3.0448

*And I need to find the first 5 terms.*0456

*Well, the first term is given (the first term is 3); and I need to find a*_{2}.0458

*a*_{2} is going to equal 2, times the previous term; so, given whatever n + 1 is, I am going to take away 1 from that and get the previous term.0464

*So, that is going to give me 2 times 3 minus 3, equals 6 - 3, which equals 3.*0481

*I found the second term; I need to find the third term.*0489

*So, I am going to take 2 times the previous term (the value of that is 3), minus 3.*0493

*You are starting to see a pattern here: again, we have 3.*0499

*Well, I know what is going to happen if I put 3 back into this formula, but I will go ahead and do it anyway.*0503

*2 times 3, minus 6, equals 6 - 3, is 3; I have one more term to find...again, 2 times 3, minus 3, equals 6 - 3, equals 3.*0508

*So, this is actually a sequence where this formula generates the same value over and over and over.*0521

*So, I am just going to end up with a sequence that looks like this; and that is actually going to go on forever.*0526

*This is kind of an interesting formula where we ended up with the same value over and over and over in this sequence.*0532

*Example 2: this time we are going to work with iterates--we are asked to find the first three iterates of this function.*0543

*And the function is f(x) = 4x - 7, and x*_{0} = 3.0549

*To find the first iterate, I am going to go ahead and find f(x*_{0}, which equals 4.0555

*And then, I am going to insert 3 here.*0567

*Actually, just to clarify: this is equal to f(3), because x*_{0} is 3.0569

*So, we are going to write this as 4(3) - 7 = 12 - 7; that equals 5.*0575

*That is the first iterate; the second iterate is going to be f(f(x*_{0})), which, in this case, is going to be f(5).0582

*That equals 4(5) - 7; that is going to give me 20 - 7, which is 13.*0592

*Next, f(f(f(x*_{0}))): I know that this, in here, is 13, so I just take f(13) = 4(13) - 7.0600

*4 times 13 is 52, minus 7 gives me 45; so I found the first three iterates, and those are 5, 13, and 45.*0615

*Example 3: Find the first five terms--they give me the first term; they give me the second term;*0639

*and they give me a formula that is a recursive formula, because it depends on the previous two terms.*0645

*So, a*_{n} = 2(a_{n - 1}) + 3(a_{n - 2}).0653

*I need to find the first five terms: the first term is given; the second term is given; I just need to find a*_{3}, a_{4}, and a_{5}.0660

*Let's start with a*_{3}: a_{3} is going to equal 2 times the term that came just before,0672

*or a*_{2}, plus 3 times the term before that, which is a_{1}.0682

*This is going to be equal to 2 times 3, plus 3 times 2, equals 6 + 6; that is going to give me 12.*0688

*So, a*_{3} is 12; a_{4} is going to be equal to 2 times a_{3}, plus 3 times a_{2},0699

*which equals 2 times...a*_{3} is 12, plus 3 times a_{2}, which is 3.0708

*Of course, 2 times 12 is 24, plus 3 times 3 (is 9)...24 + 9 is 33: a*_{4} is 33.0717

*a*_{5} is equal to 2 times a_{4}, plus 3 times a_{3}; this is going to be equal to 2 times 33, plus 3 times 12.0730

*That is going to give me 66, plus 3 times 12 (is 36), and that adds up to 102.*0742

*So, I found the first 5 terms; I was actually given these 2; and then, I used a recursive formula to find the third, fourth, and fifth terms.*0751

*Find the first three iterates: f(x) = x*^{2} + 3x - 4; and then, x_{0}, or "x naught," is equal to 1.0763

*The first iterate will be f(x*_{0}), which equals f(1); that gives me 1^{2} + 3(1) - 4, equals 1 - 3 - 4.0772

*So, I have, let me see...actually, that is plus right here...so that is 1*^{2} + 3 - 4, so 1 + 3 - 4; that is 4 - 4; the first iterate is 0.0790

*The next iterate: f(f(x*_{0})): this is going to be equal to...since f(x_{0}) is 0, it is just f(0).0805

*0*^{2} + 3(0) - 4 is 0 + 0 - 4; so that is going to give me -4.0817

*Next, I want f(f(f(x*_{0}))); this is going to be equal to f(-4) = (-4)^{2} + 3(-4) - 4.0829

*This equals 16 - 12 - 4; well, -12 and -4 are -16, so this is 16 - 16 = 0; so the first three iterates are 0, -4, and 0.*0847

*That concludes this lesson of Educator.com; thanks for visiting, and see you soon!*0869

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be covering the binomial theorem.*0002

*We are going to start out by talking about what we mean when we say we are going to expand a binomial.*0007

*So, if you expand a binomial, this is what happens.*0013

*I have (a + b)*^{n}; and I am going to expand that for n = 1, 2, 3, 4.0019

*I am going to let n equal 1; and I am going to then get (a + b)*^{1}, which is just a + b.0027

*If I let n equal 2, then I have(a + b)*^{2}; that is going to give me, if you will recall, a^{2} + 2ab + b^{2}.0041

*Let n equal 3: you get (a + b)*^{3}, which is a little bit more complicated to work out.0057

*But if you figure that all out, (a + b)(a + b)(a + b), you would end up with a*^{3} + 3a^{2}b + 3ab^{2} + b^{3}.0064

*Letting n equal 4, (a + b)*^{4} is going to give you a^{4} + 4a^{3}b +0079

*6a*^{2}b^{2} + 4ab^{3} + b^{4}.0091

*One more: let n equal 5: this is going to give me (a + b)*^{5}.0099

*This equals a*^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b...0108

*actually, this is going to be b*^{3}...+ 5ab^{4} + b^{5}.0127

*All right, before we go on to talk about Pascal's triangle, let's note a few things about this expansion.*0136

*And these will be summed up on the next slide.*0144

*A few things that you will notice are that the first term is a*^{n}, and the last term is b^{n}.0146

*The second thing is that the number of terms equals n + 1.*0157

*You will also notice a pattern: let's look at n = 4, (a + b)*^{4}.0179

*The first term is a*^{n}, so it is a^{4}; in each subsequent term, the power that a is raised to (the exponent) decreases by 1.0185

*So, I started out with a*^{4}; then here it's a^{3}; a^{2}; a; and then a is gone.0195

*So, the exponent for a decreases by 1 for each term; b does the opposite.*0203

*There is no b here; this would be b*^{0}...it is 1; and then in my next term, I get b to the first power, b^{2}, b^{3}, b^{4}.0213

*So, with each term, exponents for a are decreasing; exponents for b are increasing.*0223

*The next thing that you might notice is the coefficients; let's look at the coefficients.*0230

*The coefficient of the second term is equal to n; again, this is all summed up on the next slide, but this is just to introduce it here.*0234

*So, if I am looking at n = 3 (that is (a + b)*^{3}), and I expand that; then I get that the second term has a coefficient of 3.0242

*The second term here has a coefficient of 4; the second term here has a coefficient of 5.*0252

*Pascal's triangle--we are getting to that now.*0258

*I can take the coefficient of these terms and use them to create an array that is called Pascal's triangle.*0260

*Looking at it, it actually starts out with 1; so if we let n equal 0, that is just going to give me (a + b)*^{0}, which is going to be 1.0266

*Then, in my next row, I am going to have 1, 1, and 1; those are my coefficients.*0277

*In the next row, they are 1, 2, 1; in the next row, 1, 3, 3, 1.*0289

*Now, I can look up here to get the next row; but I don't even need to, because each number is the sum of the two numbers just above it.*0302

*So, 1 + 1 forms 2; 1 + 2 gives me 3; 2 + 1 gives me 3.*0314

*So, all I have to do, actually--these outside numbers are always 1; but I could just say 1 + 3...this is going to be 4.*0325

*3 + 3 is equal to 6; 3 + 1...that is equal to 4.*0334

*So, I look up here to verify that; and indeed, it is 1, 4, 6, 4 1.*0340

*In the next row, I would add 1, and I add 1 + 4; I am going to get 5; 4 + 6 gives me 10;*0345

*6 + 4 gives me 10; 4 + 1 gives me 5; and again, I have a 1 out here.*0352

*And that matches up with what I have for the expansion of (a + b)*^{5}.0357

*So, Pascal's triangle is an interesting array that comes from the coefficients of the expansion of this binomial.*0363

*Something else to note is the symmetry around the middle term; you might have seen that up here, but it is even more obvious right here.*0371

*The middle term is 2; it is flanked by 1's; here I have two middle terms, because there is an even number of terms;*0377

*so the two middle terms are 3; they are flanked by 1's; now I have a middle term of 6; next to it are 4's; outside of that, 1's.*0384

*Two middle terms are 10; the next terms are 5; the next terms are 1.*0392

*And that is helpful, because if you are looking for the coefficients, you don't need to find all of them; you only need to find half of them.*0397

*And then, you can use this symmetry to know what the other half are.*0403

*Summing up these properties of the expansion of (a + b)*^{n}: there are n + 1 terms.0409

*Let's use an example that we just talked about, (a + b)*^{5}, which gave me a^{5} +0418

*5a*^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} +0427

*5ab*^{4} + b^{4}, to illustrate this.0437

*There are n + 1 terms; so here, n = 5; therefore, I am going to have n + 1 terms, equals 5 + 1, so I am going to have 6 terms.*0442

*And I do have 1, 2, 3, 4, 5, 6 terms.*0455

*The first term is going to be a*^{n} (that is a^{5}), and the last term is b^{n}, which should be b^{5} right here.0458

*In successive terms, the exponent of a decreases by 1, while that of b increases by 1.*0470

*I start out with a*^{5}, then a^{4}, to the third, to the second, and then a.0475

*b goes the opposite way: I start with b, b*^{2}, b^{3}, b^{4}, and b^{5}.0482

*I pointed out before that the second term has a coefficient that is equal to n.*0490

*Something I also pointed out in the last slide is that there is symmetry around the middle term or terms,*0498

*which tells me that, if I figure out these middle coefficients, well, I already know that the first and last coefficients are 1.*0504

*And I know that the second (and, by symmetry, second-to-last) coefficients are equal to n.*0511

*So then, all I would have to figure out is one of these; and I know that the other one is the same.*0519

*Something else to be aware of is that the sum of the exponents in each term is n.*0523

*Let's look right here: for this term, 10a*^{3}b^{2}, if I add up 3 and 2, that is going to give me 5.0528

*So, the sum of the exponents is equal to 5.*0537

*Or looking at this, 10a*^{2}b^{3}, I get the same thing.0540

*The sum of the exponents of each term is equal to n.*0544

*The next concept that we will learn requires that you remember how to work with factorials.*0552

*So, I am going to go ahead and review factorials: factorials are products, and the symbol that looks like an exclamation point is actually read as "factorial."*0557

*This is a product: n! is equal to n times n - 1 times n - 2 times n -3...and you continue on that way until you get all the way to 2, and then finally 1.*0567

*For example, 4 factorial: here, n = 4; this is going to be equal to 4, so here n = 4; n - 1...4 - 1 is 3, so that is 4 times 3.*0586

*4 - 2 gives me 2; 4 - 3 gives me 1; and this equals...4 times 3 is 12, times 2, times 1, is just 24.*0599

*So, 4! is just the product of 4 times 3 times 2 times 1.*0613

*If you are working with fractions that involve factorials in the numerator and denominator, you can often do a lot of canceling out to make things easier.*0618

*If you were working with 8!/6!,that gives you 8 times 7 times 6 times 5 times 4, and then on down to 1.*0625

*In the denominator, you have 6 times 5 times 4 times 3 times 2 times 1.*0637

*So, instead of multiplying all of this out and dividing, it makes a lot more sense to start canceling out all of these common terms.*0644

*1 through 6; 1 through 6; this just leaves me with 8 times 7, or 56.*0657

*Now, the reason we reviewed it is to work with this formula.*0665

*First, we are talking about the binomial theorem, and then the binomial formula.*0668

*The binomial theorem is really what we just discussed; and it is the idea that, when you expand (a + b)*^{n},0672

*you are going to get an equation with the properties we just discussed.*0682

*And those properties are that the first term is going to be a*^{n};0688

*the second term is going to have some coefficient; and it is going to be raised to the power a*^{n - 1}.0692

*b is going to be to the first power; then you are going to get another term; and you are going to have a total of n terms.*0699

*You are going to have another term; it is going to have a coefficient that is going to be a*^{n - 2}.0707

*b is going b*^{2}; it is going to increase by 1.0711

*Then, you are going to get another coefficient, a*^{n - 3}; b is going to become b^{3}.0714

*And that is going to go on and continue on until you get ab*^{n - 1}, b^{n}, and so on.0721

*Excuse me, plus...that is a*^{n - 1}, times b, plus b^{n}; the last term will be b^{n}; the first term is a^{n}.0731

*Each of these terms can be given in the general form a*^{n - k}b^{k}.0751

*And we will look at some examples in a few minutes, and talk about what this means, as we relate it to the binomial formula.*0760

*This is the binomial formula; this is pronounced "n choose k."*0769

*Now, why are we even looking at this--what is this for?*0781

*Think about how we can find these coefficients.*0785

*We can use Pascal's triangle, but that could be really impractical and quite a bit of work,*0788

*if you have to go through that whole large array, if you have a really big expansion.*0793

*And if you just find the coefficient for a particular term, you are going to have to go through a lot of work to get there.*0799

*So, as usual, we have a formula that gets us directly there, without having to go through all that work.*0804

*This formula will give me the coefficients for a term that has the exponent a*^{n - k}b^{k}.0812

*So, the binomial formula allows me to find this coefficient.*0823

*Let's use an example of (a + b)*^{7}: I can find the coefficient, using this, of a particular term;0829

*or I can expand the whole thing and find all of the terms.*0839

*Let's look at what this expansion would look like: it would look like this.*0841

*I know that the first term is going to be a*^{7}; recall that the second coefficient is going to be equal to n.0845

*So this one I also know; it is 7; and then, I know that a...the exponent is going to decrease by 1; and b is going to appear, raised to the first power.*0851

*I also know that I am going to have 8 terms, because I am going to have n + 1 terms, or 7 + 1.*0866

*So, I have another term here; I don't know the coefficient, but I do know that it is going to be a*^{5}b^{2}.0872

*I have another term with an unknown coefficient: a*^{4}b^{3}.0878

*I have another term, again with an unknown coefficient: a*^{3}b^{4}.0884

*Another term is a*^{2}b^{5}; another term is ab^{6}; plus b^{7}.0890

*I have 8 terms, so I am going to have these two middle terms the same; these two terms will be the same;*0900

*the second-to-last terms will be the same; so I know this is 7; and these two outer terms will be the same, 1 and 1.*0907

*I may be asked to find this expansion; I found this much; now, in order to finish it out, I need to find this coefficient and this coefficient.*0917

*If I have those, I have the rest, because I would have the first half, and then I just reflect for the second half.*0927

*Let's use this formula to find the coefficient for this third term.*0935

*In order to use this formula, I need to know two things: I need to know n, and I need to know k.*0950

*n is easy--we know n equals 7; k...if you look up here, k is the exponent of b for that term.*0956

*So, I am looking for the coefficient that goes right here, where n is 7 and k is 2.*0969

*Once I have that, it is a matter of using the formula: n!...this is going to be 7!/k!, which is 2!...times (7 - 2)!;*0978

*this equals 7! divided by 2!, times 5!.*0995

*I am going down here to complete this: 7 times 6 times 5 times 4 times 3, and so on.*1002

*In the denominator, I have 2!, 2 times 1, times 5 times 4 times 3 times 2 times 1 (factorials).*1010

*Again, I am going to cancel out common factors to make this much easier to work with.*1023

*This leaves me with 7 times 6, divided by 2; that is simply 42/2, or 21.*1030

*Therefore, the coefficient for the third term is actually 21.*1043

*Since I know that, and I know that we are going 1, 7, 21, starting from this end*1052

*(we are also going 1, 7, 21), I could do the same thing to find the fourth term coefficient.*1057

*The fourth term coefficient would have, again, an n equal to 7; but this time, k would be equal to 3.*1067

*Note one thing: instead of having to write all this out, to figure out what k is and everything,*1077

*you can just be aware that the term number equals k + 1.*1082

*Let's look at the fourth term: the term number is 4--that equals k + 1; k = 3.*1092

*For the fourth term, k = 3; so that is 1, 2, 3, 4; I know n = 7; and since this is the fourth term, I quickly know that k = 3.*1101

*So, you can use the binomial theorem, and specifically the binomial formula, to find a specific term of the expanded form of (a + b)*^{n}.1117

*And I talked, in the last section, about how you can use that formula to find the whole expansion.*1126

*Or you may just need to find a certain coefficient within the expansion.*1133

*Recall that the binomial formula is equal to n!, divided by k!, times (n - k)!, where n is this n,*1140

*and k is the value of the exponent for the b in the term that you are looking for the coefficient of.*1158

*As an example, we are going to do something slightly more complex.*1167

*Instead of just a + b, we are actually going to say a - 2b; and we are going to expand that--it is going to be to the fifth.*1169

*So, we have to account for this -2--that the second term here is not just b; it is -2b.*1178

*I know that the first term is going to be a*^{5}.1188

*I know that the second term is going to have a coefficient of n, so it is a coefficient of 5, and that a is going to decrease by 1.*1193

*I have to be very careful and not just write b; I have to write -2b.*1200

*Then, I have an unknown coefficient; but I do know that I am going to get a*^{3}, because the exponent that a is raised to will decrease by 1.1206

*And here, I just have to say that b is really -2b; that is going to be squared; it is going to increase by 1.*1216

*I have another unknown coefficient, and it becomes a*^{3} and goes down to a^{2}; -2b squared goes up to -2b cubed.1226

*Then, there is another coefficient here, and I get a*^{3}, -2b to the fourth, and then finally...1237

*actually, this is just a; it goes from a*^{2} to a; and then finally, a is gone, and I have -2b^{5}.1247

*By symmetry, I know the coefficient of these outer terms, right here, is actually 1; you can think of that as 1.*1256

*The two outer terms, then, have coefficients of 5.*1266

*All right, so back to finding a specific term: let's say that I want to find the coefficient for this third term.*1270

*n is going to equal 5; so in order to find the coefficient for this third term, I need to use the binomial formula; I need to have n, and I need to have k.*1298

*k...I could just say it is one less than the term number, so k is 2.*1307

*Or I could look up here and say it is whatever power this is raised to, which is 2.*1311

*Next, I insert these values into the formula; that is going to give me 5!, divided by 2!, times (n - 2)!.*1318

*This is equal to 5! divided by 2!, times 3!, equals 5 times 4 times 3 times 2 times 1, divided by 2! (2 times 1), times 3 times 2 times 1.*1334

*Cancel out common factors; that leaves me with 5 times 4, divided by 2 times 1, or 20 divided by 2, is 10.*1354

*Therefore, coefficient for the third term is equal to 10.*1368

*I can go back up and put this in here, but there is some simplifying that needs to be done.*1374

*So, the third term equals 10a*^{3}, (-2b)^{2}; this equals 10a^{3}...-2 times -2 is 4b^{2}.1379

*This equals 10 times 4, a*^{3}, b^{2}, which equals 40a^{3}b^{2}.1398

*It is a little bit more complicated than just putting the 10 in there.*1408

*And then, something to be careful of, as well: then I can say that by symmetry, this is also 10.*1411

*But I need to work with what is in here, which is actually a*^{2}(-2b)^{3}, so it is somewhat different.1416

*So, let's look at what this fourth term is: 1, 2, 3,...the fourth term.*1425

*Again, I am going to use that coefficient, 10, but here I have a*^{2}(-2b)^{3}.1431

*So, it is 10a*^{2}...-2 times -2 is 4, times -2 is -8; so it is times -8b^{3}.1439

*I can rewrite this as 10 times -8a*^{2}b^{3} = -80a^{2}b^{3}.1454

*When I simplify that fourth term, this is what it is going to look like.*1466

*That was using the binomial formula to find a specific term of the expanded form.*1471

*In the first example, we need to expand (x + 2)*^{4}.1481

*The general form is (a + b)*^{n}; here, I do the same thing, just realizing that a is equal to x, and b is equal to 2.1486

*I am going to get something that looks like this: (a + 2)*^{4}; I am going to have five terms.1497

*I am going to have n equal 4; so, there are going to be 5 terms.*1502

*The first term is going to be a*^{2}; in this case, it is x, so it is x^{2}.1508

*Excuse me, not squared; to the n--it is going to be a to the n, which is 4, so it is actually going to be x*^{4}.1517

*In the next term, I have the coefficient; and the coefficient of the second term is equal to n.*1525

*So, that is going to be 4, times x...this exponent is going to decrease by 1...times 2.*1531

*The next term: I am going to have some unknown coefficient; x is going to go from cubed to squared;*1543

*2 is going to go from being the first power to the second power.*1555

*I am going to have another term, and then x*^{2} is just going to become x; 2 is going to go from being squared to being cubed;1560

*and then, finally, x drops out, and I get, for my last term, 2*^{n}, or 2^{4}.1572

*By symmetry, I know that I am going to have a middle term (I have an odd number of terms--I have one middle term);*1581

*the two next to it are going to have the same coefficient--it is going to be 4.*1587

*I can simplify this later on, because I am going to have to deal with this 2*^{3} and this 2, and multiply that times the coefficient.1591

*But right now, let's just leave it like this.*1600

*I am almost done with my expansion; for this middle term, there are two ways I could go, because this is not that large of an expansion.*1603

*I could use the binomial formula, and we are going to get more practice with that in a minute.*1612

*But this time, I am actually going to use Pascal's triangle.*1615

*Remember that it starts out with (a + b)*^{0}, and you end up with 1 as the only coefficient, because this equals 1.1617

*Then, I get (a + b)*^{1} = (a + b); that gives me 1 and 1 as coefficients.1632

*Once I have those two, I can find the rest, because I know I have 1's out here.*1639

*And to find the other terms, I add the two above: 1 + 1 gives me 2.*1645

*So, for (a + b)*^{2}, these are my coefficients; then I want (a + b)^{3}.1651

*And what I am finally looking for, to find this coefficient, is when (a + b) is expanded to the fourth.*1659

*So, I have a 1 out here and a 1 out here; 1 + 2 gives me 3; 2 + 1 gives me 3;*1667

*so the coefficients for this expansion (that is going to have 4 terms) are going to be 1, 3, 3, 1.*1674

*Finally, the one I am looking for: I have a 1 and a 1, which I have up here;*1680

*1 + 3 is 4; 1 + 3 is 4, which I already knew; this gives me my middle coefficient of 6.*1685

*Now, I could have used the binomial formula to find this; but I wanted*1695

*to just do it a little bit differently this time, since it was a fairly small expansion, and use Pascal's triangle.*1699

*Now, I have the expansion, but it is not really simplified; we need to simplify these various terms.*1706

*So, let's take...the first term is already simplified; the second term equals (here is the second term) 4 times x*^{3} times 2; that equals 8x^{3}.1712

*The third term equals (that is 1, 2...the third term) 6x*^{2}, 2^{2}, equals 6x^{2} times 4, equals 24x^{2}.1733

*That is the third term; the fourth term equals 4x times 2 cubed; that is 4x times 8; 8 times 4 is 32, so that is 32x.*1750

*The fifth term equals 2*^{4}; 2^{3} is 8, times 2 again gives me 16.1764

*Rewriting it all here, the expansion of (x + 2)*^{4} is1774

*x*^{4} + (second term) 8x^{3} + (third term) 24x^{2} + (fourth term) 42x + (the final term) 16.1779

*Again, I was able to expand this using my properties that I am aware of for the expansion of a binomial*1794

*to find what these exponents are, how many terms there are going to be, and what this coefficient and this coefficient are.*1802

*I was left with just finding one coefficient that I could have used Pascal's triangle for.*1810

*That is what I did; I found the 6 (or I could have used the binomial formula).*1814

*Once I had these coefficients, all I did was to simplify; and this is the expansion, right down here.*1818

*Here, I am asked to find the fourth term of the expansion of (a + 2b)*^{8}.1829

*So, I am asked to find this specific term; and we know that we can use the binomial formula, n!/k! times (n - k)!.*1834

*I know that I have n; n = 4; what is k?*1847

*Well, the term number equals k + 1; the term number is 4, so this means 4 = k + 1.*1853

*This gives me k = 3; so I have n = 4; k = 3; I can find the coefficient.*1866

*I need to find the coefficient of this fourth term; it is going to have a coefficient.*1875

*For the first term, it is going to be a raised to some power; for the second component of it, it is going to be 2b (not just b), raised to some power.*1884

*How do I know what these powers are?*1902

*Well, recall that it is going to be equal to (let's give ourselves some more room)...*1904

*when you find a term used in a binomial formula, you also know that it is in the form n - k, *1910

*and then the second part is raised to the k power, because that is what k is--*1919

*it is whatever you are raising b to; and in this case, b = 2b.*1926

*I also know that the sum of the exponents has to equal n; the sum of the exponents of a term equals n.*1934

*This makes sense, because n - k + k needs to equal n; the sum of the exponents must equal n, and that works out; so this makes sense.*1945

*So, the fourth term: I need to find the coefficient using the binomial formula.*1954

*I know what the power a is going to be raised to is, n - k; and I know that the power that 2b is going to be raised to is k.*1960

*Let's go ahead and find the coefficient: the coefficient of the fourth term is going to be equal to n!, which is 4!...*1967

*excuse me, n is actually 8--a correction on that: we are finding the fourth term, but n is right here, so n is actually 8.*1981

*So, it is n! (that is 8!), divided by k! (that is 3!); n (which is 8), minus 3, factorial, equals 8!/3! times 5!.*1993

*That gives me 8! divided by 3 times 2 times 1, times 5!.*2013

*I do my canceling to make things easier to work with.*2026

*This leaves me with 8 times 7 times 6, divided by 3 times 2 times 1.*2032

*6 is equal to 3 times 2, so I have more common factors and more canceling.*2040

*That just leaves me with 8 times 7, divided by 1, which is 56; that was the hardest part here--finding the coefficient.*2045

*The coefficient of this fourth term is equal to 56.*2053

*To write that out completely, I am going to write it out as 56a*^{n - k}; that is 8 - 3;2058

*times (2b) raised to the k power (k is 3); this equals 56a*^{5}(2b)^{3}.2071

*Simplifying further: 56a*^{5}...2 cubed is 8, times b^{3}.2087

*If you multiply out 56 times 8, you will get 448; a*^{5}, b^{3}.2095

*This is the fourth term of this expanded form of (a + 2b)*^{8}.2103

*It is a little bit complicated: the first thing I had to do is find the coefficient.*2111

*And I did that by using my binomial formula and knowing that n = 8 and k = the term number, plus 1.*2115

*The term number plus 1 is k, so k is one less than the term number: k is 3.*2128

*I was able to find, then, using this formula, that the coefficient is 56.*2133

*The second thing I had to do is figure out what power I should raise a to, and what power I should raise 2b to.*2139

*Well, when I am using the binomial formula, I am finding the coefficient for a term in this form, which is a*^{n - k}2b^{k}.2145

*So, I knew my n and my k; so I put my coefficient here and said, "OK, I have a*^{8 - 3}(2b)^{3}," using this.2153

*Simplify, and then multiply to get 448a*^{5}b^{3}.2163

*We are asked to expand (2x + 3y)*^{5}...this binomial to the fifth.2172

*And the general form is (a + b)*^{n}.2178

*So, here we actually have that a is equal to 2x, and b is equal to 3y; so that makes things a little bit more complicated at the end.*2181

*We have to take these into account when we simplify our final answer.*2190

*Since this is the fifth power, I have an n equal to 5.*2197

*I am going to have a total of 6 terms.*2202

*It is going to look something like this at the end: (2x + 3y)*^{5} =...my first term is 2x raised to the n power,2205

*which is the fifth; plus the coefficient, and then 2x is going to decrease to the fourth power, and I am going to have my 3y here.*2214

*Plus...some other coefficient; 2x becomes cubed, and 3y increases from an exponent of 1 to an exponent of 2.*2225

*Another coefficient is going to bring me to 2x*^{2}3y^{3};2241

*another coefficient...now I just have 2x, and 3y*^{4}; and then, finally, 3y^{5}.2250

*And I expect to have 6 terms--let's verify that I am not missing anything: 1, 2, 3, 4, 5, 6.*2261

*Down here at the bottom, I am going to start building up what our final answer is going to look like, because I have some simplifying that I need to do.*2270

*The first term, I can figure out: I already know that it is 2x*^{5}.2281

*2*^{3} is 8, times 2 is 16, times 2 is 32; so that is 32x^{5}.2286

*The second term: recall that the coefficient for the second term is equal to n--that is 5.*2293

*Also, remember that symmetry is involved: since there are 6 terms (that is an even number of terms), I am going to have two middle terms.*2301

*These two are going to have the same coefficient; the two next to those will have the same coefficients; and the two outer have the same coefficients.*2309

*It is not the same value, once I simplify them; but the same coefficient.*2319

*So, since these two are the same, and these two are the same, this term also has a coefficient of 5.*2323

*Let's work this one out up here; this is the second term, and I know that it is equal to 5 times (2x)*^{4} times 3y.2332

*That is...2 times 2 is 4; 2 cubed is 8, times 2 is 16; so that is times 16x*^{4}, times 3y.2346

*Multiplying 5 times 16 times 3 is going to give you 240x*^{4}y; I am going to put that in here; that is 240x^{4}y.2358

*I don't know what I have for these two middle ones, but it is going to be something here, and it is going to be x*^{3}y^{2}.2373

*Something is here, then x*^{2}y^{3}; I can figure this one out:2384

*right here, this is the third term; it equals 5 times 2x, times (3y)*^{4}.2394

*5 times 2x is just 10x; 3 times 3 is 9, times 3 is 27, times 3 is going to give me 81.*2406

*So, that is times 81y*^{4}; so this one is easy, because it is just 81 times 10, is 810xy^{4}.2419

*So, that was...actually not the third term: correction--we don't know the third term; the first, second, third, and fourth...*2436

*the two middle terms we don't know; we do know the fifth term--this is the fifth term.*2448

*All right, so the fifth term I can fill in: 810xy*^{4}.2456

*The sixth term, right here, I can also fill in: the sixth term equals (3y)*^{5}.2465

*3*^{3} is 27; 3^{4} is 81; 3^{5}, if you figure it out, is 243; 243y^{5}.2475

*Just using the properties of binomial expansions, I got most of the terms.*2489

*And since there is symmetry, the two middle terms will have the same coefficient inserted up here.*2496

*And then, I can work out the rest and finalize it.*2505

*So, I only need to find one coefficient, using the binomial formula.*2507

*The binomial formula, recall, is n!/k!, times (n - k)!.*2511

*I know that n = 5; for the first, second, third term, k is equal to 2; you could just say "Third term: 3 - 1 is 2";*2521

*or you could look up here and say, "OK, 3y is raised to the second power, so k = 2."*2532

*Therefore, I am going to have 5!/2!, times (5 - 2)!; this equals 5!/2! times 3!,*2537

*which equals 5 times 4 times 3 times 2 times 1; 2 times 1; 3 times 2 times 1.*2559

*1, 2, and 3 all cancel; this gives me 5 times 4, divided by 2; that is 20/2, or 10.*2568

*So, the missing coefficient here is 10; by symmetry, the missing coefficient right here is also 10; I am almost done.*2578

*So, for the first, second, third term, I have my coefficient; but I need to simplify that a bit.*2588

*My third term equals 10 times (2x)*^{3} times (3y)^{2}.2595

*This is going to be equal to 10 times...2 cubed is 8x*^{3}; 3 squared is 9y^{2}.2605

*This is going to give me 80 times 9x*^{3}y^{2}, which is just 720x^{3}y^{2}.2617

*That is my third term, so I am going to put a 720 right here.*2625

*The fourth term: I also have a 10 right here.*2630

*What is slightly different, though: it is 10 times (2x)*^{2}(3y)^{3}.2635

*This is going to give me 10 times 4x*^{2}; 3 cubed is 27, so it is y^{3}.2642

*This is 40 times 27...let's put all of the constants together...x*^{2}y^{3}.2651

*40 times 27 actually comes out to 1080; you can multiply that out for yourself; x*^{2}y^{3}.2663

*The fourth term, therefore: I am going to put a 1080 right here.*2673

*So, this is quite a bit of work; but we actually only needed to find one coefficient using the binomial formula,*2676

*because by symmetry, these two middle ones were the same (the two middle coefficients that we will put in the blank);*2682

*and I knew that the second and the second-to-the-last had coefficients equal to n.*2690

*Once I found that a 10 went into these blanks, all I needed to do was work out what each of these terms would simplify to,*2695

*to end up with the expansion that I found down here.*2703

*We are asked to find a specific term this time, not the entire expansion.*2710

*I just want to find the fourth term of (x - 3y)*^{7}, using the binomial formula, n!/k!, times (n - k)!.2713

*All right, n = 7; what does k equal? Well, it is the fourth term; k is going to be one less than that; 4 - 1...I am going to have k = 3.*2725

*This is going to give me the coefficient for a term in this form.*2737

*It is going to have a coefficient, and then it is going to have x raised to the n - k power, times -3y, raised to the k power.*2741

*This is going to become x*^{7 - 3}, (-3y)^{3}.2757

*Therefore, I am going to get some coefficient (that I am going to find in a moment), times x*^{4}, times (-3y)^{3}.2768

*Let's go ahead and find the coefficient: this missing coefficient equals n!, 7!, divided by k!, times (7 - 3)!;*2775

*that equals 7!/3!, times 4!, equals (let's write this out)...divided by 3 times 2 times 1, times 4 times 3 times 2 times 1.*2793

*Do some canceling out: 1's, 2's, 3's, and 4's are the same.*2812

*This leaves me with 7 times 6 times 5, divided by 3 times 2 times 1.*2819

*6 has common factors here of 3 and 2; I can just cancel those out.*2825

*So, this is 7 times 5, divided by 1; so just 7 times 5 equals 35.*2832

*I know that my coefficient is, then, 21 that is what is going to go in this blank.*2838

*So, I am going to end up with...the fourth term is going to be 35 (I am going to put 35 right here in this blank)*2842

*x*^{4}(-3y)^{3}, which equals 35x^{4}...-3 squared is 9, times -3 again is -27.2849

*So, it is -27y*^{3}; so it is time to get out your calculators and figure out 35 times -27, or multiply it out, if necessary.2864

*And you will find that that equals -945x*^{4}y^{3}.2875

*OK, we found the fourth term of this expansion by using the binomial formula to find the coefficient,*2881

*and also, an awareness that the exponent of this x is going to be n - k; the exponent for the -3y is going to be k.*2887

*We figured out the coefficient (it is 35), and then simplified, using our rules for working with exponents.*2896

*That concludes this session on the binomial theorem here at Educator.com; thanks for visiting!*2905

*Welcome to Educator.com.*0000

*For today's Algebra II lesson, we are going to be discussing relations and functions.*0002

*And recall that some of these concepts were discussed in Algebra I, so this is a review.*0008

*And if you need a more detailed review, check out the Algebra I lectures here at Educator.*0013

*Beginning with the concept of the ***coordinate plane**: the coordinate plane describes each point as an ordered pair of numbers (x,y).0021

*The first number is the x-coordinate, and the second is the y-coordinate.*0031

*For example, consider the ordered pair (-4,-2): this is describing a point on the coordinate plane with an x-coordinate of -4 and a y-coordinate of -2.*0036

*Or the pair (0,2): the x-value would be 0, and the y-value would be 2; this is the point (0,2) on the coordinate plane.*0057

*Or (3,5): x is 3; y is 5.*0073

*Also, recall that the quadrants are labeled with the Roman numerals: I, and then (going counterclockwise) quadrant II, quadrant III, and quadrant IV.*0080

*In the coordinate pairs in the first quadrant, the x is positive, as is the y.*0097

*In the second quadrant, you will have a negative value for x and a positive value for y, such as (-2,4)--that would be an example.*0103

*In the third quadrant, both x and y are negative; and then, in the fourth quadrant, x is positive; y is negative.*0114

*And we will be using the coordinate plane frequently in these lessons, in order to graph various equations.*0127

*Recall that a relation is a set of ordered pairs.*0135

*The domain of the relation is the set of all the first coordinates, and the range is the set of all the second coordinates.*0140

*A relation is often written as a set of ordered pairs, using braces to denote that this is a set,*0149

*and then the ordered pairs, each in parentheses, separated by a comma.*0158

*Sometimes, the relation is represented as a table; so, -2, 1; -1, 0; 0, 1; and 1, 2.*0172

*We will be doing some graphing of relations also, in just a little bit.*0192

*So, as discussed up here, the domain is the set of all first coordinates.*0196

*And the set of all first coordinates here would be {-2, -1, 0, 1}.*0207

*The range is the set of the second coordinates; so the range right here--all the y-values--is {1, 0, 1, 2}.*0219

*However, you don't actually need to write the 1 twice; so in actuality, it would be written as such.*0232

*It is OK to repeat the values if you want, but usually, we just write each value in the domain or range once; each is represented once.*0245

*Functions are a certain type of relation: so, all functions are relations, but not all relations are functions.*0261

*A function is a relation in which each element of the domain is paired with exactly one element of the range.*0272

*For example, consider the relation shown: {(1,4), (2,5), (3,8), (4,10)}.*0280

*Each member of the domain corresponds to exactly one element of the range.*0298

*We don't have a situation where it is saying {(1,4), (1,6), (1,8)}, where that member of the domain is paired with multiple members of the range.*0305

*Another way, again, to represent this as a table--another method that can be used--is mapping.*0314

*And mapping is a visual device that can help you to determine if you have a function or not,*0321

*by showing how each element of the domain is paired with an element of the range.*0327

*A map would look something like this: over here, I am going to put the elements of the domain, 1, 2, 3, and 8;*0333

*over here, the elements of the range: 4, 5, 8, and 10.*0347

*And then, using arrows, I am going to show the relationship between the two.*0355

*So, 1 corresponds to 4 (or is paired with 4); 2 to 5; 3 to 8; and (this should actually be 4) 4 to 10.*0360

*OK, so as you can see, there is only one arrow going from each element of the domain to each element of the range.*0374

*And that tells me that I do have a function.*0383

*Let's look at a different situation, using a table form: let's look at a second relation.*0386

*In this one, I am going to have (-2,2), (-3,2), (-4,5), and (-6,7).*0392

*And I am going to go ahead and map this: -2, -3, -4, and -6: these are my elements of the domain.*0403

*For the range, I don't have to write 2 twice; I am just going to write it once; 5, and 7.*0416

*OK, -2 corresponds to 2; -3 also corresponds to 2; -4 corresponds to 5; and -6 corresponds to 7.*0422

*This is also a function, so both of these are relations, and they are also functions.*0436

*It is OK for two elements of the domain to be paired with the same element of the range; this is allowed.*0445

*What is not allowed is if I were to have a situation where I had {1, 2, 3}, {4, 5, 6}; and I had 1 paired with 4, and 1 paired with 5.*0452

*So, if you have two arrows coming off an element of the domain, then this is not a function.*0468

*Here are two examples of relations that are also functions.*0476

*There is a specific type of function that is called a one-to-one function.*0482

*And a function is one-to-one if distinct elements of the domain are paired with distinct elements of the range.*0486

*In the previous example, we saw a situation where we did have a one-to-one function, and another situation where we did not.*0494

*OK, so to review: the ordered pairs in that first function that we just discussed were (1,4), (2,5), (3,8), and (4,10).*0501

*OK, and we can use mapping, again, to determine what the situation is with this relation (which is also a function).*0519

*The domain is {1, 2, 3, 4}; and the range is {4, 5, 8, 10}.*0529

*When I put my arrows to show this relationship, you see that distinct elements of the domain are paired with distinct elements in the range.*0540

*1 is paired with 4; they are each unique--each pair is unique.*0552

*Looking at the other function that we discussed: the pairs are (-2,3), (-3,2), (-4,3)...slightly different, but the same general concept...slightly different, though.*0558

*OK, here I have -2, -3, -4, and -6; over here, in the range, I have 3, 2...I am not going to repeat the 3--I already have that...and then 7.*0587

*-2 corresponds to 3; -3 corresponds to 2; -4 also corresponds to 3; -6 corresponds to 7.*0605

*This is still a function; OK, so these are both functions: function, function.*0620

*However, this is a one-to-one function; this is not one-to-one.*0628

*They are both functions, since each element of the domain is paired only with one element of the range.*0639

*But in this case, it is not a unique element of the range: these two, -2 and -4, actually share an element of the range.*0648

*In other words, this is unique; it is a one-to-one correspondence.*0656

*OK, we can graph relations and functions by plotting the ordered pairs as points in the coordinate plane, as discussed a little while ago.*0662

*There are a couple of types of graphs that you can end up with.*0673

*The first is discreet, and the second is continuous; let's look at those two different types.*0676

*Consider this relation: OK, so if I am asked to graph this relation, I am going to graph each point:*0682

*(-4,-2): that is going to be right here; (-2,1)--right here; here, (0,2), 0 on the x, 2 on the y.*0698

*This is a discrete function--discrete graph--discrete relation.*0714

*This actually is both a relation and a function; so it is a discrete relation or a discrete function.*0721

*And the reason is because I have a set of discrete points; they are not connected.*0727

*And I can't connect them, because I haven't been given anything in between, or a way to know if or what lies in between these.*0734

*I can't just connect them when I don't know; there could be a point up here, or actually this is just the entire relation.*0741

*So, I can just work with what is given.*0747

*OK, a different scenario would be if I am given a relation y=x+1.*0749

*And I can go ahead and plot this out, if I say, "OK, when x is -1, -1+1 is 0; when x is 0, y is 1; when x is 1, 1+1 is 2; when x is 2, y is 3."*0759

*OK, so I am going to go ahead and plot this out.*0779

*When x is -1, y is 0; when x is 0, y is 1; when x is 1, y is 2.*0781

*Let's remove this out of the way.*0790

*When x is 1, y is 2; when x is 2, y is 3.*0793

*Now, I have a set of points, because these are the points I chose.*0798

*But because I am given this equation, there is an infinite number of points in between.*0801

*I could have chosen an x of .5 to get the value 1.5 here, to fill that in--and on and on, until this becomes continuous and forms a line.*0807

*So, this is a continuous function: the graph is a connected set of points,*0821

*so the relation (or the function in this case, since we do have a function) is continuous relation or continuous function,*0834

*because I have a line; whereas this, which is a set of points, is a discrete relation.*0841

*OK, one visual way to tell if a relation is a function is using the vertical line test.*0849

*And a relation is a function if and only if no vertical line intersects its graph at more than one point.*0856

*This is most easily understood through just working through an example.*0866

*Consider if you were given the following graph.*0870

*OK, the vertical line test: what you are seeing is, "Can you put a vertical line*0874

*somewhere on the graph so that it intersects the graph at more than one point?"*0880

*And I can: I put a vertical line here, and it intersects this graph at 1, 2, 3 places.*0887

*Over here, it only intersects at one place; that is fine; but if I can draw a vertical line **anywhere* on the graph0897

*that intersects at more than one place, then we say that this failed the vertical line test.*0903

*And when something fails the vertical line test, it means that it is not a function.*0912

*The reason this works is that, if two or three or more points share the same x-value,*0919

*then they are going to lie directly above or below each other on the coordinate plane.*0926

*For example, looking right here, I have x = 3; x is 3; y is 0.*0931

*Then, I look right above it, up here: again, x is 3, and y is...say 2.1--pretty close.*0940

*Then, I look up here; again, x is 3, and y is about 4.6, approximately.*0950

*So, when x-values are the same, but then the y-values are different, that is telling me*0959

*that members of the domain are paired with more than one member of the range; by definition, that is not a function.*0968

*OK, consider a different graph--consider a graph like this of a line--a straight line.*0975

*OK, now, anywhere that I pass a vertical line through--anywhere on this graph--it is only going to intersect at one point.*0984

*So, this passed the vertical line test.*0996

*Therefore, this line, this graph, represents a function.*1006

*OK, so the vertical line test is a visual way of determining if a relation is a function.*1014

*Working with equations: an equation can represent either a relation or a function.*1024

*If an equation represents a function, then there is some terminology we use.*1031

*And let's start out by just looking at an equation that represents a function.*1036

*The variable corresponding to the domain is called the independent variable, and the other variable is the dependent variable.*1043

*So, here I have x and y; and let's look at some values--let's let x be -1.*1050

*Well, -1 times 2 is -2, minus 1--that is going to give me -3.*1057

*When x is 0, 0 times 2 is 0, minus 1 is -1.*1063

*When x is 1, 1 times 2 is 2, minus 1--y is 1.*1071

*When x is 2, 2 times 2 is 4, minus 1 gives me 3.*1077

*So, looking at how this worked, x is the independent variable.*1084

*The value of x is independent of y; I am just picking x's, and here it could be any real number.*1099

*We sometimes also say that this is the input; and the reason is that I pick a value for x (say 0),*1109

*and I put it in--I input it into the equation; then, I do my calculation, and out comes a y-value.*1116

*So, the value of y is dependent on x; therefore, it is the dependent variable; and we also sometimes say that it is the output.*1127

*You put x in and do the calculation; out comes the value of y; so x is independent, and y is dependent.*1141

*The notation that you will see frequently in algebra is function notation.*1151

*We have been writing functions like this: y = 4x + 3; but you will often see...*1157

*instead of an equation written like this, if it is a function, you will see it written as such.*1165

*And when we say this out loud, we pronounce it "f of x equals 4x plus 3."*1170

*And we are talking about the value of a function for a particular value of x.*1179

*So, we say, "The function of f at a particular x."*1186

*Let's let x equal 3; then, we can talk about f of 3--the value of the function, the value of y,*1193

*of the dependent variable, when the independent variable, x, is 3.*1204

*And in that case, since it is telling us that x is 3, I am going to substitute in 3 wherever there is an x.*1209

*And I could calculate that out to tell me that f(3) is...4 times 3 is 12, plus 3...so f(3) is 15.*1217

*Here, x is an element of the domain, of the independent variable; f(x) is an element of the range.*1231

*So again, we are going to be using this function notation throughout the remainder of the course.*1246

*Looking at the first example: the relation R is given by this set of coordinate pairs.*1253

*Give the domain and range, and determine of R is a function.*1260

*Well, recall that the domain is comprised of the first element of each of these coordinate pairs.*1267

*So in this case, the domain would be {1, 2, 6, 5, 7}.*1275

*The range: the range is comprised of the second element of each ordered pairs, so I have 4, 3...4 again;*1291

*I don't need to write that again; 3--I have 3 already; and 5; I am just writing down the unique elements.*1303

*This is the domain, and this is the range.*1309

*Now, is R a function? Well, I can always use mapping to just help me determine that.*1311

*And I am going to write down my members of the domain, and my elements of the range.*1320

*And then, I am going to use arrows to show the correspondence between each:*1330

*1 and 4--1 is paired with 4; 2 is paired with 3; 6 is also paired with 4; 5 is paired with 3; and 7 is paired with 5.*1333

*Now, I am looking, and I only see one arrow leading from each element of the domain.*1350

*There is no element of the domain that is paired with two elements of the range.*1354

*So, in this case, this is a function; so, is R a function? Yes, this relation is a function--R is a function.*1359

*So, always double-check and make sure you have answered each part.*1373

*I found the domain; I found the range; and I determined that R is a function.*1375

*OK, the relation R is given by the equation y=2x*^{2}+4; is R a function?1384

*What are the domain and range? Is R discrete or continuous?*1393

*Let's just look at some values for x and y to help us determine if this relation is a function.*1400

*If I let x equal -1, -1 times -1 is 1, times 2 is 2, plus 4 is 6.*1412

*OK, if x is 0, this is 0, plus 4--that gives me 4.*1422

*If x is 3, 3 squared is 9, times 2 is 18, plus 4 is 22.*1428

*So, as you are going along, you can see that, for any value of x, there is only one value of y; therefore, R is a function.*1434

*What is the domain? Well, I could pick any real number for an x-value that I wanted, so the domain is all real numbers.*1450

*You might, at first glance, say, "Oh, the range is all real numbers, as well"; but that is not correct, because look at what happens.*1467

*Because this is x*^{2}, whenever I have a negative number, it becomes positive; if I have a positive number, it stays positive, of course.1475

*Therefore, if I have, say, -1, that becomes 1; this becomes 6.*1486

*So, I am not going to get any value lower than...for y, the smallest value I will get is for when x is 0.*1493

*OK, so if x is 0, y is 4; because -1 is going to give me a bigger value--it is going to give me 6.*1505

*If I do -2, that is going to be 4 times 2 is 8, plus 4 is 12.*1510

*So, the lowest value that I will be able to get for y will occur when x is 0.*1516

*And that is going to give me a y-value of 4.*1522

*Therefore, the range is that y is greater than or equal to 4.*1525

*So, the most difficult part of this was just realizing that the range is not as broad as it looked initially.*1530

*Because this involves squaring a number, there is a limit on how low you are going to go with the y-value.*1538

*So, this is a range with a domain of all real numbers, and a range of greater than or equal to 4.*1545

*OK, in Example 3, graph the relation R given by 2x - 4y = 8.*1559

*Is R a function? Find its domain and range. Is R discrete or continuous?*1572

*OK, so graph the relation given by 2x - 4y = 8.*1583

*Let's go ahead and find some x and y values, so that we can graph this.*1590

*When x is 0, we need to be able to solve for y; when x is 0, let's figure out what y is.*1598

*0 - 4y equals 8; therefore, y equals -2 (dividing both sides by -4).*1607

*OK, when x is 2, 2 times 2 minus 4y equals 8; that is 4 minus 4y equals 8; that is -4y equals 4; y = -1.*1615

*And let's do one more: when x is -2, this is going to give me -4 - 4y = 8.*1635

*That is going to then give me, adding 4 to both sides, -4y = 12, or y = -3; that is good.*1646

*All right, so when x is 0, y is -2; when x is 2, y is -1; when x is -2, y is -3, right here.*1657

*I am asked to graph it; and I have some points here that I generated,*1686

*but I also realize that I could have picked points in between these, which would actually end up connecting this as a line.*1691

*So, I am not just given a set of ordered pairs; I am given an equation that could have an infinite number of values for x,*1702

*which would allow me to graph this as a continuous line.*1707

*Therefore, I graphed the relation...is R a function?*1713

*Is R discrete or continuous? Well, I have already answered that--seeing the graph of this, I know that this is continuous.*1720

*And let's see, the next step: is R a function?*1731

*Yes, it is a function, because if I look, for every value of x (for every value of the domain), there is one value only of the range.*1746

*So, every element of the domain is paired with only one element of the range.*1758

*It is continuous, and it is a function.*1762

*Find the domain and range: well, this is another case where I could choose x to be any real number, so it would be all real numbers--any real number.*1765

*Here, the situation is the same for the range--all real numbers.*1778

*Depending on my x-value, I could come up with infinite possibilities for what the range would be, what the y-value would be.*1784

*R is a function; its domain and range are all real numbers; and this is a continuous function.*1792

*OK, in Example 4, we are given f(x) = 3x*^{2} - 4, and asked to find f(2), f(6), and f(2k).1805

*First, f(2): recall that, when you are asked to find a function for a particular value of x,*1824

*you simply substitute that value for x in the equation; so f(2) equals 3(4) - 4, so that is 12 - 4; so f(2) = 8.*1834

*Next, I am asked to find f(6), and that is going to equal 3(6*^{2}) - 4.1852

*f(6) = 3(36) -4, and that turns out to be 108 - 4, so f(6) is 104.*1860

*Now, at first, this f(2k) might look kind of difficult; but you treat it just the same as you did with the numbers, when x is a numerical value.*1877

*Everywhere I see an x, I am going to insert 2k.*1887

*And figuring this out, 2 times 2, 2 squared, is 4; k times k is k*^{2}.1891

*3 times 4 is 12, so I have 12k*^{2} - 4; so f(2k) = 12k^{2} - 4.1902

*So again, if you are asked to find the function of a particular value of x, you simply substitute whatever is given, including variables, for x.*1910

*That concludes this lesson of Educator.com; I will see you back here soon!*1919

*Welcome to Educator.com.*0000

*In today's lesson, we will be covering linear equations.*0002

*Again, this is some review from Algebra I, which we will discuss here.*0008

*And if you need further detail, go back to the Algebra I lectures and check those out for this subject.*0011

*So, a linear equation is an equation of the form ax + by = c, where, a, b, and c are constants.*0018

*For example, 2x + 5y = 3, and here a = 2, b = 5, and c = 3.*0028

*Another example could be x - 7y = 4; here, even though it is not written, there actually is a 1 in front of the x.*0041

*Just by convention, we don't write it out when the coefficient is 1.*0052

*So, a = 1, b = -7, and c = 4.*0055

*We will talk a little bit more about what this form is.*0062

*A linear equation represents a function, which is known as a linear function; and that can be written in the form f(x) = mx + b.*0067

*And this is a very useful form of the equation that will be discussed in another segment.*0078

*And an example of this would be something such as f(x) = 4x + 2; you may also see it written as y = 4x + 2.*0084

*And here, m = 4; b = 2; and these two numbers represent certain elements that we will talk about in a few minutes.*0107

*OK, so the first form of the linear equation that we talked about just a minute ago is also called standard form.*0118

*So, a linear equation is in standard form if it is written in the form ax + by = c,*0129

*where a, b, and c are integer constants (and this is important) with no common factor.*0135

*If a common factor remains, it is not in standard form.*0140

*For example, if I have an equation 9x + 4y = 10, this is in standard form.*0144

*I have ax (9 is a), plus by (b is 4), equals c (c is 10), and there is no common factor.*0156

*Consider another possible equation: 6x + 8y = 12; this is not in standard form, because 6, 8, and 12 have a common factor.*0168

*What I need to do is divide both sides of this equation by 2 to get 3x + 4y = 6; now, it is in standard form.*0189

*Another example would be if I had 7y + 5 = -x; again, it is not in standard form.*0207

*I can add x to both sides to get x + 7y + 5 = 0, and then subtract 5 from both sides to get x + 7y = -5; this is standard form.*0218

*We can graph linear equations using the intercept method.*0246

*Recall that the point where a line crosses the x-axis is its x-intercept.*0250

*And the point where the graph crosses the y-axis is the y-intercept.*0259

*For example, if I had a line such as this, first the x-intercept is right here; the x-intercept is -2.*0265

*And at the x-intercept, y is always going to be 0; so, the coordinate pair would be (-2,0), because it crosses the x-axis, and so y is 0.*0283

*The y-intercept is right here: y equals 4, and at this point, x is going to equal 0, so (0,4) is the y-intercept.*0294

*And we can use this knowledge in order to graph a linear equation.*0308

*For example, if I have a linear equation 2x + 3y = 6, all I need in order to plot this are two points on the line.*0314

*Now, I can easily get two points by finding the x- and y-intercepts.*0328

*I know that, with the y-intercept, x will be 0; so if I let x equal 0, that is going to give me 0 times 2 is 0,*0334

*so that will be 3y = 6; 6 divided by 3--y equals 2.*0343

*OK, to find my second point, I am going to find the y-intercept (this is actually the x-intercept right here)...0.*0354

*Excuse me, this is the y-intercept right here; now I am looking for the x-intercept.*0368

*The x-intercept is the point at which y equals 0, and the line of the graph crosses the x-axis.*0375

*So, this time, I am going to let y equal 0; so I am going to have 2x, plus 3 times 0, equals 6, or 2x + 0 = 6.*0383

*So then, I end up with x = 3; this is the x-intercept.*0395

*Now, I have two points; I can plot the line.*0400

*At the y-intercept, x is 0; y is 2; the y-intercept is right here, at (0,2); the x-intercept is right here at (3,0).*0405

*Now, I have two points; and when I have two points, that means I can connect them to form a line.*0422

*And I did that by finding my x-intercept at (3,0) and my y-intercept at (0,2).*0431

*The technique is to find the x- and y- intercepts by letting x equal 0, and putting that into the equation,*0449

*and solving for y to get the y-intercept, and letting y equal 0*0458

*to find the x-intercept, and then using those two points to plot the line.*0467

*In the first example, is the function linear? Well, recall that a linear function can be written in standard form.*0475

*And standard form is ax + by = c: I look at what I have here, and I have ax (so it seems like I am going along OK),*0482

*minus by (or that would be the same as plus -by, so that is fine), equals c; but then I have this over here.*0492

*And this is not part of standard form.*0500

*And you could attempt to get rid of that; maybe you say, "OK, I will multiply both sides by x to get rid of it."*0502

*But then, see what happens: all right, let's see what happens if I try to multiply both sides by x to get rid of this x in the denominator.*0508

*OK, I end up with 2x*^{2} - 3xy = 4x + 1; so I am no better off.0518

*I still don't have it in standard form; there is no way for me to get this in standard form; therefore, this is not a linear function.*0530

*It is not in standard form; I can't get it into standard form.*0546

*OK, is this function linear? Well, recall that we talked earlier about the form g(x) = mx + b.*0550

*And remember, g(x) is the same idea as f(x): you can use different letters, so it is still just that we are looking for this form for a linear function.*0566

*Well, I am looking at this, and I have what looks like an mx and a b, but the problem is that I also have this.*0580

*And this is not a part of the form for a linear function--this x*^{2} term--therefore, this is not a linear function.0588

*Once again, I can't get it into the right form for a linear function.*0598

*OK, in Example 3, we are asked to write the equation in standard form.*0604

*And recall that standard form is ax + by = c.*0608

*And I am dealing with some fractions, and I need to get rid of those in order to get this in standard form.*0613

*I can do that by multiplying first both sides of the equation by 3; this is going to give me 3y/6 =...here, the 3's will cancel out; -7 times 3--that is -21.*0620

*I can simplify this to y/2 = x - 21; so, I am still left with a fraction.*0637

*I am going to multiply both sides of the equation by 2 to eliminate that fraction.*0644

*The 2's cancel out, so I have y = 2x...-21 times 2; that is -42.*0652

*Next, I am going to subtract 2x from both sides to get the x on the left side of the equation.*0665

*And just switching around the order of the x and y to match this: -2x + y = -42.*0677

*And you could stop here, or you could multiply both sides by -1; and that is more conventional, to have the a term be positive.*0688

*So, that is going to leave me with 2x - y = 42.*0698

*I started out with this equation, and I wanted to get it into this form.*0707

*And I could achieve that by first multiplying by 3 to get rid of this fraction, then (I ended up with this)*0711

*multiplying by 2 to get rid of the y fraction, then simply subtracting 2x from both sides to get the 2x (the ax)*0721

*on the left side of the equation, and finally just a little more work to clean this up, multiplying both sides by -1.*0730

*This is the equation in standard form, where a equals 2, b equals -1, and c equals 42.*0738

*Example 4: Find the intercepts and graph the equation: 4x - 5y = 20.*0748

*We already talked about how, with the x- and y-intercepts, you can find those, and then you will have two points, and you can graph an equation.*0756

*So, first, in order to find the y-intercept, I am going to let x equal 0; this will give me the y-intercept, the point at which the line crosses the y-axis.*0766

*So, to find the y-intercept, I am going to substitute 0 for x.*0779

*This is going to give me 0 - 5y = 20, or -5y = 20; divide both sides by -5 to get y = -4.*0788

*OK, next I want to find the x-intercept: to find that, I am going to let y equal 0.*0801

*I am going to go back and do this again, this time letting y equal 0.*0810

*This is for the y-intercept, and this is for the x-intercept.*0821

*This is going to give me 4x - 0 = 20, or 4x = 20, which is x = 5.*0824

*Now that I have my x- and y-intercepts, I actually have two points on the line.*0834

*My first point is (0,-4), the y-intercept; my second point is at (5,0), and that is the x-intercept.*0839

*OK, connecting these two points gives me the graph of this equation.*0859

*We use the intercept method in order to graph, finding the x- and y-intercepts,*0869

*plotting those out on the coordinate plane, and then using those to form a line.*0874

*That concludes this lesson for Educator.com; and I will see you here again soon.*0880

*Welcome to Educator.com.*0000

*In today's lesson, we will be discussing slope.*0002

*Recall the definition of slope: the slope, m, of the line passing through a point (x*_{1},y_{1}),0007

*and another point (x*_{2},y_{2}), is given by:0015

*the slope equals the change in y, over the change in x (the change in the y coordinates, over the change in the x coordinates).*0020

*Therefore, if you have any two points on a line, you can use that to find the slope.*0030

*For example, imagine that you are given the graph of a line that looks like this, and you are asked to find the slope.*0037

*Well, I have a point right here at x is 0, so my (x*_{1},y_{1}) point is going to be x is 0, y is 2.0053

*I have a second point over here, and I am going to call this (x*_{2},y_{2}); and here, x is -2, y is 0.0065

*So, I am using the intercepts; but I could have used any other point on this line.*0076

*OK, an important point is that I could have assigned either one of these as...*0081

*I could have said this is (x*_{1},y_{1}), and this is (x_{2},y_{2}).0086

*It doesn't matter, as long as you are consistent (I can't say (x*_{1},y_{2}), (x_{2},y_{1})).0090

*As long as I am consistent, and I follow that consistency when I am subtracting, I am fine.*0096

*So, I arbitrarily assigned this as (x*_{1},y_{1}), and this point as (x_{2},y_{2}).0103

*So, I found two points on the line; now, my y*_{2} value is 0, and I am going to subtract my y_{1} value (which is 2) from that.0107

*The x*_{2} value is -2; my x_{1} value is 0, so minus 0; so here, the slope equals -2 over -2, or 1.0120

*So, I could have used any two points on this line, and found the difference between those x and y coordinates--the change in y over the change in x.*0138

*So, what this is telling me is that...we say m = 1, but what we really mean is m = 1/1; think about it that way, and say that.*0147

*For every increase in y by 1, I am going to increase x by 1; increase y by 1, increase x by 1.*0156

*So, I may be given a line and asked to find the slope; or I may directly be given a set of points and asked to find the slope.*0167

*So, if I were given two points on the line, but not given the line itself, or a graph, or anything, I could easily find the slope.*0175

*This is telling us the vertical change over the horizontal change.*0182

*OK, some cases to keep in mind when you are dealing with slope: a horizontal line has a slope of 0.*0187

*And let's think about why this is so: consider a horizontal line up here at, say, 4; OK, so this is 4.*0198

*And if I look at any point--for example, right here--here, x is 1; y is 4; and then maybe I look over here at this point; here x is 3; y is still 4.*0212

*Now, recall that the slope formula is the change in y over the change in x.*0229

*So, I have my two points; and I am going to call this (x*_{1},y_{1}), (x_{2},y_{2}).0239

*Again, I could have done it the other way around.*0246

*All right, so I have y*_{2} (that is 4), minus y_{1};0249

*and then I have my x*_{2} (which is 3), minus x_{1} (which is 1).0255

*It actually doesn't matter what I have down here, because, since I have a horizontal line, the y-values everywhere here are going to equal 4.*0261

*So, 4 - 4 is going to be 0; therefore, the slope is 0.*0273

*So, a horizontal line has a slope of 0, because the y-values are the same at every point, so the difference in y is going to be 0.*0281

*OK, vertical lines are also a special case: and they have an undefined slope.*0291

*For example, let's look at a vertical line right here at x = -3.*0298

*OK, if I look right here at (I could pick any point)...I am going to pick this point, x is -3; y is 1.*0306

*Then, I will pick another point right here; let's say x is -3; y is -2.*0315

*So, I have my (x*_{1},y_{1}), and my (x_{2},y_{2}).0322

*Again, slope is change in y over change in x; change in y is y*_{2} (that is -2), minus 1, over change in x (-3 - -3).0330

*This is going to give me -3 over...-3 minus -3 is -3 plus 3; that is over 0.*0349

*And this is not allowed; this is undefined.*0359

*We say that a vertical line has an undefined slope because, since the change in x is 0,*0367

*because x is the same at every point, what you end up with is a denominator that is 0, and that is undefined; that is not allowed.*0372

*So, for a horizontal line, the slope is 0; for a vertical line, slope is undefined.*0379

*A couple of other things to keep in mind regarding slope: one is that in a line that rises to the right (such as this one), m is a positive number.*0393

*So, it is increasing to the right; the values are increasing.*0407

*If you have a line that falls to the right, such as this one, here the slope is positive; in this case, the slope is negative.*0413

*Rises to the right--slope is positive; falls to the right--slope is negative.*0428

*A horizontal line has a slope of 0; a vertical line has an undefined slope.*0433

*Parallel lines: two lines are ***parallel** if and only if they are vertical or have the same slope.0439

*Let's first check out the case where the lines are not vertical, but they are still parallel.*0446

*If I have two lines that are parallel, it is going to look like this, for example.*0451

*These are going to have the same slope; so if this is my line 1 and line 2, m*_{1} will equal m_{2}--the slopes will be equal.0459

*And that is because they are changing at the same rate--these two lines are changing at the same rate, so they never intersect.*0468

*Now, we set vertical lines aside; we put them separately; we don't say they have the same slope.*0476

*And remember that the reason is because a vertical line has an undefined slope.*0483

*So, I can't say these two have the same slope, because their slope is undefined.*0492

*However, we recognize that they are still parallel; so we say that two lines are parallel*0496

*if they are vertical (that is a separate case), or if they have the same slope; but both of these represent parallel lines.*0501

*Perpendicular lines: two non-vertical lines are ***perpendicular** if and only if the product of their slopes is -1.0512

*So, "perpendicular" would tell me that these two lines intersect at right angles.*0522

*And if this is my line 1 and line 2, and then I have a slope m*_{1} and m_{2}, what this is stating0535

*is that the product of m*_{1} and m_{2} is -1.0544

*Let's look at an example: let's say I have an m*_{1} that is equal to 4, and I say that it is perpendicular to a line L_{2}.0552

*So, this is line 1, and this is line 2--the perpendicular line.*0569

*I can find the slope of the perpendicular line, because I know that m*_{1} times m_{2}0575

*is -1 (since these are perpendicular lines), and I am given m*_{1}.0584

*So, 4 times m*_{2} equals -1; therefore, the perpendicular line would have a slope of -1/4.0591

*So, the perpendicular line has -1/4 as the slope, and that is the negative reciprocal of m*_{1}.0606

*Knowing the relationship between parallel lines and the slopes of parallel lines and the slopes of perpendicular lines*0618

*can be helpful in answering problems and graphing lines; and we will see that right now with the examples.*0624

*OK, in this first example, find the slope of the line passing through the points (-9,-7) and (-6,-3).*0633

*Well, the definition of slope is the change in y over the change in x.*0643

*And I could choose to assign this either way; but I am just going to go ahead*0651

*and say this is (x*_{1},y_{1}), and this is going to be my (x_{2},y_{2}).0654

*And it doesn't matter which way you assign, as long as you are consistent.*0660

*So, I could have called this (x*_{2},y_{2}), and it would have been fine; we would have gotten the same answer.0663

*OK, y*_{2} is -3, and y_{1} is -7, so minus -7; over x_{2} (which is -6), minus x_{1} (which is -9).0668

*Simplifying this: this is -3; a negative and a negative gives me a positive; and that gives me 4/3.*0686

*So, the slope of a line passing through these points is 4/3, just using my slope formula.*0697

*OK, Example 2: Graph the line passing through the point (-2,-1) with a slope of -2/3.*0705

*Well, I am given the slope, and I am given a point on the line.*0717

*So, let's start out with the point that I am given, which is (-2,-1); that is right here.*0722

*Remember that the slope is the change in y over the change in x.*0730

*So, if the change in y is -2, then I am starting out here at -1; I am going to go to -2, -3.*0735

*For every 2 that y is decreased in value, x is going to increase by 3: 1, 2, 3; so, I am going to end up with this point right here.*0744

*You could have also looked at this as this, and I could say that for every 2 y is increased, x is decreased by 3: 1, 2, 3, right here.*0756

*The same thing--you could look at it either way.*0771

*OK, so now, I am able to graph this line because I had a starting point right here, and I have the slope.*0777

*So, I know, starting from here, how much the x is going to change and how much the y is going to change.*0783

*So then, I simply connect these lines to give my graph.*0789

*And notice that this line is decreasing to the right: and that fits with what I have, which is a negative slope.*0800

*So, simply plot the point you are given; and then use the change in y over the change in x to find additional points on the line.*0806

*OK, graph the line passing through (1,-3) and parallel to the graph of this equation.*0820

*Well, I have my point; x is 1; y is -3; so, I have my starting point.*0829

*I need my slope; and I am not directly given the slope; however, I am given a parallel line.*0835

*And recall that parallel lines have the same slope.*0843

*That means that, if I find the slope of this line, I am going to have the slope of the line that I am looking for.*0853

*In order to find the slope, I need to be able to find the change in y over the change in x; so I need two points on this line.*0860

*So, the slope is (y*_{2}-y_{1})/(x_{2}-x_{1}).0870

*Thinking of an easy way to find a couple points on this line, I am going to use the intercept method.*0876

*And I could find these points and graph, or I could just find these points and plug them into my slope equation.*0883

*First, I am going to let x equal 0 to find the y-intercept.*0890

*OK, so this gives me -3y = 6, or y = -2.*0901

*When x is 0, y is -2; when y is 0, let's go ahead and plug that in; 2x - 3(0) = 6, so this is 2x - 0 = 6; 2x = 6.*0913

*Divide both sides by two; that gives me x = 3.*0933

*OK, so when x is 0, y is -2; when x is 3, y is 0.*0937

*Now, I can find the slope of the sign, because this can be my x*_{1};0946

*this can be my y*_{1}; this can be my x_{2}; this can be my y_{2}; so let me find the slope.0953

*This is y*_{2}, which is 0, minus y_{1}, which is -2.0959

*This is x*_{2}, which is 3, minus 0, which is x_{1}.0967

*This is going to give me 0 minus -2 (0 plus 2), over 3 minus 0; so the slope is 2/3.*0975

*Therefore, the slope of this line is 2/3, and since parallel lines have the same slope, the slope of the line I am looking for is also 2/3.*0988

*Now, I was given this point, (1,-3); now I know that the vertical change is going to be an increase in y by 2;*1003

*and there is going to be a change in x--an increase by 3--1, 2, 3; so I have another point.*1011

*Now that I have this second point, I can go ahead and graph this line.*1019

*And this is the line I was asked to graph, so I have completed what I have been asked to do.*1027

*But just to look at the idea that this is indeed a parallel line, I am going to go ahead and plot this other line, as well.*1032

*This is the line we were asked to find; this other line--when x is 0, y is -2; when x is 3, y is 0.*1038

*And you can plot that out and see that, visually, this does look like a parallel line.*1047

*And these two are changing at the same rate: the slope of both of these is 2/3.*1052

*I was able to graph the line passing through this point and parallel to the graph of this line*1059

*by finding the slope of this line, and then realizing that my parallel line is going to have the same slope, which is 2/3.*1068

*Example 4: Graph the line passing through the point (-2,1) and perpendicular to the graph of this equation: 3x + 4y = 12.*1077

*OK, I am given a point; and let's see, let's call this -2, -4, -6, -2, -4, -6, 2, 4, 6, 8; OK.*1090

*-2 would be right here; and then 1 would be right about there; so that is my point.*1110

*But I also need the slope: recall that the slopes of perpendicular lines--the product of those slopes is equal to -1.*1116

*So, the product of perpendicular lines' m*_{1} and m_{2} (I have two perpendicular lines;1127

*the slope of one is m*_{1}, and the other is m_{2})--their product is -1.1133

*So, I am going to call this first line line 1, and its slope is going to be m*_{1}; I am going to find m_{1}.1137

*My line I am going to call line 2; and its slope is going to be m*_{2}.1147

*I need to find the slope of this line, recalling that slope is the change in y over the change in x.*1153

*So, in order to find this change, I need to find two points on this line.*1160

*I am going to go for something easy, so I am going to let x equal 0, and I am going to plug that in here.*1167

*I am just going to find the intercepts as my two points; but I could have found any two points on that line to get the slope.*1172

*This is going to give me y = 3; when x is 0, y is 3.*1185

*For my second point, I am going to let y equal 0; that is 3x + 4(0) = 12; so, if 3x = 12, x = 4.*1189

*So now, I have two points; two points means I can find the slope.*1206

*I am going to say that this is my (x*_{1},y_{1}), (x_{2},y_{2}), so I can keep track of everything.1210

*y*_{2} is 0; y_{1} is 3; so 0 - 3, over x_{2} (which is 4), minus 0.1219

*OK, I have found m*_{1}: I have found the slope of this line.1233

*And I know that the line I am looking for, which is going to have a slope m*_{2}, is perpendicular to this line.1239

*So, I am going to go right here and say that m*_{1}...I know that is equal to -3/4; and I am looking for m_{2}.1246

*m*_{1} times m_{2} equals -1; so now, all I have to do is substitute this in.1258

*-3/4 times m*_{2} equals -1; I am going to multiply both sides by -4/3 to move this over to the right.1265

*And that is going to give me -1, times -4/3, or m*_{2} = 4/3.1277

*OK, now I found this slope, and I need to graph the line.*1285

*I have my starting point here at (-2,1); and the change in y is going to be 4, so that is going to be 1, 2, 3, 4,*1291

*for an increase in x of 3; so that is 1, 2, 3, right there.*1300

*Increase in y by 4: 1, 2, 3, 4; increase x by 3: 1, 2, 3--so right about there; OK.*1308

*I can go ahead and plot this line out.*1325

*Reviewing what we did: we were given a point that the line passes through, and we needed the slope in order to graph it.*1329

*We were told that this line is perpendicular to the line described by this equation.*1335

*Therefore, we found the slope of this line; and we did that by finding two points in the line (the x- and y-intercepts) and plugging that into the slope formula.*1340

*Once I found the slope of this perpendicular line, I wanted to find the slope of my line.*1351

*And I did that by recalling that the product of the slopes of perpendicular lines equals -1.*1356

*So, -3/4 times the slope of the perpendicular line is -1.*1363

*I figured out that the line I am looking for has a slope of 4/3, and then I just took my point that I was given and increased y by 4, and then x by 3.*1368

*That concludes this lecture about slope on Educator.com.*1382

*Welcome to Educator.com.*0000

*In today's lesson, we are going to talk about writing linear functions.*0002

*And in particular, we are going to discuss two forms of these functions.*0006

*The first form is the slope-intercept form; and this is a very useful form of the equation, because it can help you to graph a linear equation.*0012

*The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept.*0022

*Recall that the y-intercept is the point at which the graph of the equation (the line) intersects the y-axis.*0033

*For example, if we look at an equation in this form, and it is given as y = -2x + 1, then the slope is -2, and the y-intercept, or b, is 1.*0043

*This information alone allows me to graph the line.*0060

*Before, we talked about graphing the line by finding a couple of points.*0063

*And we, in particular, used the intercept method, where we found the x-intercept and the y-intercept, and graphed that.*0067

*This time, I am going to use a slightly different method.*0073

*So, here I have the y-intercepts: this is the y-coordinate where the graph of the line is going to cross the y-axis.*0076

*That is going to be at y = 1; x will be 0, and y is 1.*0088

*Now, I also have the slope; the slope is the change in y, over the change in x.*0094

*And this is written as -2; but you can think of it in your mind as -2/1.*0100

*For every 2 that y is decreased, x is increased by 1.*0105

*And I am already thinking of what I expect this graph to look like.*0110

*Recall that, if the slope is negative, the line is going to decrease going from left to right--it is going to go this way.*0113

*If the slope is positive, the line is going to increase going from left to right.*0120

*I am starting right here, and I am going to decrease y by 2--1, 2--for every increase in x by 1.*0125

*So now, I have another point: 1, 2, and then increase x by 1.*0133

*And I now have plenty to go ahead and graph this with.*0139

*So, you can see how this form of the equation is very helpful in getting information about the line and actually graphing the line.*0152

*The second form of these linear equations that we are going to look at today is point-slope form.*0161

*The point-slope form of the equation of a line is y - y*_{1} = m (x - x_{1}).0169

*And you will see that this is related to slope and the slope formula.*0177

*So, since the slope is (y*_{2} - y_{1})/(x_{2} - x_{1}), you can look and see that that is pretty familiar.0182

*Now, previously we talked about having two points: (x*_{1},y_{1}), and the other (x_{2},y_{2}).0191

*Well, when we are working with the point-slope form, we have one point (x*_{1},y_{1}),0199

*and then the other point could be anywhere on the line; it is just another point (x,y) that isn't specified.*0205

*So, look at how you could manipulate this to be similar to this,*0212

*if I said, "OK, the slope is some point on the line, minus a given point, over some point on the line, minus the given point."*0215

*Now, I am going to multiply both sides of this by (x - x*_{1})--both sides of the equation.0226

*These cancel out; and that is going to give me (x - x*_{1}) times m equals (y - y_{1}).0243

*A little bit of rearranging: I am going to put the y portions on the left side of the equation, and the x and the slope on the right side of the equation.*0255

*And you see, that gives me the point-slope form.*0267

*And this is useful, because imagine if I am given some facts about a line, and I am told that the slope equals 4, and that this line passes through a certain point.*0271

*And I am told that it passes through the point (-2,6).*0287

*Knowing this, I can write the equation for this line in point-slope form.*0290

*So, point-slope form: y - y*_{1} = the slope times (x - x_{1}).0295

*So, y is any other point on this line.*0301

*y*_{1} is 6; slope is 4; x is another x point on this line that goes along with this y coordinate, minus x_{1}, which is -2.0306

*Simplifying this: a negative and a negative is a positive.*0319

*What is helpful about this point-slope form is: with a little bit of work, I can put this into the slope-intercept form.*0329

*Recall that the slope-intercept form (this is the point-slope form of the equation) is y = mx + b.*0338

*I am going to multiply this out; this is 4x + 8; and I want to isolate y by adding 6 to both sides.*0358

*So now, I have it in slope-intercept form; this is a very useful form of the equation.*0374

*And again, this can allow us to graph the equation.*0384

*Parallel and perpendicular lines: first, recall that parallel lines have the same slope.*0388

*So, if I have two parallel lines, line 1 and line 2, and the first one has a slope of m*_{1}, and the second one has a slope of m_{2}, those are equal.0402

*Perpendicular lines: the product of the slope of two perpendicular lines is equal to -1.*0415

*The slope-intercept form and point-slope form can be used to solve problems involving parallel and perpendicular lines.*0429

*And the reason is: if I am told that a line is parallel to another line, I have the slope.*0435

*With a little bit more information, I can write the equation in slope-intercept form.*0442

*Or I may be given the equation in slope-intercept form and told that a line is parallel to that line.*0446

*The same with perpendicular lines: by having these forms of the equation, which involve slope,*0453

*and knowing the relationship between two lines and their slopes, I can write the equation for the second line.*0458

*I can graph the lines, and it is all about knowing the relationships between these two lines.*0464

*For example, if I am told that the graph of a line is parallel to the line described by the equation y = 1/2x - 4,*0471

*and I am also told that the line I am looking for passes through a point (4,-3),*0499

*I can graph the line I am looking for; I can also write an equation for the line I am looking for.*0514

*So, the graph of a line is parallel to the line y = 1/2x -4, and the line I am looking for passes through a certain point.*0521

*Well, since parallel lines have the same slope, and I am looking at this, and it is in slope-intercept form, which is y = mx + b, I now have the slope.*0529

*So, the slope equals 1/2; since these lines are parallel, I also have the slope of the line I am looking for, and it is 1/2.*0540

*The line I am looking for passes through (4,-3); so that is (4,-3), and I am going to plot that out: (4,-3).*0549

*And I have the slope, so I know that when I increase y by 1, I increase x by 2; therefore, I can plot the line.*0564

*OK, and this is as expected, because it has a positive slope, so it is increasing as it goes to the right.*0577

*The same holds true for perpendicular lines: for example, if I am told that the graph of a line is*0598

*(this stands for perpendicular) perpendicular to the graph of the line defined by the equation y = 1/4x + 6,*0612

*and the line passes through some point (say (1,2)), I can graph this line.*0627

*And the reason I can graph it is that I know that this slope is 1/4, and I recall that the two slopes are related by this formula.*0638

*So, if I am given a point on a line, as well as the knowledge and relationship between that line and a parallel or perpendicular line,*0648

*I have the point; I can find the slope; I can graph the line.*0659

*OK, first example: Find the equation in slope-intercept form of the line with the slope 2/3 and passing through (2,-4).*0665

*Slope-intercept form is y = mx + b; and I am given slope, so I am given m = 2/3; so let's start from there: y = 2/3 x + b.*0678

*Well, in order to write this out, I also need to find b; and b is unknown.*0694

*However, I am given an x value and a y value; and since I have m, x, and y, I can solve for b.*0699

*So, substituting in -4 for y, and 2 for x, now I can solve for b.*0707

*So, -4 = 2 times 2...that is 4, so that is 4/3, plus b.*0718

*Subtract 4/3 from both sides...equals b...and you could really think of this as -1 and 1/3; it might be easier to look at it that way.*0734

*This is -5 and 1/3...equals b.*0746

*OK, going back to the beginning here: y = mx + b, so y equals...m is given as 2/3; x; and then b is -5 and 1/3.*0749

*Using the facts that I was given, I am able to write this in slope-intercept form.*0765

*I was given the slope, and I was given a point on a line.*0773

*The point on the line, and the slope: by plugging those into this equation, I could find the y-intercept; and therefore, I could write this out in slope-intercept form.*0776

*Example 2: find the equation in slope-intercept form of the line passing through these two points.*0786

*Slope-intercept form is y = mx + b, so I need to have the slope, and I need to have the y-intercept, in order to write this.*0793

*The slope is the change in y, over the change in x; and I am given two points, so I can find the change in y over the change in x.*0802

*I am going to call this (x*_{1},y_{1}), and this (x_{2},y_{2}).0813

*OK, so m = y*_{2} (which is -3) minus -7, over -6 minus -2.0819

*-3...and that negative and negative becomes a positive, so plus 7, minus 6--a negative and a negative--that is plus 2.*0833

*-3 + 7 is 4; -6 + 2 is -4; I have the slope now--the slope is -1.*0841

*My next thing is to find b, which is the y-intercept; so, I have y = mx (-1x; we usually just write this as -x,*0853

*but we are writing it out right now as -1x) + b.*0864

*I need to solve for b, and I can do that, because I have some x and y values I can substitute in.*0868

*You could choose either set of coordinates, either ordered pair.*0874

*I am going to go ahead and choose this first one: y is -7; it equals (instead of writing -1, I am just going to write) -x (which is -2) + b.*0880

*Solving for b: -7 equals...a negative and a negative is a positive, so that is 2 + b.*0898

*Subtracting two from both sides, b equals -9.*0905

*In order to write this equation, I needed to have my slope, which I do.*0910

*And so, I have that m equals -1, and b equals -9.*0916

*So now, I can go ahead and write this as y = -x (that is mx, where m is -1) + b (and that is -9): y = -x - 9.*0922

*This is the equation for this line, written in slope-intercept form.*0942

*Example 3: Find the equation in slope-intercept form of the line passing through the point (-2,-3), and parallel to the graph of y = 3x - 7.*0951

*Slope-intercept form is y = mx + b: the first thing I need to do is to find the slope.*0964

*I am not directly given the slope; however, I am told that this line is parallel to the line described by this equation.*0972

*Parallel lines have the same slope, so if I know this slope, I know the slope for the line I am looking for an equation of.*0981

*Well, this is in slope-intercept form; therefore, y = mx + b, so I have the slope.*0994

*The slope of this line is 3, and the slope of the parallel line (which is my line) is also 3.*1005

*So now, I have y = 3x + b; and in order to fully have this in slope-intercept form, I need to have b.*1012

*Well, I have a point on this line, (-2,-3); so I am going to substitute in; I am going to let x equal -2 and y equal -3, and then solve for b.*1021

*OK, so y is -3; x is -2; and I have the slope: so 3 times -2 plus b; that gives me -3; 3 times -2 is -6; plus b.*1041

*Adding 6 to both sides, b equals 3.*1061

*I have the slope; I have the y-intercept; I can write this out in slope-intercept form: y = (slope is 3) 3x...and b is 3.*1068

*OK, so this one was a little bit more complicated than before, because they didn't directly give us the slope or two points on the line.*1083

*But what they did give us is the fact that this line is parallel to the line described by this equation.*1089

*Knowing that means that I have this slope (which is 3), and since my line is parallel, it is the same slope.*1096

*Once I have the slope, I just take the point on that line, substitute that in for x and y, and solve for b.*1104

*So, to write it in this form, I am essentially given m; I can figure out b; and this is the equation in slope-intercept form.*1111

*Find the equation in slope-intercept form of the line passing through (2,-1) and perpendicular to the graph of 2x - 3y = 6.*1122

*In order to write this in slope-intercept form, y = mx + b, I need to have the slope of this line.*1132

*I am not given the slope directly; however, I am told that it is perpendicular to the graph of this line.*1139

*Recall that the slopes of perpendicular lines are related by this equation.*1146

*So, if you have the slope of two lines that are perpendicular, and you take their product, it is equal to -1.*1152

*So, let's let the slope of this line equal m*_{1}; and m_{2} is the slope of my line, the line I am looking for.1158

*Let's go ahead and figure out m*_{1}: what I need to do is write this equation in slope-intercept form,1172

*and that will give me the slope of this line, which in turn will give me the slope of the line I am looking for.*1180

*First, I am going to subtract 2x from both sides; then, I am going to divide both sides by -3.*1190

*That is going to give me y = -2x/-3 + 6/-3.*1201

*Simplify: the negatives cancel out, and I will get 2/3x - 2.*1209

*OK, since this is in slope-intercept form, this is m; this is the slope; so m*_{1} equals 2/3.1217

*I now want to find m*_{2}, which is the slope of the perpendicular line, or the line that I am looking for.1226

*m*_{1} times m_{2} equals -1; and I am given m_{1}--I am given that m_{1} is 2/3.1235

*So, it is 2/3 times m*_{2} equals -1.1243

*If I multiply both sides by 3/2, I can isolate m*_{2}, and it is -1 times 3/2.1248

*Therefore, the slope of the line that I am looking for is -3/2.*1257

*So now, I have the slope: I have that, for my line, y = -3/2x + b.*1265

*I have slope; I need the y-intercept; well, I am also given a point on this line, which means I have an x and y value to substitute here.*1274

*So, y is -1; x is 2; the 2's cancel out, and that is going to give me -3 plus b.*1282

*I am going to add 3 to both sides, which is going to give me b = 2.*1298

*So now, I have the slope; I have the y-intercept; so I can write the equation for this line in slope-intercept form: slope is -3/2; b is 2.*1304

*Again, we approached this by realizing that the slope of the line we are looking for is related to the perpendicular line*1322

*by the equation m*_{1} times m_{2} (the product of the slopes of the perpendicular lines) = -1.1331

*So then, I went about looking for the slope of this line; rewriting it in slope-intercept form gave me y = 2/3x - 2.*1338

*So, I know that this slope is 2/3; once I have this slope, I can find my slope: m*_{1} times m_{2} equals -1.1347

*So, that is 2/3 times m*_{2} equals -1; m_{2} equals -3/2.1357

*I go back to this form y = mx + b, now knowing the slope of my line.*1363

*Substitute in x and y values and the slope to solve for b; b equals 2.*1369

*I have b; I have the slope; that allows me to write this equation in slope-intercept form.*1376

*That concludes this lesson of Educator.com.*1383

*Welcome to Educator.com.*0000

*In today's session, we are going to talk about several special functions.*0002

*The first one we are going to discuss is the ***step function**; and the step function makes a unique-looking graph.0007

*It is a function that is constant for different intervals of real numbers.*0013

*And the result is a graph that is a series of horizontal line segments, so they look like steps; and that is where the name "step function" comes from.*0018

*The best way to understand this is through an example.*0028

*So, for example, if apples were sold at a price of a dollar per pound, and the price is such that*0030

*you are charged $1 for each pound, or any part of a pound--in other words, they round up in order to determine the price;*0053

*if you had a pound and a half of apples, they are going to charge you $1 for the pound, and then another $1 for the half pound.*0070

*So, they charged you a dollar for a pound, and then any part of a pound is considered a full pound in terms of pricing.*0079

*So, if x is the number of pounds, and y is the cost, then let's see what kind of values we get and what the graph looks like.*0088

*OK, if I have .8 pounds, I am going to get charged $1: they are going to round up.*0104

*If I have a full pound, I am going to get charged $1; if I have a little bit over a pound--I have 1.2 pounds--*0113

*I am going to get charged $1 for the first pound, and then that .2 is going to be another dollar, so $2.*0124

*1.4 pounds--again, $2, and on up...2 pounds--a dollar for the first pound and a dollar for the second pound.*0135

*2.5--$2, and then the .5 is another $1, so that bumps me up to $3; until I hit 3 pounds, it is $3, also.*0145

*3.8: $1 for the first pound, $1 for the second pound, $1 for the third pound, and another $1 for that .8, so $4.*0155

*So, you can see that the function is constant for different intervals.*0166

*So, for this first interval, from a little bit above 0, all the way to 1, the y-value is constant; it is $1.*0170

*For the next interval, which is just above 1, all the way to 2, including 2, it is going to be $2.*0180

*Once I get above 2, up to and including 3, it is $3; and so on.*0188

*So, this is constant for different intervals of real numbers.*0193

*Looking at what the graph looks like: any part of a pound up to and including a dollar for that first pound is a dollar.*0198

*Now, 0 pounds of apples are going to be $0; so I am not going to include 0.*0207

*But even .1...the slightest part of a pound is going to be $1, so I put an open circle to indicate that 0 is not included;*0213

*but just above that is, all the way up to and including 1; since 1 is included (1 is also a dollar), I am going to put that as a closed circle.*0224

*So here, I have pounds; and here, I have the cost on the y-axis in dollars.*0233

*Now, once I get just above 1 (say 1.1 pounds), they are going to charge me $2--not including 1, but just above it--open circle.*0243

*All the way up to 2...at 2 pounds, I will also be charged $2.*0254

*Once I hit just above 2, I am going to be charged $3, all the way to 3 and including 3.*0260

*And on...so, you can see how this looks like a series of steps, and how this is a result of the fact that the function is constant for different intervals of x.*0270

*This function is the same for this entire interval; then it is the same for the second interval; and on.*0287

*So, this is a step function.*0292

*A second type of function that you will be working with is an ***absolute value function**.0295

*And these functions have special properties: looking first at f(x) = |x|, just the simplest case, here f(x) is just going to be the absolute value of x.*0301

*When x is 0, f(x) is also 0; when x is 1, the absolute value is 1; and so on for positive numbers.*0317

*Now, let's look at negative numbers: for -1, the absolute value is 1; -2--the absolute value of x is 2; and on.*0329

*The result is a certain shape of graph: when x is 0, f(x) is 0; x is 1, f(x) is 1; x is 2, f(x) is 2; and on.*0343

*Now, for negative numbers: -1, f(x) is 1; -2, f(x) is 2; -3, f(x) is 3; and it is going to continue on like that.*0358

*So, absolute value graphs are v-shaped; so we are going to end up with a v-shaped graph.*0372

*Depending on the function, the graph can be shifted up, or it can be shifted to the right or to the left; and let's see how that could happen.*0385

*Let's now let f(x) equal (here f(x) equals |x|)...let's say f(x) equals the absolute value of x, plus 1.*0393

*OK, so we are given x, and the absolute value of x is 0; we are adding 1 to them, so this is going to become 1.*0404

*The absolute value of x is 1; 1 + 1 is 2; 3; 4; the absolute value of -1 is 1; 1 + 1 is 2.*0412

*The absolute value of -2 is 2; add one to that--it is 3; and add 1 to 3 to get 4.*0425

*OK, so it is the same as this, except increased by 1: each value of the function has been increased by 1.*0432

*So now, let's see what my graph is going to look like.*0439

*Right here, I have the graph for f(x) = |x|; now, I am going to look at this graph.*0443

*When x is 0, f(x) is 1; when x is 1, f(x) is 2; when x is 2, f(x) is 3; so you can see what is happening.*0451

*And then here, I have x is 3, f(x) is 4; negative values--when x is -1, f(x) is 2; when x is -2, f(x) is 3; when x is -3, f(x) is 4.*0465

*OK, I am drawing the line through this: this is the graph of f(x) = |x| + 1.*0492

*So you see that this graph, the v-shaped graph, is simply shifted up by 1.*0509

*And again, you can also shift this from side to side; and we will see an example of that later on.*0513

*So, in the absolute value function, it is very important to find both negative and positive values of the function.*0519

*So, assign x 0; assign it some positive values; and it is very important to find what f(x) will be when x is negative,*0526

*because if I didn't--if I picked only positive values--I would end up with half of a graph.*0537

*So, to get the entire v-shape, choose negative and positive values for x.*0541

*The third special function that we are going to discuss is called a ***piecewise function**.0548

*And a piecewise function is a function that is described using two or more different expressions.*0553

*The result is a graph that consists of two or more pieces.*0560

*Just starting out with one that consists of two pieces: the notation is usually like this--one large brace on the left:*0564

*f(x) equals x + 2 for values of x that are less than 3; and f(x) equals 2x - 3 for values of x that are greater than or equal to 3.*0573

*So you see that, for different intervals of the domain, the function is defined differently.*0590

*So, it is a function that is described using two or more different expressions.*0597

*So, for the part of the domain where x is less than 3, this is the function.*0600

*For the interval of the domain that is greater than or equal to 3, this is the function that I am going to use.*0605

*Let's see what happens: let's first use this part of the function where f(x), or y, equals x + 2.*0613

*And for the domain, remember that x is going to be less than 3.*0625

*So, I will go ahead and start out with 2: when x is 2, f(x) (or y) is 4.*0630

*When x is 1 (I have to remain at x-values less than 3), f(x) is 3; when x is 0, f(x) is 2.*0636

*Just picking a negative number: when x is -4, f(x) is -2.*0648

*I am going to graph that here; OK, when x is 2, f(x), or y, is 4; x is 1, y is 3; x is 0, y is 2; x is -4, y is -2.*0656

*OK, now, this is for values of x that are less than 3; 3 is about here.*0689

*Therefore, anything just below 3, but not including 3, is going to be part of this graph.*0698

*So, we are going to use an open circle here to indicate that 3 is not going to be included as part of this function--the domain of this function.*0703

*So, it is going to begin at just below 3 and continue on indefinitely; that is the first piece of the graph.*0716

*The second piece of the graph is for x such that x is greater than or equal to 3; and here, y is going to be 2x - 3.*0723

*So, it is greater than or equal to 3, so I am first going to let x equal 3.*0735

*3 times 2 is 6, minus 3 is 3; getting larger--when x is 4, it is 2 times 4 is 8, minus 3 is 5; when x is 5, 5 times 2 is 10, minus 3 is 7.*0738

*Now, this does include 3--this section of the graph--this piece; so, when x is 3, y is 3, right here.*0755

*When x is 4, y is 5, right here; when x is 5 (let me shift that over just a bit), y is 7; OK.*0768

*Now, if you look, this actually did end up including all possible values of x (all real numbers),*0801

*because when x is less than 3, I use this function; and then, as soon as x becomes 3 or greater, I shift to this other function.*0807

*So, you can see how there are two pieces to the graph; and you actually can have situations where there are more than 2.*0816

*You could be given, say, f(x) is 4x + 3, 2x + 7, and x - 1, and then given limits on the domain for each of those.*0822

*So, there are at least two pieces; however, there can be more.*0836

*In Example 1, we have a greatest integer function.*0840

*Before we start working in this, let's just review what we mean by the greatest integer function.*0845

*So, when you see this notation with the brackets, let's say that you have a number in here, such as 4.7.*0848

*What this is saying is that this value is equal to the greatest integer that is less than or equal to 4.7; so that is 4.*0855

*It is the greatest integer less than or equal to whatever is in here; if it was 2.8, it would be 2.*0866

*Be careful with negative numbers: let's say I have -3.2--the temptation is to say, "Oh, that is equal to -3";*0875

*but if you look at it on the number line, -3.2 is right about here; OK, so if I have -3.2,*0882

*and I am trying to find the greatest integer that is less than or equal to 3.2, it is going to have to be something over here--smaller.*0893

*So, it is actually going to be -4; so just be careful when you are working with negative numbers.*0903

*Whatever is inside here--whatever that value is--the function is equal to the greatest integer that is less than or equal to this value in here.*0907

*Understanding that, you can then find the graph; so let's find a bunch of points for this,*0920

*so we make sure we know what is going to happen with various situations.*0927

*When x is 0, this inside here is going to be 2; the greatest integer less than or equal to 2 is 2.*0935

*When x is .6, then you are going to get 2.6 in here; the greatest integer less than or equal to 2.6 is 2.*0945

*.8--I get 2.8; again, I round down to 2.*0953

*All right, so when I hit 1, 1 plus 2 is 3, and the greatest integer less than or equal to 3 is 3.*0958

*Slightly above 1: that is going to give me 1.2 + 2 is 3.2; the greatest integer less than or equal to 3.2 is also 3.*0967

*OK, so you can get the idea of what this is going to look like.*0976

*And that continues on; and then, when we hit 2, 2 + 2 is 4; the greatest integer is going to be 4.*0980

*For negative numbers: let's take -.5: -.5 and 2 is 1.5; the greatest integer less than or equal to 1.5 is going to be 1.*0988

*Now, notice: I have a negative number for x, but this did not come out to be a negative number; so that is different from the case I was discussing there.*1005

*Let's go a little bit bigger--let's say -3: -3 and 2 is -1, and that is going to be -1.*1014

*Let's say I take -3.5: -3.5 and 2 is going to equal -1.5: again, just thinking about that to make sure you have it straight,*1024

*I have -1.5; so I have 0; I have -1; I have -1.5; I have -2; the greatest integer less than or equal to this is actually -2.*1037

*OK, now plotting this out: when x is 0, f(x) is 2; when x is slightly above 0 (it's .6), f(x) is 2; .8--it is 2, all the way up until I hit 1.*1053

*At 1, f(x) becomes 3; therefore, 1 is not included in this interval.*1075

*So, you can already see that this is going to be a step function, because we have intervals.*1082

*For different intervals of the domain, we have that same value for the range.*1089

*All right, for values between 1 and 2, f(x) will be 3; once we hit 2, I have to do an open circle, because at 2, the value for f(x) jumps up to 4.*1095

*OK, so you can see what this is going to look like; and that pattern is just going to continue.*1112

*Let's look over here at negative numbers: when x is slightly less than 0, then you are going to end up with an f(x) that is 1.*1116

*So, for values slightly less than 0, but not including 0, this is what you are going to end up with.*1137

*OK, looking, say, when x is -3: when x is -3, f(x) will be -1.*1145

*But when we go slightly more negative than that, when x is -3.5, f(x) is going to be -2; it is going to be down here.*1161

*So, the steps on this side are going to have the open circle on the right.*1175

*And I am going to jump down, and it is not going to include -2, because -2 and 2 is 0;*1184

*so -2 is going to be right here for the x-value, and the f(x) will be 0.*1193

*But as soon as I get to a little bit bigger than -2, the greatest integer is going to be down here.*1199

*OK, and so, we continue on like that with the steps; and you can see how this is a step function.*1207

*You just have to be very careful and pick multiple points until you can see the pattern*1220

*where for a certain interval of the domain, the range is a particular value.*1225

*OK, so that was a step function, and it involved the greatest integer function.*1234

*Example 2: now we are working with absolute value.*1240

*g(x) equals the absolute value of x, minus 3.*1244

*And we already know that the shape of this graph is going to be in a v.*1248

*But we don't know exactly where that v is going to land, so let's plot it out.*1252

*When x is 0, the absolute value of x is 0; minus 3--that gives me -3.*1258

*When x is 1, the absolute value is 1; minus 3...g(x) is -2.*1264

*When x is 2, the absolute value is 2; minus 3 is going to give me -1.*1269

*Now, let's pick some negative numbers for x, because that is really important to do with an absolute value graph.*1277

*When x is -1, the absolute value is 1, minus 3 gives me -2; you can already see that my v shape is going to occur.*1283

*When x is -2, the absolute value is 2; minus 3 is -1.*1291

*The absolute value of -3 is 3; minus 3 is 0; so this is enough to go ahead and plot.*1299

*x is 0; g(x) is -3; x is 1, g(x) is -2; x is 2, g(x) is -1; over here with the negative values,*1305

*when x is -1, g(x) is -2; when x is -2, g(x) is -3; when x is -3, g(x) is 0.*1316

*So, you can see that I have a v-shaped graph, and compared with my graph that would look like this,*1325

*that would have the v starting right here, it has actually shifted down by 3; that is an absolute value function.*1338

*Here you can see that you are given a piecewise function, because there are two different pieces.*1348

*And this could also be written in this notation.*1357

*There are two different sections to the graph; and we see that the function is defined differently for different intervals of the domain.*1360

*Starting with if x is greater than 2 (this is going to be for x-values where x is greater than 2): f(x) is going to be x + 1.*1371

*When x is 3, f(x) is 4; when x is 4, f(x) is 5; when x is 5, f(x) is 6; OK.*1386

*When x is 3, f(x) is 4; when x is 4, f(x) is 5; and it is going to go on up.*1404

*And that is going to go all the way, until just greater than 2.*1416

*2 is not going to be included on this graph, because it is a strict inequality; so I am going to use an open circle, and this is going to continue on.*1422

*Now, for x less than or equal to 2, I have a different situation: I am looking at f(x) is -2x.*1430

*OK, so when x is 2, 2 time -2 is -4; when x is 1, 1 times -2 is -2; when x is 0, f(x) is 0; when x is -2, that is -2 times -2, which is positive 4.*1443

*So, starting with x is 2: when x is 2, f(x) is -4; and that is including the 2.*1464

*When x is 1 (these are values less than or equal to 2, so I am getting smaller), f(x) is -2.*1476

*x is 0; f(x) is also 0; when x is -2, f(x) is up here at 4; OK, so I have a steep line going right through here.*1494

*So, you can see: this is a piecewise function consisting of two pieces; and here, one picks up where the other leaves off.*1509

*For values greater than 2, this is my graph; for values of x less than or equal to 2, this is my graph; so this is a piecewise function.*1515

*OK, this time, in Example 4, we have both greatest integer and absolute value in this function.*1525

*Recall that, for the greatest integer function, what that is saying is that whatever is inside this bracket--let's say it's 1.2--*1532

*it is asking for the greatest integer less than or equal to 1.2; in that case, this would be 1.*1540

*Or if I had 4.8, it would be 4.*1545

*For negative numbers, like -3.2, the greatest integer less than or equal to -3.2 is -4.*1551

*OK, now, since this is a bit complicated, it is helpful just to take it in stages.*1560

*So, I am going to look first at what the greatest integer of x is; and then, I am going to look for the absolute value of what that is.*1567

*If x is .2, the greatest integer less than or equal to .2 is 0; the absolute value of 0 is 0; so this is the function that we are looking for.*1578

*And the same would hold true of .5: round down to 0; the absolute value is 0.*1590

*When we hit 1, the greatest integer less than or equal to 1 is 1, and the absolute value of that is 1.*1596

*1.2: again, we are going to go down to the greatest integer that is less than or equal to 1.2, which is 1; and the absolute value is 1.*1608

*The same for 1.8, and all the way up until 2; once we hit 2, the greatest integer less than or equal to 2 is 2; the absolute value is 2.*1621

*So, that is working with positive numbers, greatest integer, and the absolute value; it is the same; OK.*1629

*So, let's go to negative: for -.4, the greatest integer that is less than or equal to -.4...I am looking, and I have 0, and 1,*1636

*and -1, and -.4 is about here; so I am going to go down to -1; the absolute value of that is 1.*1650

*Here you can see that the greatest integer is not the same as the absolute value.*1659

*Or for -1, the greatest integer less than or equal to -1 is -1; the absolute value is 1.*1664

*-1.8: the greatest integer that is less than or equal to -1.8...I am going to go down to -2; and the absolute value is 2.*1674

*For -2, the greatest integer less than or equal to -2 is -2; the absolute value is 2.*1686

*So, you see that there are intervals here--intervals of the domain end up with the same value for the function.*1694

*So, I am going to have a step function.*1704

*But remember that absolute value graphs are v-shaped, so I am going to end up with a v-shaped step function.*1709

*Let's plot these out: for 0, the greatest integer of 0 would be 0, and then the absolute value would be 0.*1715

*So, with 0, we are going to include it; and for all values up to but not including 1, the function is going to equal 0.*1726

*Once we get to 1, I have an open circle, because it is not included.*1736

*When x is 1, f(x) is 1; so I am going to jump up here.*1740

*All the values between 1 and 2, but not including 2, will have an f(x), or a y-value, that is 1.*1746

*As soon as I hit 2, open circle: I am going to jump up, and once I hit 2, f(x) is 2, all the way up to, but not including, 3.*1754

*And it is going to go on that way: and you see now, we have the step function, and it is v-shaped like absolute value.*1771

*Let's look over here on the negative side of things.*1778

*For -.4, somewhere in here, it is going to equal 1; -1 is also equal to 1; so here, on the left side, I have a closed circle, and an open circle on the right.*1782

*It is the opposite of what I had over here.*1799

*When I get to less than -1, my value for f(x) is going to jump up to 2; this is a closed circle;*1802

*I get slightly less than, but not including, -1; it is going to jump up to 2.*1815

*-2: my value is also 2, and everything in between; and then, when I get to just slightly more negative than -2, like -2.1, it is going to jump up to 3.*1822

*You can see how this is v-shaped, and it is a step function.*1841

*The step function comes from it being the greatest integer function; the v shape comes from that absolute value.*1847

*And you also just had to be careful how you are doing the open and the closed circles; OK.*1853

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be graphing inequalities.*0002

*Recall that, in order to graph a linear inequality, you have to take a series of steps.*0008

*And the graph of a linear inequality is a shaded region in the coordinate plane.*0014

*And those shaded regions are known as half-planes; the boundary of the region, or the half-plane, is a straight line,*0019

*which is the graph of the corresponding linear function.*0027

*The region, or the half-plane, containing the solution set can be determined by using a test point.*0032

*OK, so the steps in order to graph a linear inequality are to first graph the corresponding linear function; and that is a straight line.*0040

*And that line is dashed if the inequality you are working with is a strict inequality (less than or greater than).*0058

*The line is solid if you are dealing with less than or equal to, or greater than or equal to; and we will talk in a minute about what that means.*0071

*After you have graphed the line, then you end up with the coordinate plane divided into two half-planes.*0079

*You use a test point to determine the region containing the solution set.*0097

*So, I am going to go ahead and do an example now: but if you need more review on this, check out the Algebra I video on this topic.*0104

*All right, the easiest way to understand is to look at an example: let's look at the inequality y - 3x < -2.*0112

*My first step is to graph the corresponding linear function, which would be y - 3x = -2.*0123

*So, I am going to go ahead and find some x-values and some y-values--some coordinate pairs on this line.*0132

*When x is 0, y is -2; so that gives me the y-intercept.*0140

*When x is 1, this is going to give me -3; add 3 to both sides--y will be 1.*0145

*When x is 2, -3 times 2 is -6; add that to both sides, and y is 4.*0155

*All right, I have enough to graph the line: x is 0, y is -2--the y-intercept; then I have another point at (1,1) and another point at (2,4).*0164

*Because this is a strict inequality, I am going to use a dashed line.*0177

*This line divides the coordinate plane into two half-planes: an upper and lower half-plane.*0187

*One of these half-planes contains the solution set for the inequality; the way I figure out which one is to use a test point.*0193

*Pick a point that is not on the boundary line, and that is easy to work with.*0201

*(0,0) is very easy to work with; so I am going to take my test point, and I am going to go back to my inequality and plug it in.*0206

*I am going to see what happens: this gives me 0 - 0 < -2; 0 is less than -2--that is not true.*0216

*What this means is that this is not part of the solution set.*0230

*If this is not part of the solution set, it is in the half-plan that is not contained in the solution set.*0234

*Therefore, the solution set is actually in the lower half-plane.*0240

*And you could verify that by checking a point down here.*0245

*The idea is to graph the corresponding linear function to find the boundary line.*0249

*Now, I mentioned that this is a strict inequality, so I am using a dashed line.*0253

*And what that means is that the boundary line is not part of the solution set.*0256

*If it were a solid line (because I had less than or equal to), then the boundary line would be part of the solution set.*0262

*But actually, in this, the solution set is just below that boundary line.*0268

*I determined which half-plane was correct by using a test point.*0272

*Since the test point ended up giving me an inequality that is not true, that is not valid,*0276

*I knew that that test point did not lie in the half-plane of the solution set, so I shaded in the lower half-plane.*0281

*Graphing absolute value inequalities: the procedure for these inequalities is similar to the procedure for linear inequalities.*0290

*So, we are going to use the same technique.*0298

*Recall that the technique involved first graphing the line for the corresponding equation.*0302

*Graph the corresponding equation: in this case, it is going to be an absolute value equation.*0309

*And then, use a test point to determine the half-plane containing the solution set for the inequality.*0318

*All right, let's take an example: y ≤ |x-3|.*0338

*The corresponding equation would be y = |x-3|, so I need to graph that to find my boundary.*0347

*So, some values for x and y: when x is -2, -2 minus 3 is -5; the absolute value is 5.*0357

*When x is -1, this becomes -4; the absolute value is 4; when x is 0, this becomes -3; the absolute value is 3.*0368

*Let's jump up to 3: 3 minus 3 is 0--the absolute value is 0.*0381

*4 minus 3 is 1; the absolute value is 1; 5 minus 3 is 2--the absolute value is 2.*0389

*OK, plotting these points: first, (-2,5), (-1,4), (0,3), (3,0), (4,1), (5,2); OK, I have my typical v-shaped graph of absolute value.*0399

*And you notice that I made this a solid line; and the reason that is a solid line is because I had less than or equal to.*0439

*So, this boundary is actually going to be part of the solution set.*0447

*OK, so I graphed the corresponding equation, and I determined that it is a solid line.*0451

*Now, I need to determine if the solution set is down here, or if it is above the boundary; is it above or below the boundary line?*0455

*So, I am going to use a test point of (0,0): that is going to be my test point.*0462

*I go back to the inequality, and I see what happens when I put my test point values in there.*0468

*This gives me 0 ≤ |-3|; that would say that 0 is less than or equal to 3; that is true.*0477

*Since that is true, what that tells me is that this is part of the solution set.*0489

*(0,0) is part of the solution set, and that means that this lower half-plane where (0,0) is, my test point, contains the solution set.*0495

*So, as I mentioned, this is very similar to the technique for graphing linear inequalities.*0505

*First, graph the corresponding equation; and use a solid line if it is less than or greater than or equal to.*0510

*Use a dashed line if it is a strict inequality (just greater than or less than).*0520

*So then, once you have your boundary line, find a test point that is off the boundary line, and not so close to it as to cause any confusion.*0525

*And choose one that is simple: (0,0), (1,1)--very easy to work with.*0533

*Plug that into the absolute value inequality; and if you come out with a true inequality,*0538

*like this one, you know that you are in the correct area with the solution set.*0544

*If you come out with a statement or inequality that is not valid, you shade in the other region.*0548

*OK, the first example: we have an inequality that we are asked to graph: y ≤ 2x - 3.*0557

*My first step is going to be to graph the corresponding linear equation, in order to determine the boundary.*0567

*Some values for x and y: when x is 0, y is -3; when x is 1, 1 times 2 is 2, minus 3 is -1.*0581

*When x is 2, 2 times 2 is 4, minus 3 is 1.*0594

*OK, so plotting this out: (0,-3), (1,-1), (2,1); I have plotted this boundary line, and I look up here,*0599

*and I see that it is less than or equal to; so I am going to make this a solid line.*0615

*It is going to be a solid line: this line will be part of the solution set.*0621

*So, first I graph the boundary; next, I am going to look at a test point.*0626

*OK, I can again use this very simple test point of (0,0), the origin.*0638

*Going back to the inequality to look at my test point, y ≤ 2x -3; the test point is (0,0), so y is 0; x is 0.*0646

*This is saying that 0 is less than or equal to -3; that is not true--this is not in the solution set.*0664

*Well, the origin is up here; this means that this whole area is not part of the solution set.*0678

*So, I am going to go to the lower half-plane and shade that in to indicate that this shaded region, including the boundary line, is the solution set.*0684

*Again, the first step is to take the corresponding linear equation and graph it.*0692

*The second step: find a test point that is easy to work with and away from the boundary line, and plug that in.*0698

*If you come up with an inequality that is not true, like this, shade the other half.*0705

*If it is true, shade the half containing the test point.*0711

*Example 2: We are graphing another inequality, 2x + y > 6.*0716

*First, I am going to find the corresponding equation: 2x + y = 6.*0724

*I am going to keep this simple: I am going to use the intercept method in order to graph this.*0733

*I am going to let x equal 0 to find the y-intercept: when x is 0, y is 6.*0738

*Then, I am going to let y equal 0 to find the x-intercept: when y is 0, then I am going to end up with 2x + 0 = 6, or 2x = 6; x = 3.*0745

*So, I have my x and y intercepts; it is going to give me...when x is 0, y is 6, right about there; when x is 3, y is 0.*0759

*OK, then I look, and I have a strict inequality, so I am going to have a dashed line;*0771

*this line is going to be dashed, and is not going to be part of the solution set.*0777

*So, I found my boundary line; and that is my first step; my second step is going to be to work with my test point.*0788

*The test point is right here: it is the origin, (0,0).*0798

*Going back and looking at the inequality: 2 times 0, plus 0, is greater than 6; 0 plus 0 is greater than 6.*0803

*Well, 0 is not greater than 6, so this is not true--not in the solution set.*0817

*The origin is not part of the solution set, so the lower half-plane is not the correct half-plane; I am going to shade, instead, this upper half-plane.*0830

*Again, graph the linear equation corresponding to the inequality.*0839

*Since this is strict inequality, plot that line using a dashed line, since this boundary is not part of the solution set.*0844

*Find a test point; the origin is a really good one, as long as it is not on the boundary line or very close to the boundary line.*0853

*Substitute (0,0) into the inequality; and you come up with an inequality that is not true.*0859

*Example 3: Graph 3x - 4 > y.*0870

*First, graph the boundary line: the corresponding equation is 3x - 4 = y.*0876

*To graph that, I am going to need some x and y values; when x is 0, y is -4; when x is 1, that gives me 3 - 4 = y, so y equals -1.*0889

*One more value: when x is 2, y is 2; OK.*0908

*So, I have enough information to plot my line: when x is 0, y is -4; (1,-1), (2,2).*0918

*The next issue: strict inequality--that means this is a dashed line.*0935

*OK, so now, I have this boundary; I have an upper and a lower half-plane; and I need to figure out where my solution set is.*0948

*The origin is well away from the boundary line, so I can use that as my test point.*0956

*So, test point is (0,0): the inequality is 3x - 4 > y; and this gives me 0 - 4 > 0, which comes out to -4 > 0.*0963

*And that is, of course, not true; this is not in the solution set, because this inequality did not hold true.*0986

*Therefore, I am going to go to the other half-plane, and I am going to shade that in.*0997

*I graphed the boundary line using a dashed line, since it is a strict inequality.*1007

*And then, I took a test point, substituted those values into the inequality, and determined that the origin (0,0) is not part of the solution set.*1011

*So, my other half-plane contains the solution set.*1022

*Example 4: we are working with an absolute value inequality where y is less than the absolute value of x - 2, plus 1.*1027

*I am starting out with just thinking that I expect this to look like a v, because absolute value graphs are v-shaped.*1036

*I am graphing the corresponding equation, y = |x - 2| + 1.*1046

*And with absolute value inequalities or equations, I like to get more values,*1051

*because I want to make sure that I don't end up missing half of my v.*1055

*So, when x is -2, that is going to give me -4; the absolute value of that is 4; plus one is 5.*1061

*-1 minus 2 is -3; the absolute value is 3, plus one is 4.*1070

*0 minus 2 is -2; the absolute value is 2, plus one is 3.*1077

*Now, some positive numbers: 1 minus 2 is -1; the absolute value is 1; plus one is 2.*1085

*2 minus 2 is 0, plus one is 1; 3 minus 2 is 1; the absolute value is 1, plus one is 2.*1094

*4 minus 2 is 2; the absolute value is 2, plus one is 3; and now I have enough values to work with.*1111

*So, (-2,5), (-1,4), (0,3), (1,2), (2,1), (3,2)...so you see the v forming, so I know I have enough points,*1119

*because I have both parts of my v...(4,3).*1145

*It's a strict inequality, so I am using a dashed line.*1150

*And you see that it is the typical v shape, but it is shifted over.*1161

*And I have a -2 here, and so this is actually shifted over to the right by 2; and plus 1, so it is shifted up by 1.*1165

*I found the boundary; again, I am fortunate--I can easily use (0,0) as the test point--the origin.*1173

*Going back to the inequality: this is going to be 0 < |0 - 2| + 1.*1187

*So, that is 0 < |-2| (the absolute value of -2 is 2) + 1, so this gives me 0 < 3.*1201

*0 actually is less than 3; therefore, the origin is part of the solution set, so this is a true statement, and the origin is part of the solution set.*1219

*(0,0) is part of the solution set, and that means that I am in the correct half-plane, and I am going to go ahead and shade that in.*1240

*OK, so we handled this just with a very similar technique to linear inequalities.*1257

*Graphing out the absolute value equation, I got my v-shaped graph with a dashed line for the strict inequality.*1263

*I used a test point, (0,0), substituted that into the inequality, and found that I had a true statement--a true inequality, 0 < 3.*1270

*Since that is true, I knew that the origin lay within the region that contains the solution set.*1279

*So, I went ahead and shaded that region to note that my solution set is located there.*1289

*That concludes this lesson of Educator.com; see you next lesson!*1297

*Welcome to Educator.com.*0000

*In today's Algebra II lesson, we will be discussing solving systems of equations by graphing.*0002

*Recall that a ***system of equations**, for our purposes, is defined as two equations and two variables.0009

*Later on, we will be looking at systems of equations involving more than two variables.*0016

*But right now, we are just going to stick to this definition.*0021

*For example, a system of equations could be something such as 2x - y = 7, when it is considered along with another equation, such as x + 3y = -4.*0025

*And a solution to a system of equations would be a value for x and a value for y--a set of values that satisfies both of the equations.*0040

*The first technique we are going to discuss in solving systems of equations is solving by graphing.*0053

*In addition, later on, we will be talking about how these systems can be solved algebraically.*0059

*So, one way to solve is to graph each equation: the solution for the system is the point of intersection of the two graphs.*0065

*And we will do some examples in a minute; but for now, imagine that you are given an equation,*0075

*and you figure out some points, or one of the points in the slope, and it turns out that this is the graph of the line described by the equation.*0081

*And you are given another equation as part of that system; and you go ahead and graph that one out.*0098

*And it comes out to a line that looks like this.*0105

*The solution for the system is the point of intersection, so this is the solution.*0110

*The x- and y-coordinate for this point are the solution.*0117

*Now, here you can see a weakness in this method: and that is that, if the solution doesn't land right on an integer, it is difficult to get an exact value.*0121

*And that is one advantage to the algebraic methods.*0136

*But if you graph carefully, and the solution does land on, say, (3,4), then you could get an exact solution.*0138

*There are several types of systems of equations; we call a system ***independent** if it has exactly one solution.0150

*For example, just as I showed you in the previous slide, sometimes you will graph out a system, and usually,*0159

*in the problems you will be doing, you will see a single point of intersection, which is the solution--the single solution.*0167

*And this set of lines describes an independent system.*0175

*Other times, you will go along, and you will graph out the system of equations.*0185

*You have graphed the first equation, and perhaps the graph looks like this.*0191

*Then, you go to graph the second equation, and it turns out that it describes the same line.*0196

*So, if the system of equations--both equations within that system describe the same line, then there would be an infinite number of solutions.*0202

*And the reason is that every single point along this line, every single set of (x,y) coordinates is an intersection of the two systems.*0216

*The line...every single point along this line (and there is an infinite number of points along the line)*0226

*is the set of solutions, the set of (x,y) values where the two lines intersect, since they are the same line.*0233

*Here, we call this a ***dependent** system; here we have one solution; here an infinite number of solutions; and this is a dependent system.0239

*The third possibility is that there are no solutions; and we call this an ***inconsistent** system.0254

*So, thinking about a situation where there would not be any solution, where there is no point of intersection: it would be a set of parallel lines.*0263

*Drawing this over here for clarity: here is my x and y axis; and if I graphed the first line, the first linear equation,*0273

*and then I went ahead and graphed the second one, and those two turned out to have the same slope, those would be parallel lines.*0284

*They are never going to intersect; therefore, we can see on the graph: there is no solution.*0292

*So, no solution--that is an inconsistent system.*0298

*So, there are three possibilities: one solution--intersection at one point; an infinite number of solutions--*0306

*all the points along the line; or no solution, because it is a set of parallel lines that do not intersect.*0312

*The first example gives us a system of equations with two variables: x + y = 3 and x - y = 1.*0322

*So, graphing this first line, I am going to start with x + y = 3; and I find a few x and y values, some ordered pairs, so I can do the graphing.*0331

*First, I am going to let x equal 0; well, if x is 0, I can see easily that y is 3.*0343

*So, that gives me the y-intercept; next, I am going to let y equal 0 to find the x-intercept.*0351

*So, this would be 0; x would be 3; and just one more point to make it a better graph*0361

*(because it is especially important when you are not just graphing one line--you are actually looking*0367

*for a solution to a system of equations--to have a really good graph, so you can find a point of intersection accurately):*0372

*so, when x is 1, x plus y equals 3; so when x is 1, y equals 2.*0378

*So, I am going to graph this line: first, I have the y-intercept at 3; I have the x-intercept at 3; and another point in between those--x is 1; y is 2.*0387

*And this gives me a better sense of the slope of the line than if I had just done two points.*0399

*OK, the second equation--graphing that: x - y =1: again, I am finding a few values for x and y--a few sets of values.*0406

*When x is 0, that would give me 0 - y = 1, so - y equals 1; therefore, y is -1; the y-intercept is -1.*0422

*Now, finding the x-intercept, let y equal 0; if y is 0, x is 1.*0434

*One final point: I am going to just let x equal 2; so that would be 2 - y = 1; and that is going to give me -y = -1, so y = 1.*0442

*Again, I have three sets of points to graph.*0457

*So, the y-intercept is at -1; the x-intercept is at 1; and one more point: when x is 2, y is 1.*0460

*OK, drawing this line, I can immediately see that I have a single point of intersection, right here.*0476

*This is the solution; and this occurs at (2,1).*0487

*So, the solution for this is x = 2, y = 1.*0493

*And therefore, this is an independent system, since it has one solution.*0501

*OK, and you can always check your work by substituting x and y values back into these equations.*0515

*I have x + y = 3, so that is 2 + 1 = 3; and that checks out.*0528

*I have x - y = 1, so 2 - 1 = 1, and that checks out; so you can easily check and make sure that both of these are valid solutions for both of the equations.*0534

*The second example is a little bit more complicated, but using the same system.*0551

*And this time, instead of just finding (x,y) values, I am going to graph these out by putting both equations into slope-intercept form.*0557

*So, the first one is 2x - 4y = 12, so I am going to subtract 2x from both sides, and then I am going to divide both sides by -4.*0565

*And that is going to give me y = 1/2x - 3.*0577

*Slope-intercept form is very helpful when you are graphing: this is -2, -4, -6, -8, -10, 2, 4, 6, 8, 10.*0582

*OK, here I have a y-intercept at -3; and I know the slope--the slope is 1/2.*0597

*That tells me, when I increase y by 1, I am going to increase x by 2; if I increase y by another 1, I am going to increase x by 2.*0606

*So, you can see the slope of this line right here; that gives me enough to work with.*0620

*I am going to do the same with the second equation, 4x - y = 10, putting this into slope-intercept form, y = mx + b,*0638

*subtracting 4x from both sides, and then dividing both sides by -1 to give me y = 4x - 10.*0646

*So here, I have a y-intercept at -10 and a slope of 4; I am going to increase y by 4 (that is 2, 4), and increase x by 1.*0656

*Increase y by 2, 4; increase x by 1; increase y by 4 (2, 4); increase x by 1; OK.*0669

*So, this is a much steeper line; the slope is much steeper--the slope of 4; so I am going to go ahead and draw it like that.*0687

*And the point of intersection is right here, and again, we have an independent system with a single solution.*0695

*And it is at (2,-2), so that is my point of intersection: x is 2; y is -2.*0708

*Again, you could always check your answers by substituting either or both equations with these x and y values, and ensuring that the solutions are valid.*0723

*So again, I have an independent system; it has one solution--one set of valid solutions.*0733

*OK, this next example: again, solve by graphing; and my approach is going to be to put these into y-intercept form and use that to graph.*0748

*First, 2x - 3y = 6, so -3y = -2x + 6; divide both sides by -3; that is going to give me...-2/-3 is 2/3x, and then 6/-3 is -2.*0758

*So, this line has a y-intercept of -2 and a slope of 2/3; so increase y by 2; increase x by 3.*0781

*OK, that is my first equation; the second equation--again, putting it into the slope-intercept form, which is an easy way of graphing:*0799

*This gives me 6y = 4x - 12; dividing both sides by 6 gives me 4/6x - 2.*0808

*And I am going to go ahead and simplify this to 2/3x - 2.*0823

*So, you may already see that these are the same equation--these are going to describe the same line.*0828

*Even if you didn't notice it right away, when you start graphing, you are going to see: the y-intercept is -2; the slope is 2/3.*0835

*You are going to end up with the same line.*0842

*Because this is the same line, they intersect at every single point; these two equations--their graphs intersect at every single point,*0844

*which is an infinite number of points, along the line; they don't intersect everywhere, but everywhere on this line.*0852

*Therefore, this is a dependent system, and there is an infinite number of solutions; all points along this line are solutions.*0857

*And this is known as a dependent system.*0871

*So initially, you looked at this; you might not have recognized that these are actually describing the same line.*0877

*But once you started to plot it, either by putting it in the y-intercept form or by finding points along these lines,*0882

*you would have quickly realized that this is the same line.*0890

*In this last example, again, there is a system of equations that we need to solve by graphing.*0896

*Again, using the method of slope-intercept form for graphing: 4x - 2y = 8; subtract 4x from both sides, and then divide both sides by -2 to give me y = 2x - 4.*0905

*OK, the y-intercept is -4; the slope is 2; increase y by 2; increase x by 1; increase y by 2; increase x by 1; and so on.*0925

*I have three points; that is plenty to graph this line...so that is my first line.*0942

*The second line: again, slope-intercept form: add 6x to both sides--it gives you 3y = 6x + 12.*0947

*Now, divide both sides by 3 to get y = 2x + 4.*0961

*So here, I have a y-intercept of 4 and a slope of 2; so increase x by 2; increase y by 1.*0968

*So, I am going to get a line looking like this; and what you may quickly realize is that these are parallel lines.*0978

*One thing you could have noted is that the slope, m, equals 2 for this line, and it equals 2 for this line.*0991

*Since these are parallel lines, they are never going to intersect; and this is an inconsistent system, and there is no solution.*0998

*And what we say here is: you could say the solution is just the empty set; there is no solution.*1008

*This is described as an inconsistent system.*1016

*OK, that concludes this lesson of Educator.com, describing graphing to solve systems of equations.*1024

*And I will see you next time.*1032

*Welcome to Educator.com.*0000

*In today's lesson, we will be talking about solving systems of equations algebraically.*0002

*In the previous lesson, we talked about solving systems of equations by graphing.*0008

*However, that method has some limitations; therefore, we are going to talk about a few other methods of solving.*0013

*And this is some review from Algebra I; so again, if you need more detail--more review on these concepts--check out our Educator.com Algebra I series.*0020

*First, we are going to review solving by substitution, first jotting down a system of equations:*0030

*2x + 3y = 12, and the second equation in the system is x + 4y = 1.*0036

*In this method, you are going to solve one equation for one variable in terms of the other variable.*0047

*Then, substitute the expression for the variable in the other equation.*0053

*Let's look at what this means: first step: solve one equation for one variable in terms of the other one.*0059

*The easiest thing to do is to find a situation where you have a variable that has a coefficient of 1; and I have that right here.*0065

*So then, I can solve for x in terms of y pretty easily.*0071

*So, x + 4y = 1, so I am going to solve for x in terms of y; this gives me x = -4y + 1.*0075

*The second step is to substitute this expression for the variable in the other equation.*0088

*So, I am going to substitute -4y + 1 for x in the first equation; and that is going to create*0094

*an equation that has only one variable, and then I will be able to solve that.*0101

*It is important that you go back into the other equation to substitute in.*0105

*This is going to give me 2(-4y + 1) + 3y = 12; then, I can just solve for y.*0111

*2 times -4y; that is -8y, plus 2 times 1--that is just 2--plus 3y, equals 12.*0124

*Combine like terms: -8y and 3y gives me -5y, plus 2 equals 12; subtract 2 from both sides to get -5y = 10.*0135

*Dividing both sides by -5, I get y = -2.*0150

*Once I have one variable solved, I can easily solve for the other.*0155

*So, I just go back to either one of these (and this one is easier to work with), and I am going to substitute in...I am going to let y equal -2.*0159

*x + 4(-2) = 1; this is going to give me x + -8 = 1, or x - 8 = 1; adding 8 to both sides, I get x = 9.*0170

*Here, y equals -2 and x equals 9.*0192

*This method of substitution works really well when the coefficient of one of the variables is 1.*0197

*So, use substitution when a variable in one of the equations has a coefficient of 1.*0203

*So again, we looked at this system of equations, saw that this variable had a coefficient of 1,*0232

*then solved for this variable x in terms of the other variable y to get x = -4y + 1.*0240

*And then, I went ahead and substituted this 4x in the other equation; that gave me one equation with one variable, and I can solve for y.*0247

*Once you have y, you can substitute that value into either equation and then solve for x.*0260

*The second method is solving by elimination: in elimination, you add or subtract the two equations to eliminate one of the variables from the resulting equation.*0267

*And this system works well when you have variables, either the two x's or the two y's, that have either the same coefficient or opposite coefficients.*0277

*By "opposite coefficients," I mean the same number with opposite signs, such as 2 and -2 or 3 and -3.*0286

*For example, if you were asked to solve this system, you look and see that there is no variable with a coefficient of 1.*0293

*Therefore, substitution is not the ideal--it is not the easiest way to go.*0306

*But what you see is that you have one variable--you have y--that has the same coefficient.*0312

*So, adding is certainly not going to help me; if I add these together, I will get 6x + 10y = 8.*0318

*What I need to do is subtract, because my goal is to get one variable to drop out.*0324

*So, I am going to subtract the second equation from the first.*0328

*And just to help me keep my signs straight, I rewrite that as adding the opposite; so I am adding -4x, -5y, and -3.*0333

*OK, this is going to give me 2x - 4x is -2x; the y's are going to drop out: 5y - 5y gives me 0, so I can write + 0, but they just drop out.*0352

*And then, 5 minus 3 equals 2; this is -2x = 2; I can easily solve for x, and x equals -1.*0365

*Once I have one variable, I can go to either original equation, and then substitute in order to solve for the other variable.*0376

*I know that x equals -1, so I am going to substitute that up here.*0385

*Add 2 to both sides (that is going to give me 5y = 7), and divide both sides by 5 to get y = 7/5.*0395

*So, solving by elimination, I was able to come up with the solution that x equals -1 and y = 7/5, which will satisfy both of these equations.*0403

*Again, this works well; use elimination when the same variable (meaning both x's or both y's)*0412

*have the same or opposite coefficients (the same coefficient, but opposite signs).*0426

*Sometimes, you look, and you see that there is no variable with a coefficient of 1; and then you say, "OK, I will use elimination."*0442

*But then, you realize that none of the variables have the same or opposite coefficients.*0449

*In that case, you can still use elimination, but you are going to have to take an extra step before you do the elimination.*0453

*And the extra step is to multiply one or both equations so that one of the variables has the same or opposite coefficient in the two new equations.*0459

*So, your goal is to create a situation where one of the variables has the same or opposite coefficients.*0469

*Then, you just use elimination, as we did previously.*0476

*If you had a system of equations, 5x + 4y = 1, and 6x + 3y = 3, you see that neither of these*0480

*has a variable with a coefficient of 1, and neither set of variables has the same or opposite coefficients.*0489

*But if I look at these two, 4y and 3y, the least common multiple is 12; so what I want to do*0497

*is multiply each of these by something, in order to end up with a coefficient of 12 in front of the y.*0505

*Sometimes you will be lucky, and all you will have to do is multiply one of the equations by a number to get the opposite or same coefficient.*0511

*Other times, like this, you are going to have to multiply both.*0519

*So, for the first equation, I am going to multiply by 3; and this is going to give me 15x + 12y = 3.*0522

*The second equation I am going to multiply by 4: this is going to give me...4 times 6x is 24x + 12 y = 12.*0542

*OK, I now see that I have the same coefficient for y; so rewriting this over here, I need to subtract.*0559

*So, I am going to subtract the second equation from the first.*0576

*And to keep everything straight, I like to go about this by adding the opposite to make sure that I keep my signs straight.*0581

*So, I am going to add -24x - 12y, and then that is going to be a -12.*0589

*OK, -24x - 12y - 12: add these together: I get 15 - 24x--that is going to give me -9x; the y's drop out, and then I have 3 - 12; that is -9.*0603

*Divide both sides by -9 to get x = 1.*0618

*From here, I substitute back; I will choose this top equation, 5x + 4y = 1, and I know that x equals 1.*0622

*This is 5(1) + 4y = 1; so that is 5 + 4y = 1, or 4y = -4, so y = -1.*0631

*So, x equals 1; y equals -1.*0645

*Again, this is just the first step in elimination; and you use it in a situation like this,*0648

*where you want to use elimination, but you don't already have a set of variables with the same or opposite coefficients.*0653

*So, you figure out what the least common multiple is, and multiply one or both equations in order to achieve that.*0660

*From there, you just proceed as we did before, solving by elimination.*0667

*There are several different possibilities that can occur when you are solving systems of equations algebraically.*0675

*Usually, in the problems that you will see, what will happen is what just happened previously,*0683

*that I showed you, where you will end up with a value for x and a value for y that satisfy both equations.*0688

*However, there are times when that doesn't occur.*0695

*You can be going along, doing substitution, doing elimination, and things are going fine;*0699

*and then you end up with something like this: c = d.*0705

*Your variables drop out, and you end up with an equation that is saying that a constant is equal to a different constant--for example, 4 = 5.*0710

*Well, that is not true; so when you see this, this is an equation that is never true.*0719

*And what this tells me is that the system of equations is inconsistent.*0727

*So, this is a situation where there is no solution.*0742

*If you start seeing a situation where the constants drop out, and then you see something like 4 = 5*0752

*or 9 = 10--something that is not true--then you know you have an inconsistent system.*0758

*The other possibility is that you could have a system that is dependent, or always true.*0764

*In that case, you are going along; you are doing elimination; you are doing substitution;*0773

*and then you see that you end up with variables dropping out, and a constant that equals a constant--the same constant--for example, 3 = 3.*0778

*Well, that is always true; so if you end up with an equation--the sum or difference of the two equations--*0788

*the system of equations--that is always true, this system is dependent, and it has an infinite number of solutions.*0795

*Recall, when we were solving systems of equations by graphing: this is analogous to the situation where you would end up with the same line.*0812

*So, if you have two equations (a system of equations), you graph both equations,*0819

*and you find out that they are the same line, well, then, there is an infinite number of solutions--all points along those lines.*0823

*Here, no solution would be like parallel lines, where they never intersect; there is no solution to that system.*0829

*OK, the first example is: Solve algebraically for the system of equations.*0839

*As soon as I see that I have variables (or even one variable) with a coefficient of 1, I recognize that substitution would be a really good method to use.*0844

*So, use substitution when a variable has a coefficient of 1.*0853

*So, I am going to solve for x in terms of y: x + y = 5, so x = 5 - y.*0856

*Now, I have solved for x in terms of y; and now I am going to substitute this value into the other equation.*0867

*2x + 3y = 13: I am going to substitute this expression for x, so 2 times (5 - y), plus 3y, equals 13.*0875

*Or, 10 - 2y + 3y = 13; this is 10 + y = 13; then subtract 10 from both sides to get y = 3.*0889

*Once I have this, I can plug y into this simple equation up here: x + y = 5.*0906

*Substitute in; let y equal 3; and I get x = 2.*0916

*And this is very easy to check; you can see that, if x is 2 and y is 3, this equation holds true.*0924

*And you could do the same--plug it back into that first equation, put these values in for x and y, and just verify that these solutions are correct.*0929

*OK, so again, we are solving by substitution because you had a situation where you had a variable (actually, two variables) with a coefficient of 1.*0942

*That is the easiest method to use.*0950

*Here, I am solving this system of equations algebraically; and I don't have any variables with a coefficient of 1, so I am not going to use substitution.*0954

*But if I look; I have variables (y) that have the opposite coefficient, -3 and 3.*0963

*What that means is that, if I add these equations, the y's are going to drop out; then, I can just solve for x.*0970

*Rewrite this right here and add: I am going to add these two.*0977

*2x + 4x: that gives me 6x; the y's drop out: -3y + 3y is 0; and then 0 + 0 is 9.*0988

*I am going to divide both sides by 6; this is going to give me 9/6, and that simplifies to 3/2.*1004

*I have x = 3/2; I am going to pick either one--I will go ahead and pick the top one: 2x -3y = 0.*1011

*And let x equal 3/2; substitute that for x.*1018

*The 2's cancel out; this gives me 3 - 3y = 0, or -3y = -3; if I divide both sides of the equation by -3, I will get y = 1.*1028

*So, the solution is x = 3/2, y = 1; again, I could always check these solutions*1042

*by substituting the values in here and making sure that this equation holds true.*1051

*And this was a perfect setup to use elimination, because I already had variables that had the opposite coefficients.*1056

*I simply added these equations together; the y dropped out, allowing me to solve for x, and then substitute 3/2 in for x.*1063

*OK, Solve algebraically: again, there are no variables with a coefficient of 1, so I am going to go to elimination.*1075

*But these do not have the same or opposite coefficients, and the y's do not have the same or opposite coefficients.*1083

*So, this time I am going to need to use that extra step; I am going to need to use multiplication*1090

*in order to create a situation where I have variables with the same or opposite coefficients.*1095

*And I see that, if I multiply the top equation by 3, I will get 6x; and if I multiply the bottom equation by 2, I will get 6x.*1102

*So, I am going to go ahead and do that--multiply the top equation by 3: that is going to give me 3 times 2x - 3y = -16.*1112

*And this comes out to 3(2x) is 6x; 3 times -3...that is -9y; and 3 times -16 is -48.*1129

*The second equation I am going to multiply by 2.*1144

*So, 2 times 3x, plus 5y, equals 14; OK, 2 times 3x gives me 6x; 2 times 5y is 10y; and 2 times 14 is 28.*1147

*I have the same coefficient here; so what I need to do is subtract.*1165

*I am going to subtract, and again, I am going to convert this so that I am adding the opposite.*1171

*Rewrite the top equation the same, and the bottom as adding the opposite (adding -6x - 10y - 28).*1185

*OK, when I do that, the 6x's drop out, and that is going to give me -19y equals...-48 and -28 comes out to -76.*1196

*Divide both sides by -19, and -76 divided by -19 actually equals 4, so it divided very nicely and evenly; so I ended up with y = 4.*1215

*Now that I have that, I am going to substitute y = 4 into either equation; I am going to choose the top one.*1227

*So, 2x - 3y = 16; let y equal 4; so, 2x - 3(4) = -16, or 2x - 12 = 16.*1233

*Add 12 to both sides to get 2x = -4, and divide both sides by 2 to get x = -2.*1251

*So, the solution is that x equals -2 and y equals 4.*1259

*So, when I looked at this, I saw that I could use elimination, if I got the x's to have the opposite coefficient*1264

*by multiplying the top equation by 3 to give me 6x, and the bottom equation by 2 to give me 6x here, as well.*1274

*So, I did that; and then, these have the same coefficient, so I subtracted and solved for y.*1283

*Once I had a value for y, I substituted that value into one of the equations and then solved for x to get the solution.*1291

*Here, I have 2x + 3y = 8, and -4x - 6y = -16.*1303

*Since there is no variable with a coefficient of 1, I am going to use elimination.*1311

*But again, I am going to have to do a little work to get the opposite coefficients.*1314

*And I see here that all I have to do is multiply the top by 2, and that will give me 4x; that is opposite coefficients.*1319

*I don't have to do anything to the bottom equation.*1329

*So, that is 2 times 2x, plus 3y, equals 8, and this is going to give me 4x + 6y = 16.*1330

*That is the opposite coefficient, so I am going to add this new equation and this equation: + -4x - 6y - 16.*1343

*And you may have already realized that each one of these terms is opposite.*1354

*And so, here is what is going to happen: 4x - 4x is 0; 6y - 6y is 0; 16 - 16 is 0; so I end up with 0 = 0.*1360

*And I didn't make any mistakes--I did everything correctly--but then all my variables went away.*1373

*And what this tells me is that I have a dependent system, and it has an infinite number of solutions.*1377

*If I were to graph this, I would see that I have an infinite number of solutions--that these are intersecting at every point along the line.*1386

*So, we call this a dependent system.*1399

*If I had been going along, and all of the variables dropped out, and then I got something that wasn't true,*1405

*like 4 = 2 or 4 = 0, then I would have a situation where it is an inconsistent system and there are just no solutions.*1410

*But here, this is always true; so I have an infinite number of solutions.*1418

*So today, we covered solving systems of equations algebraically.*1425

*And that concludes today's lesson for Educator.com; I will see you again.*1430

*Welcome to Educator.com.*0000

*Today, we will be covering solving systems of inequalities by graphing.*0002

*In order to solve a system of inequalities by graphing, you need to graph the solution set of each inequality.*0009

*So, previously we discussed techniques; and you can review that lecture about how to find the solution set of an inequality by graphing.*0015

*Here, all you are going to do is graph each inequality.*0024

*And the solution of the system of the inequalities put together is the intersection of the solution sets of the inequalities.*0027

*So, what this is, really, is the overlap--the area of overlap--the points that are in common between each of the solution sets.*0035

*For example, if you are given a system of inequalities: y ≤ x - 3, and y > -2x + 1,*0044

*recall the techniques for graphing an inequality to determine the solution.*0056

*First, you want to graph the corresponding linear equation to find the boundary line of the solution set.*0063

*Then, use a test point to determine the half-plane containing the solution set.*0082

*Let's go ahead and do that, and then talk about how to find, then, the solution set.*0099

*This is just graphing each one--finding the solution set for each one; and then we have to talk about how to find the solution set for the system.*0105

*So, I am going to start with this one; and that is y ≤ x - 3; and the corresponding linear equation would be y = x - 3.*0111

*So, this is in slope-intercept form, y = mx + b, so I can easily graph this, because I know that the y-intercept is -3 and the slope is 1.*0126

*Since the slope is 1, increase y by 1; increase x by 1; increase y by 1; increase x by 1; and on.*0139

*Now, looking back here, this says y is less than or equal to -3; and what that tells me is that I am going*0148

*to use a solid line here for the boundary line, because the line is included in the solution set; so use a solid line to graph this.*0158

*The next thing is to determine which half-plane--the upper or lower half-plane--this solution set is in.*0174

*And I want to use an easy test point, and that is the origin (0,0), as my test point.*0179

*And it is well away from the boundary line, so I can use that.*0185

*If it were near or on the boundary, I would want to pick a different test point.*0189

*I am going to take my test point, (0,0), and substitute those values in to the inequality.*0194

*So, 0 is less than or equal to 0 minus 3; 0 is less than or equal to -3.*0200

*Well, that is not true; since that is not true, that means that the solution set is not in this upper half-plane.*0208

*(0,0) is not part of the solution set; therefore, the solution set is in this lower half-plane.*0216

*OK, that is the solution set for this inequality; now let's work on the other one.*0223

*Here, y is greater than -2x + 1; the corresponding linear equation is y = -2x + 1.*0229

*So, here I have the y-intercept at 1 and a slope of -2.*0240

*So, decrease y by 2...1, 2; increase x by 1; decrease y by 2; increase x by 1; and so on.*0246

*Now, since this is a strict inequality, I am going to use a dashed line, because the boundary line here is not part of the solution set.*0256

*OK, so here is my dashed line; next, I need to find a test point.*0267

*And again, I am going to use (0,0); I am going to use that test point.*0281

*Actually, let's use a different one, because that is a bit close; so let's go ahead and pick one*0293

*that is farther away, so we don't have any chance of causing confusion.*0297

*Let's use something up here...(2,1); (2,1) would be good for the test point.*0301

*So, my test point is going to be (2,1): I am going to insert (2,1) into that inequality.*0308

*1 is greater than -2; x is 2; plus 1; this is my test point; and 1 is greater than -4 plus 1; 1 is greater than -3.*0316

*And this is true; since this is true, that means that the test point is in the correct half-plane where the solution set lies.*0331

*So, for this line, this is the upper half-plane, and this is part of the solution set; so I am going to go ahead and shade that in.*0340

*OK, so far, I had just been going along, doing what I usually do when I would find the solution set for an inequality.*0356

*But remember, we are looking for the solution set for a system of inequalities.*0364

*So, this lower half-plane is the solution set for this inequality; this half-plane is the solution set for this inequality.*0369

*Now, I am looking for the intersection between the two solution sets, and that is right here, in this quadrant right here.*0376

*Darken that in even more; I am going to go ahead and use another color to emphasize that.*0382

*OK, so the area bounded by this line and including that line--that is the boundary line for the system of equations.*0396

*And then, the boundary line over here is the dashed line; so this line is not part of the solution set.*0406

*The technique: graph each of the inequalities; find their solution sets; and then find the area that is the intersection of the two solution sets.*0414

*You may come across a situation where there is no solution to a system of inequalities.*0427

*And this is because the two inequalities may not have any points in common.*0433

*Remember: the common points in each of the solution sets comprise the solution set for the system of inequalities.*0438

*If the two inequalities have no points in common, then the solution set of the system is the empty set.*0446

*For example, let's say I am given y < x - 4 and y ≥ x - 2.*0454

*So, I am going to go ahead and graph those, starting with y < x - 4.*0462

*I need to find the boundary line, so the corresponding linear equation is y = x - 4.*0467

*So, - 4 is the y-intercept, and the slope is 1.*0474

*So, increase x by 1, and increase y by 1, with each step.*0480

*Since this is a strict inequality, I am going to use a dashed line.*0487

*I have my boundary line; now I need to use a test point, and I am going to use this (the origin) as the test point.*0495

*And I am going to substitute (0,0) into the inequality and see what happens.*0503

*This tells me that 0 is less than -4; and that is not true.*0510

*Since that is not true, this is not part of the solution set; the solution set is actually below this line--the lower half-plane.*0516

*OK, that was my first inequality; now, my second inequality is y ≥ x - 2.*0525

*The corresponding linear equation is y = x - 2.*0532

*Here, I have a y-intercept of -2 and a slope of 1; so increase x by 1 and y by 1 each time.*0538

*Now, here I am going to use a solid line; this line is part of the solution set.*0551

*And a test point: I can use (0,0) again--that is well away from this boundary line.*0561

*Substituting these values into this inequality gives me 0 ≥ -2; and that is true: 0 is greater than -2.*0567

*So, this upper half-plane describes the solution set.*0578

*Now, you can see what happened here; and you may have already noticed that these two lines have the same slope.*0585

*Since they have the slope, and they are parallel lines, that means they are never going to intersect.*0592

*Since this solution set is above this line, and this one is below the line, there are not going to be any points in common.*0598

*These two lines will go along forever, and the points above and below them will never intersect.*0605

*So, this is a situation where there is no solution; we just say it is the empty set.*0611

*So here, we saw a situation where there is no solution, because there are no points in common.*0622

*OK, the first example is very straightforward: x ≥ 2; y > 3.*0628

*Starting with the x ≥ 2: the corresponding linear equation would be x = 2, and that just tells me that, if x equals 2,*0635

*no matter what the y-value is, it is just saying x is 2; so that is going to be a vertical line.*0650

*And since this is greater than or equal to, I am going to make this a solid line.*0657

*Now, you could certainly use a test point; or you could just look at this and say,*0674

*"Well, it is saying that x is greater than or equal to 2, which means that it is going to be values to the right."*0678

*You could certainly say, "OK, I want to do a test point at (0,0); 0 is greater than or equal to 2--not true, so this is not part of the solution set."*0685

*Or, like I said, you could just look back here and say, "OK, it is telling me that the values of x are greater than or equal to 2."*0697

*So, this half-plane contains the solution set.*0704

*The second inequality, y > 3, has a corresponding linear equation of y = 3.*0711

*So, if y equals 3, that is going to be a horizontal line right here.*0720

*And it is a strict inequality, so I am going to make that a dashed line, making it a different color so it shows up.*0725

*So, this blue line is that y = 3; OK.*0735

*Again, I could either just go back and say, "All right, this is y > 3, so those points would be up here";*0743

*or I can always use my test point, (0,0), and say 0 is greater than 3--that is not true, so I know this is not part of the solution set.*0750

*So, I need to go up here; OK.*0761

*So again, the solution set for the system of inequalities is the area of intersection of the two solution sets.*0774

*For this first inequality, the solution set is over here.*0782

*For the second inequality, the solution set is up here; and I can see the area of intersection is right up here.*0786

*And it is to the right of this solid vertical line, and it includes the points on the line.*0793

*And it is above the dashed blue line, and it does not include the points on the line.*0799

*We are just graphing each inequality and finding this area of intersection.*0805

*Here it is slightly more complicated, but the same technique.*0811

*Again, this is a system of inequalities, starting with the first one: y > x + 1.*0815

*The corresponding linear equation is y = x + 1.*0822

*The y-intercept is 1, and the slope is 1.*0828

*So, increase y by 1; increase x by 1; increase y by 1; increase x by 1.*0835

*And this is also going to be a dashed line, since it is a strict inequality; this boundary line is not part of the solution set.*0842

*Take a test point, (0,0); substitute back into that inequality to give 0 is greater than 0 + 1; so that says 0 is greater than 1.*0858

*Well, that is, of course, not true--which means that this half-plane where the test point is, is not the solution set.*0874

*Instead, it is the upper half-plane; so I am going to shade that in to indicate that this is the solution set for the first inequality.*0884

*The second inequality is y = -x + 2; well, the inequality is y > -x + 2; the corresponding linear equation is y = -x + 2.*0893

*So, that gives me a y-intercept of 2 and a slope of -1.*0907

*Decrease y by 1; increase x by 1; decrease y by 1; increase x by 1; and on down.*0914

*Again, it is a strict inequality, so we have another dashed line; I am making this a different color so it stands out as a separate line.*0923

*This blue line is the boundary line for the solution set of the second inequality.*0941

*OK, again, I am using (0,0) as my test point.*0949

*I am substituting into that inequality: 0 is greater than -0 + 2, so 0 is greater than 2.*0956

*And again, this is not true; so looking at this line, this point is not part of the solution set.*0966

*So, this lower half-plane is not the solution set.*0976

*I am going to go to the upper half-plane and shade that in.*0979

*OK, therefore, the solution set for the system of inequalities is right up here in this corner.*0996

*It is the points above the black line (but not including this line) and above the blue line (but not including that boundary line).*1004

*The technique, again: graph the first inequality (which was right here); we found the solution set in the upper half-plane.*1014

*We graphed the second inequality and found its solution set in the upper half-plane.*1022

*And then, we noted that this area up here is the overlap between the two inequalities; and that contains the solution set for the system.*1027

*OK, Example 3: 2x - y > 4 is my first inequality.*1041

*And one thing that I see right away is that this is not in the standard form we usually use.*1050

*So, I am going to work with this to put it in a standard form, where y is isolated on the left; and that is going to give me -y > -2x + 4.*1055

*Now, I have to divide by -1; and recall that when you divide an inequality by a negative number,*1067

*then you have to reverse the direction of the inequality symbol.*1073

*So, this is going to become y < 2x - 4; and it is absolutely crucial that, if you are multiplying*1079

*or dividing by a negative number, you immediately reverse the inequality symbol; or you won't end up with the correct solution set.*1089

*OK, so that is my first inequality; I am going to go ahead and do that same thing with the second inequality before I work with either one of these.*1096

*So, here I have 4x + 2y ≥ -2, so I am going to subtract 4x from both sides.*1107

*And then, I am going to divide both sides by 2; and since 2 is a positive number, I can just keep that inequality symbol as it was.*1118

*This is -4 divided by 2, so that gives me -2x - 1.*1125

*OK, looking at this first inequality: the corresponding linear equation, now in standard slope-intercept form, is y = 2x -4.*1138

*So, by putting it in that form, we made it much easier to graph.*1153

*The y-intercept is at -4, and the slope is 2.*1158

*So, increase y by 2; increase x by 1 for the slope.*1162

*Looking back, this is a strict inequality, so I need to use a dashed line, indicating that the boundary is not part of the solution set.*1169

*And then, a test point--I need to determine where the solution set lies--in the upper or lower half-plane.*1189

*I am using the origin as my test point, substituting (0,0) in for x and y.*1198

*This gives me 0 - 0 > 4, or 0 is greater than 4; and that is not true.*1208

*Since that is not valid, that is not part of the solution set; so this is not the correct half-plane.*1216

*I am going to shade the lower half-plane instead.*1221

*OK, over here, the corresponding linear equation is y = -2x - 1.*1228

*So, the y-intercept is -1; the slope is -2x.*1237

*So, when I decrease y by 2, I am going to increase x by 1; decrease y by 2; increase x by 1.*1243

*Again, I have to check here, and I see that I am going to use a solid line for this boundary line,*1258

*because the line is part of the solution set.*1264

*And then, I need a test point here; and I want to pick a test point that is not quite so close to this boundary line,*1278

*just in case my graphing wasn't perfect, so I am going to select (1,2) right up here as my test point, just to be safe.*1285

*Substitute those values into this inequality right here, or back up here--either way.*1296

*I am going to go ahead and use this one, because it is simpler: 2 ≥ -2(2) - 1.*1308

*2 is greater than or equal to -4 minus 1; so, 2 is greater than or equal to -5; and that is true.*1319

*So, my test point is in the half-plane containing the solution set.*1333

*I am going to shade this upper half-plane.*1340

*Once I have done that, I can see that the solution set for the set of inequalities, the system of inequalities, is right here.*1348

*It is the area bounded by this line and including the line, and then the area bounded by the dashed line,*1356

*right here, but not including it; so, this lower right section of the graph.*1363

*For this one, we had to take an extra step, just to make it easier--put this in standard form.*1372

*And then, we graphed the corresponding linear equation to find the boundary line.*1376

*We used the test point to find that the lower half-plane was the solution set for this first inequality.*1381

*The same technique for the other inequality: and the upper half-plane turned out to contain the solution set.*1388

*So, my area of intersection is right here; that is the solution set for the system.*1395

*OK, Example 4: y < 2x + 1; y ≥ 2x + 3.*1403

*We are going ahead and starting out with the first inequality, y < 2x + 1.*1412

*I am finding my boundary line with the corresponding linear equation, y = 2x + 1.*1424

*Here, the y-intercept is 1; the slope is 2; so increasing y by 2 means increasing x by 1, and continuing on.*1429

*I am checking back and seeing that I need a dashed line, because this is a strict inequality.*1446

*So, I am graphing out this line with a dashed line; so this line will not be part of the solution set--the points on the line.*1452

*The second inequality is y ≥ 2x + 3; the linear equation is y = 2x + 3.*1463

*I am going to go ahead and graph this out; and this tells me that the y-intercept here is 3.*1475

*And the slope is 2: increase y by 2, increase x by 1; or decrease y by 2, decrease x by 1; and so on.*1484

*So, this is going to give me another line, right next to this one.*1499

*But this time, I am actually going to use a solid line.*1504

*So, this is a solid line, because this is greater than or equal to.*1509

*Now, let's look at some test points for each.*1517

*For this first one, let's use a test point of (0,0) right here for this line and substitute in.*1520

*0 is less than 2 times 0 plus 1; that gives me 0 < 0 + 1, or 0 is less than 1.*1528

*And that is true; so I have this as part of my solution set right here, so it is the lower half-plane.*1537

*I found the solution set for this inequality.*1550

*For this one, I am also going to use (0,0); that is well away from that line--that is a good test point.*1552

*Substitute in: 0 is greater than or equal to 2 times 0, plus 3; 0 ≥ 0 + 3; 0 is greater than or equal to 3.*1560

*That is not true; so, for this line, (0,0) is not part of the solution set; so it is the upper half-plane.*1572

*Now, it is a little tough to draw: but if you look here, one thing that you will see is that these lines have the same slope.*1585

*Because of that, I know that these are parallel lines, so I know that these two lines are parallel and that they will never intersect.*1593

*Since the points are all below this parallel line in that half-plane, and above this line (they are parallel to it), these solution sets are never going to intersect.*1602

*In this case, the solution is the empty set; there are no points in common, since these are parallel lines.*1614

*That concludes this session on solving systems of inequalities by graphing at Educator.com; see you again!*1625

*Welcome to Educator.com.*0000

*Today we are going to be going on to talk about solving systems of equations in three variables.*0002

*In previous lessons, we talked about how to solve systems with two variables.*0007

*So now, we are going to go up to systems involving three variables.*0012

*In order to solve these systems, you are actually going to use the same strategies you used for solving systems with two variables.*0017

*Recall those techniques: substitution, elimination, and multiplication.*0027

*This time, though, a solution is an ordered triple of values.*0031

*So, you will end up having three variables: for example, (x,y,z).*0035

*And the solution would be something like (5,3,-2), where x is 5, y is 3, and z is -2.*0040

*Now, just to work out an example to show you how to approach these using the same techniques that you already know:*0056

*looking at this system of equations, I have three equations, and I have three total variables.*0082

*And the idea is to work with the three equations so that you get one of the variables to drop out.*0088

*Once you get one of the variables to drop out, you will be left with a system of two equations*0096

*with two variables, and you already know how to work with that.*0100

*So, the technique would be to first just consider two of the equations together; I am going to call these equations 1, 2, and 3, so I can keep track of them.*0103

*I am going to first consider equations 1 and 2.*0113

*And when I look at these, I see that the two z's have opposite coefficients.*0116

*And you will recall that elimination works really well in that situation.*0125

*So, I am going to take 2x + 3y - z = 5 (that is equation 1) and equation 2: 3x - 2y + z = 4, and I am going to add those.*0129

*This will give me 5x, and then 3y - 2y is going to give me y; the z's drop out; 5 and 4 is 9.*0146

*So now, I have a new equation, and I will just mark this out so I can keep track of it.*0158

*So, once you have gotten a variable to drop out, work with two different equations to get the same variable to drop out.*0167

*So, I got z to drop out; and what I want to do is work with two different equations--I worked with 1 and 2.*0175

*I could work with 1 and 2, or I could work with 2 and 3.*0182

*And I am actually going to work with 2 and 3; and I want to get z to drop out.*0187

*Equations 2 and 3: looking at this, how am I going to get z to drop out?*0193

*Well, in order to do that, I could use elimination; but I am first going to have to multiply this second equation by 2.*0197

*So, this is equation 2; and it is 3x - 2y + z = 4; and I am going to multiply that by 2 to give me 6x - 4y + 2z = 8.*0207

*Now, I am going to take equation 3 (this is equation 2, and I am going to take equation 3): I want to make sure that I am working with two different equations.*0229

*So, I have 1 and 2, and 2 and 3 (or I could have done 1 and 3--either way).*0237

*-4x + y + 2z = 3: my goal is to get the z's to drop out.*0243

*In order to do that, I am going to have to subtract: I need to subtract 3 from 2.*0256

*I want to be very careful with my signs here, so I am going to change this to adding the opposite.*0267

*This is going to give me 6x + 4x, which is 10x; -4y and -y is -5y; 2z and -2z drops out 8 minus 3 is 5; OK.*0276

*At this point, what I am left with is a system of two equations with two unknowns.*0297

*Once you get that far, you proceed using the techniques that we learned previously (again, substitution, elimination, and multiplication).*0303

*But since you are only working with two equations with two unknowns, you are on familiar territory.*0312

*And you can solve for one of the variables and then find that value; substitute in for the other variable*0316

*and find that value; and then you can go back and find z.*0325

*And we are going to work more examples on this; but the basic technique is to work with two equations*0329

*to eliminate a variable, using either elimination or substitution, then work with two other equations*0335

*to eliminate that same variable, resulting in two equations with two variables,*0342

*allowing you to solve for one variable, then the other, and then the third.*0349

*Just as in systems with two variables, a system with three variables may have one solution, no solutions, or an infinite number of solutions.*0357

*So, recall: the solution set here, if I had three variables (x, y, and z) would be a value for x*0368

*(such as 2), a value for y (such as -4), and a value for z (such as 1) that would satisfy all three equations.*0375

*The other possibility is that there may be no solutions; recall from working with systems of equations with two variables--*0384

*you know that you are in this situation when you are using elimination, or you are using substitution;*0393

*you are going along; and then you see variables start to drop out, and you end up with an equation*0399

*where you have a constant equaling another constant, which is never true.*0406

*So, if you start seeing variables drop out, and you end up with something such as 4 = 7 (which is never true),*0413

*this tells you that there is no solution to this system of equations.*0420

*There can be an infinite number of solutions; when you are working with your equations;*0426

*you are eliminating; you are substituting; you are using your techniques; you are being careful;*0434

*you are doing everything right, and then you see variables drop out, and you get a constant equaling a constant, like 2 = 2.*0438

*Well, that is always true; and this means that that system of equations has an infinite number of solutions.*0447

*So, typically, you will get one solution: a value for x, y, and z that is the set of values that makes the equations true--that satisfies the equations.*0457

*You may end up, though, with no solutions (there are no solutions to this equation) or an infinite number of solutions for this system of equations.*0467

*OK, in the first example, we are given a system of three equations with three variables.*0479

*So again, I am going to work with two of the equations; I will number them 1, 2, and 3.*0486

*And I will work with two of them, and then a different two.*0493

*So, first, I am going to look at the second two equations; and they are very easy to work with, because I have opposite coefficients.*0500

*I could also use substitution, because I have coefficients of 1; but I am just going to use elimination.*0509

*So first, I am going to work with 2 and 3: y - z = 2, x (let me move that, so it doesn't create confusion) + 2y + z = 2.*0515

*OK, here I end up with x; all I am doing is adding these together using elimination.*0533

*x + 3y; the z's drop out; 2 + 2 is 4.*0543

*OK, so I eliminated z from this first set of equations; now, I need to work with a different set of equations.*0550

*And there are actually multiple different ways to approach this, and I am going to work with 1 and 2.*0567

*I already worked with 2 and 3, so I eliminated z; and I want to...actually, I already have...this does not have z in it, so I don't even need to proceed.*0579

*That way, I already have two equations with two unknowns; this is a particularly easy situation, compared with when all three have 3 variables.*0598

*OK, so I look up here, and I have my new equation; I can call it equation 4.*0607

*And then, I have equation 1; and these just have x and y.*0613

*So, I am just going to proceed, like I usually do, with two equations with two unknowns.*0616

*Let me rewrite these right here: 2x + y = 3, and this is x + 3y = 4.*0620

*I can use substitution; that would be fine, because I have a coefficient of 1.*0634

*So here, I am going to solve for y; and this will give me y = -2x + 3.*0640

*And then, I am going to substitute into this equation; so I have x, and I am going to substitute for y: plus 3, times -2x, plus 3, equals 4.*0651

*Working this out: x...3 times -2x is -6x; 3 times 3 is 9; it equals 4.*0668

*Here, now, I just have an equation with a single variable, so I can solve that.*0679

*First, I am combining like terms: x - 6x is -5x; plus 9 equals 4.*0684

*Subtract 9 from both sides: -5x = -5; x = 1.*0692

*OK, so the first thing I wanted to do is just get rid of one of the variables; I am just working with two variables.*0700

*And I did that by just adding these two; in the second and third equations, the z dropped out.*0705

*I was lucky, because the first equation already didn't have a z; so I had two equations with two unknowns.*0711

*And then, I just used those two; I solved by substitution, and I came up with x = 1.*0717

*Since I know that x equals 1, I can go ahead and substitute this into this equation to find y.*0726

*So, looking at equation 1: 2x + y = 3; I know that x equals 1, so that is 2(1) + y = 3, or 2 + y = 3.*0734

*Subtracting 2 from both sides, I get y = 1; so now, I have x, and I have y.*0754

*I need to find z: well, this will easily tell me what z is.*0759

*That is y - z = 2, and I know y: y equals 1, so 1 - z = 2; -z = 1, therefore z = -1.*0765

*So, the solution to this set of equations is that x equals 1, y equals 1, and z equals -1.*0780

*The hardest step was just getting rid of that third unknown (the third variable).*0791

*I did that by adding these two together: then, working with two equations with two variables, I was able to solve for x.*0797

*Once I am there, all I have to do is start substituting.*0805

*Here, I substituted x into the first equation and solved for y.*0807

*Once I got y, then I was able to substitute y into the second equation to solve for z.*0813

*So, 1, 1, -1 is the solution for this system of equations with three variables.*0822

*OK, Example 2: again, my goal is going to be to get a variable to drop out, so I am just left with two variables.*0833

*Looking at this first and second equation, considering these together, y and -y...if I add those together,*0843

*the y's will drop out, because they have opposite coefficients (1 and -1).*0852

*So, I am going to start off by adding the first two equations: this is 1 and 2.*0857

*x + y + z = 2; and then, I am going to add x - y + 2z = -1; and I am going to come up with a new equation.*0861

*This is 2x; the y's drop out; z + 2z is 3z; 2 - 1 is 1; OK, I have this.*0874

*And I worked with these first two; I now need to work with two different equations to get the same variable to drop out.*0886

*This time, I am going to pick equations 1 and 3, and I want y to drop out.*0894

*So, I have equations 1 and 3: that is x + y + z = 2, and (equation 3) that is 2x + y + 2z = 2.*0900

*Now, I need to subtract in order to get the y to drop out.*0919

*To keep everything straight, as far as my signs go, I am going to keep the first equation the same;*0925

*but for the second one, I am going to change it to adding the opposite: add -2x, -y, -2z, and -2.*0931

*OK, x - 2x is -x; the y's drop out, which is just what I wanted; z - 2z is -z; 2 - 2 is 0.*0944

*Now, I have two equations and two variables; I am rewriting these two up here to see what I have to work with.*0956

*2x + 3z = 1; now, I just use my usual methods of solving a system of equations with two variables.*0965

*Since I have coefficients here of -1, it is pretty easy to use substitution, so I am going to solve for x in this second equation,*0976

*and then substitute that value up in the first equation.*0984

*I have -x - z = 0, which would give me -x = z, or x = -z.*0989

*So, I am going to take this -z and substitute it in right here; OK, that gives me 2x + 3z = 1, and let x equal -z.*0996

*So, 2 times -z, plus 3z, equals 1; that is -2z + 3z = 1.*1014

*Combine these two like terms to get z = 1; now, I have my first value.*1025

*OK, so I know that z equals 1, so I am on my way.*1034

*And I look up here, and I see, "Well, I know that x equals -z, so that makes it very easy to solve for x."*1038

*If x equals -z, and z equals 1, then x equals -1.*1047

*So now, I have x = -1, z = 1; I am just missing y.*1059

*Well, look at that first equation: it tells me that x + y + z = 2.*1063

*The x is -1; I don't know y; and I know that z is 1; these two cancel, and that gives me y = 2.*1074

*So, -1 + 1 is 0, so I end up with y = 2.*1088

*Putting all this together up here as my solution, I end up with x = -1, y = 2, and z = 1.*1092

*That was a lot of steps; it is really important to keep track of what you are working with--especially, in the beginning,*1106

*that you work with two equations (I worked with 1 and 2) to get a variable to drop out.*1112

*I added those, and the y's dropped out; then I want to work with either 1 and 3 or 2 and 3 (two different equations) to get the y to drop out.*1118

*I chose 1 and 3; and I saw that I could get the y to drop out of 1 and 3 if I just subtracted 3 from 1; that is what I did right here.*1128

*At that point, I clearly mark out what I ended up with, which is two equations with two variables.*1140

*We wrote those up here; and I decided I was going to use substitution.*1146

*I solved for x in this second equation: x equals -z; I substituted that in right here, into the first equation.*1151

*That allowed me to have one equation with one variable, z; and I determined that z equals 1.*1162

*From there, it was much easier, because I saw that x equals -z, and I knew z; so x equals -1.*1170

*I had x; I had z; and I had three equations that I could have used,*1180

*but I picked the easiest one to substitute in x and z and solve for y to get my set of solutions.*1184

*Again, this is a set of three equations with three variables that I need to approach systematically.*1198

*And my first goal is to eliminate the same variable, so I am working with two equations with two variables.*1204

*And I see several possibilities; you could approach it differently, and you will come up with the same answer, as long as you follow the rules and the steps.*1216

*I am seeing that -y and y are opposite in terms of coefficients (-1 and 1), so I am going to add those two.*1227

*I am going to add 1 and 3; that is going to give me 5x; the y's drop out, so 5x - z; 1 - 3 is -2.*1235

*I just marked that, so I can keep track of it, because I am going to need to use it in a minute,*1260

*once I generate another equation in which y has been eliminated.*1265

*OK, I worked with the first and the third; now, I need to work with two different equations;*1270

*and I am going to work with the first and the second, and I want to eliminate y.*1276

*In order to eliminate y, I need to multiply the first equation by 2; so I will do that up here.*1282

*This is going to give me 4x - 2y + 2z = 2; that is the first equation.*1294

*Now, I am going to add it to the second equation: + x + 2y - z = 0.*1305

*Adding these together, I am going to get 5x; the y's drop out; 2z - z is z; 2 and 0 is 2.*1319

*OK, I first worked with the first and the third, and then I worked with the first and the second, to get the y's to drop out.*1330

*So now, I have two equations and two variables: x and z.*1338

*So, put these together so I can see what is going on with them: 5x - 2 = -z; 5x + z = 2.*1345

*Well, I can see that, if I add these, z will drop out; and I am just back to my usual two equations with two variables--usual techniques.*1355

*5x and 5x is 10x; the z's drop out; -2 and 2 is 0.*1365

*Divide both sides by 10; it gives me x = 0.*1373

*So, I have my first value, which makes things much easier.*1377

*Since I know that x equals 0, I can substitute into either of these to solve for z.*1384

*So, 5x - z = -2; so 5(0) - z = -2; 0 - z = -2; -z = -2; divide both sides by -1 to get z = 2.*1389

*I have my second value; OK, so I know x; I know z; I just need y.*1411

*I am going to solve for y; I could use any of these--I am going to pick the top one and solve for y, knowing that x equals 0 and z equals 2.*1417

*OK, that is 0 - y + 2 = 1, which gives me -y + 2 = 1.*1437

*Subtract 2 from both sides to get -y; if I say -1 - 2, that is going to give me -1.*1447

*Multiply all of that by -1 to give me y = 1.*1455

*OK, so the solution here is x = 0, y = 1, and z = 2.*1461

*And I approached that by seeing that I could add 1 and 3 because of -y and y, and those would drop out.*1470

*I could have added the first two and had the z drop out, and had that be my variable to eliminate; I happened to choose y.*1479

*I added those together and got this equation.*1486

*Then, I worked with the first and the second equation--a little more complicated, because to get opposite coefficients,*1490

*I had to multiply the first equation by 2.*1496

*I did that to generate this equation, which I added to the second; this is the resulting equation.*1500

*I then had two equations with two variables; I looked at those two and saw I had opposite coefficients with z.*1507

*So, when I added them together, z dropped out, and I could solve for x.*1514

*Once I determined that x is 0, I substituted 0 into this top equation for x to solve for z, and determined that z equals 2.*1519

*At that point, I just needed to solve for y, so I took this equation and substituted my value for x and my value for z, and determined that y equals 1.*1529

*So again, we are using the same techniques that we used previously, only you are working with more equations, and there is more to keep track of.*1543

*OK, in this system of equations (three equations with three unknowns), it is a little bit more complicated.*1551

*I do have one equation that has a coefficient of 1, but the rest of them have larger coefficients.*1559

*So, I am going to work with the first and second equations; and what I want to do is eliminate the z.*1566

*So, in order to do that, I am going to need to multiply this first equation by 2.*1577

*I want to work with the first and second equations; I want to get the z to drop out.*1583

*But I need to multiply this by 2 first; so let me do that right over here.*1589

*That is going to give me 6x - 4y + 2z = 8; so I am going to rewrite that right here: 6x - 4y + 2z = 8.*1600

*And that came from that first equation, multiplied by 2; and I am going to add it to the second equation.*1615

*So, + -6x + 4y - 2z = 2: now, you might have already seen what happened.*1621

*I was really just focusing on "OK, I want to get the z's to be the same or opposite coefficients," so that they would cancel out.*1635

*But what happened is: everything ended up with opposite coefficients: 6 and -6; -4 and 4; 2 and -2; OK.*1640

*So, I have 6x - 6x; that is 0; -4y and 4y--0; 2z and -2z--0; 8 and 2--10.*1654

*0 = 10: well, we know that 0 does not equal 10, so this is not true; it is never true that 0 equals 10.*1669

*So, there is no solution to this set of equations.*1678

*I could have had a situation where I got a solution.*1682

*I could have also had a situation where maybe I got 10 = 10 with a different system of equations.*1686

*If I had come up with something like 0 = 0 (that is always true) or 10 = 10, then I would have had an infinite number of solutions.*1694

*But instead, what happened is that my variables drop out; I got c = d, which is saying*1701

*that I have a constant that is equal to another constant, which is never true; so there is no solution.*1706

*So, this one turned out to actually be less work than the others.*1712

*But when this happens, you just want to be really careful that you were doing everything correctly;*1714

*you didn't make a mistake; but actually, it can end up that there simply is no solution to the system of equations.*1718

*That concludes this lesson on Educator.com on solving systems of equations with three variables; and I will see you next lesson!*1725

*Welcome to Educator.com.*0000

*Today's lesson introduces the concept of matrices.*0002

*And matrices are used throughout math and science as an approach to problem solving.*0006

*In this course, we are going to use them to solve systems of equations.*0013

*However, they are also used in fields such as physics, computers, and genetics.*0018

*First of all, defining what a matrix is: a ***matrix** is a rectangular array containing variables or constants, which is enclosed by brackets.0026

*And the plural form of the word "matrix," which I just used, is matrices; so you will hear me say that in the course.*0037

*And to give you an example of what a matrix looks like, as I said, a matrix is enclosed in brackets, and it contains variables or constants.*0045

*And starting out, we are going to be working with matrices that contain constants.*0062

*And then, towards the end of this series of lectures, we will see a matrix that involves variables.*0066

*This is an example of a matrix; and these are usually designated by capital letters, so I could call this matrix A.*0079

*Another matrix, B, might be smaller, perhaps containing only two numbers.*0087

*Another matrix, C, could have a different set of numbers.*0097

*Each variable or constant within a matrix is called an element; so, 2 is an element; 1 is an element; each of these is an element.*0108

*And later on, we will also see matrix equations, because you can perform operations with matrices.*0118

*For example, I could have a matrix equation that says A + B = C.*0126

*So, I could have some matrix, A, that when added to B equals another matrix C.*0134

*So, just as we performed operations on constants and terms, we can perform operations on matrices.*0138

*OK, starting out, it is important to understand the concept of dimensions when referring to a matrix.*0148

*And with a matrix, ***dimensions** refers to the number of rows and columns in the matrix.0154

*For example, let's say I have a matrix, and you can see, of course, that it has a certain number of rows and a certain number of columns.*0161

*Well, m refers to the number of rows; and here, that is 1, 2, 3, 4--I have 4 rows.*0179

*For this matrix, looking at columns (and we will say n here is the number of columns), I have 1, 2 columns.*0191

*Therefore, I would call the dimensions of this matrix 4 by 2; this is a 4x2 matrix.*0200

*And this is important, because the dimensions often limit what you are able to do--*0207

*which operations you can perform on a set of matrices.*0214

*So, looking at another matrix: this matrix has 1, 2 rows and 1, 2, 3, 4 columns; so this is a 2x4 matrix.*0219

*It is important--the order of the numbers is very important: here it is 4x2 (4 rows, 2 columns); here it is 2x4, because it's 2 rows, 4 columns.*0243

*OK, there are certain matrices that are special cases: a 1xn matrix has one row, and is called a ***row matrix**.0262

*For example, this is a row matrix, because it has one row (in this first term here, it tells the number of rows).*0272

*And looking at the number of columns, I have 1, 2, 3, 4, 5, 6; so this is a 1x6 matrix, and it is a row matrix, because it has only one row.*0286

*A second type of special matrix is a ***column matrix**; here, it is an mx1 matrix--it has one column.0300

*So here, it is 1 row; and here, 1 column.*0309

*For example (let me make this a little bit more spread out)...2, 3, 8, 4, 2: OK, so here, I have 1, 2, 3, 4, 5 rows, but I have only 1 column.*0312

*So, this would be a 5x1 matrix, and since it has a 1 right here, this is a column matrix.*0346

*A third type of special matrix is called a ***zero matrix**; and this zero matrix has all its elements equal to 0.0358

*For example, this would be an example of a zero matrix.*0365

*So, these are three special types of matrices: a row matrix with one row, a column matrix that has one column,*0375

*and a zero matrix, which contains elements that are all 0.*0382

*We say that two matrices are equal if they have the same dimensions, and their corresponding elements are equal.*0389

*Let's talk about what corresponding elements are, using this set of matrices as an example.*0396

*OK, so I am going to call this A, and this matrix B; and I can say that A = B, because the corresponding elements of A and B are equal.*0415

*Elements are ***corresponding** if they have the same position within a matrix.0424

*By "position," I mean the same row number and the same column number.*0428

*So, right here, this element is in row 2, column 1; its corresponding element is also going to be in row 2, column 1.*0432

*And these are the same; and my row 1, column 1 elements are the same; and so on.*0450

*For each position, all of the elements are the same--the corresponding elements; therefore, these two matrices are equal.*0456

*If I had the same numbers, but they weren't in the same positions, the two matrices would not be equal.*0463

*The same dimensions (and I do have the same dimensions--1, 2, 3 rows, 2 columns; this is a 3x2 matrix,*0471

*and the same here--1, 2, 3 by 2 matrix) and corresponding elements are all equal: 1 equals 1, 3 equals 3, and so on.*0477

*I can say that these two matrices are equal.*0487

*OK, doing some examples: first, we are asked, "What are the dimensions of this matrix?"*0493

*And remember that when we do dimensions, we say mxn, so we look first at the number of rows, times the number of columns.*0500

*So, all I need to do is count the number of rows (that is 1, 2, so m = 2, and that is the number of rows),*0513

*and columns (n = 1, 2, 3--there are three columns); therefore, this is a 2x3 matrix; the dimensions of this matrix are 2x3, rows by columns.*0523

*Example 2 (a different matrix here): What are the dimensions of this matrix?*0544

*Again, we are looking at rows by columns; I have 1, 2 rows; 1, 2, 3, 4, 5 columns.*0549

*So, this is a 2x5 matrix, because it contains 2 rows and 5 columns; always have rows first, then columns.*0568

*Write the 3x3 zero matrix: this is telling me that I am going to have 3 rows and 3 columns.*0579

*And since it is a zero matrix, all elements are zero; OK, so that is going to give me 1, 2, 3 columns and 1, 2, 3 rows.*0588

*I just need to fill in, and all elements are 0; and again, a matrix is contained within brackets.*0606

*This is a 3x3 matrix (3 rows, 3 columns); it is a zero matrix; and I have my brackets around it to indicate that it is a matrix.*0613

*This is a 3x3 zero matrix.*0622

*Example 4: Write a matrix that is both a row matrix and a column matrix.*0628

*Recall that a row matrix has one row, and then some number of columns.*0632

*A column matrix has some number of rows, but only one column.*0643

*Putting this together, this is telling me that I am going to have...*0649

*if it is both a row and a column matrix, that means that in this m position,*0653

*the rows are going to be 1; and the columns are also going to be 1.*0659

*This is a 1x1 matrix that they are asking for; and I could use any constant or variable.*0662

*This would be both a row matrix and a column matrix; it has one row, and it has one column.*0670

*Or it could be 5, or I could use a variable (such as y or x).*0675

*So, any of these examples would be a row matrix and a column matrix at the same time.*0680

*That concludes this lesson from Educator.com; and I will see you next time, when we talk more about matrices.*0689

*Welcome to Educator.com.*0000

*In today's lesson, we are going to continue on with matrices, and this time doing operations on matrices.*0002

*Just as you can perform mathematical operations on numbers (such as addition, subtraction, and multiplication), you can do the same thing with matrices.*0009

*Starting out with addition of matrices: addition is defined only for matrices with the same dimensions.*0018

*So, just to review: when we talk about dimensions, the ***dimensions** of a matrix are m times n,0026

*where m is the number of rows and n is the number of columns.*0033

*So, two matrices must have the same number of rows and columns in order for addition to occur.*0040

*If that specification is met, then you add the corresponding elements of the two matrices.*0049

*The sum matrix will have the same dimensions as the original matrices.*0055

*To illustrate this, let's say that I have two matrices, A, and then I have a matrix B; and I want to add those together.*0062

*Well, let's look at my first matrix, A; and it will be 4, 0, 3, 1, -2, -4; OK.*0074

*And I have a second matrix, B, that I am going to add (A + B).*0085

*OK, so first looking at the dimensions to make sure that I am allowed to even add these: there are 2 rows on this one and 1, 2, 3 columns.*0101

*Therefore, this is a 2x3 matrix; matrix B has 2 rows and 1, 2, 3 columns; this is a 2x3 matrix.*0109

*Since these have the same dimensions, they can be added.*0119

*The sum matrix (the resulting matrix) is also going to have the dimensions of 2x3: it has the same dimensions as these two original matrices.*0123

*Now, to add these, I am going to add corresponding elements.*0133

*Recall that corresponding elements occupy the same position in the matrix.*0136

*So, if I am looking at this position, this is 1, 2; so it is row 2, and in columns, 1, 2; so it is row 2, column 2.*0140

*The corresponding element over here is also going to be in this position of row 2, column 2.*0149

*So, all I am going to do is add corresponding elements: 4 + 5 right here, and then I am going to have 0 and -1;*0155

*over here, 3 + 2; and down here, 1 + 0; in this position, -2 + 3, and in this position, -4 + -5.*0169

*OK, if I work out the addition on that, that is going to give me 9, -1, 5, 1, (-2 + 3) is 1, (-4 + -5) is -9.*0185

*So again, first always verify that the two matrices you are being asked to add have the same dimensions.*0202

*If they do, then you just add the corresponding element and put it in that same position*0208

*to get the sum matrix, which will have the same dimensions as the two originals.*0216

*Matrix subtraction is very similar: again, matrix subtraction is defined only for matrices with the same dimensions.*0222

*To subtract two matrices, subtract the corresponding elements of the two matrices.*0230

*The result--the difference matrix--will have the same dimensions as the original matrix.*0237

*So, it's the same concept as with matrix addition.*0241

*So, looking at that situation here: if I have 3, 2, 4, -1, 0, 6, -2, 0, and this is A;*0244

*and I have a second matrix that I am going to call B, and I am asked to find A - B;*0267

*OK, this is the difference matrix, which will be A - B, over here.*0276

*OK, so first verify that these have the same dimensions: I have 1, 2, 3, 4 rows, and I have 2 columns.*0283

*Over here, I have 1, 2, 3, 4 rows and 2 columns.*0292

*My difference matrix is also going to have 4 rows and 2 columns.*0299

*OK, so I simply subtract: 3 - 6 is going to give me -3; the corresponding elements 2 - 0 is 2; 4 - -5...*0303

*if I have 4 minus -5, that is going to give me 9; -1 - 6 is going to give me -7; 0 - 2 is going to give me -2;*0317

*6 - -1 is going to give me 7 (a negative and a negative are going to give me a positive); -2 - 0 is -2, and 0 - 1 is -1.*0341

*So again, verify that the two matrices have the same dimensions, and then simply*0354

*subtract corresponding elements; and since we are doing subtraction, you need to be very careful with the signs.*0360

*OK, now we are going to talk about scalar multiplication.*0368

*And scalar multiplication is not the multiplication of one matrix times another.*0370

*Matrix multiplication, we will actually cover in a separate lecture.*0376

*This is called ***scalar multiplication**, because what you are going to be doing is multiplying a matrix by a constant.0380

*And that constant is referred to as a ***scalar**.0387

*So, for example, if I am given a matrix that looks like this, and I am asked to multiply it by 2;*0391

*well, 2 is a constant called a scalar; and let's say this matrix is called A.*0402

*If I am asked to find 2A, then I am going to multiply the scalar by the matrix.*0408

*In order to do that, you multiply each element of the matrix by the scalar.*0414

*So, each element of the matrix, you multiply by the scalar to get the result.*0419

*OK, therefore, I am going to say 2 times 2 for this position; for this second position in row 1, column 2, it is 2 times 3.*0425

*Here, it is 2 times -1; 2 times 0; 2 times 4; and finally, 2 times 1.*0437

*Doing the multiplication out will give me 4; 2 times 3 is 6; 2 times -1 is -2; 2 times 0 is 0; 2 times 4 is 8; and 2 times 1 is 2.*0451

*And as you can see up here, the scalar product matrix has the same dimensions as the original.*0468

*This original matrix had 1, 2, 3 rows and 2 columns.*0473

*The scalar product over here, 2A, has 1, 2, 3 rows and 2 columns--the same dimensions as the original.*0480

*Again, for scalar multiplication, simply take the scalar (the constant) and multiply it by each element in the original matrix.*0493

*Just as we have certain properties when we are working with operations on regular numbers and variables,*0503

*there are also properties that regulate matrix operations.*0510

*So, if A, B, and C are matrices with the same dimensions, and k is a scalar, then the following hold.*0514

*This first one, you will recognize as the commutative property of addition.*0522

*Recall that with numbers, we have the commutative property, where, if you want to add 3 + 2,*0530

*you can add it in the other order--2 + 3--and get the same result.*0535

*And the same is true if you are adding two matrices.*0539

*So, matrix addition follows the commutative property.*0542

*The next property you may recognize as the associative property for addition of matrices.*0548

*What this is stating is that I can do these operations in either order.*0562

*I can either add the two matrices A + B together, then add C to those; or I can add B + C together first, and then add A to those two.*0567

*So, I can perform these operations in either order; and the result will be the same.*0578

*Now, looking at scalar multiplication, combined with the addition of two matrices here:*0584

*this follows the distributive property, and what this is stating is that I can add matrix A and matrix B,*0591

*and then multiply this scalar by that result; or I can multiply the scalar times one matrix, multiply the scalar by the other matrix,*0604

*and then add those two together; and I will get the same result.*0615

*So, this is the distributive property, which you have seen previously.*0618

*First example: we are going to add two matrices.*0625

*First, verify that addition is allowed: do these have the same dimensions? two rows, three columns--OK, so far, so good.*0628

*1, 2 rows; 3 columns: since these have the same dimensions, then addition is allowed.*0637

*OK, so I am rewriting these down here so we can work with them more easily.*0646

*Recall that, for addition, you are going to add corresponding elements; you are going to add the elements occupying the same row and column number.*0657

*So, 2 + -4 is going to give me -2; -1 + 6 is going to give me 5; 3 + -2 is going to give me 1.*0671

*0 and -3 is going to give me -3; 6 and 0--I am going to get 6; and then, 4 + 4 is going to give me 8.*0692

*So, all I have to do to find the sum matrix is to add the corresponding elements of these two matrices to get the result.*0705

*Example 2 involves subtraction: first, verify the dimensions--1, 2, 3 rows, 1, 2, 3 columns.*0718

*So, this is a matrix with three rows and three columns, so it is a square matrix, since it has the same dimensions in both directions.*0727

*Here, I have 1, 2, 3 rows and 1, 2, 3 columns; so this is also a 3x3 matrix; so it is a square matrix, since it has the same dimensions in both directions.*0740

*Therefore, I can subtract these; when I go ahead and subtract, I am going to get a difference matrix that is also 3x3.*0755

*Beginning with these two corresponding elements: 2 - -1 (I have to be very careful, when I am working with negatives,*0772

*that this becomes 2 + 1, which is 3, so I get 3 right here); -1 - -2 is going to give me -1 + 2, which is 1.*0779

*Here, I just have 6 - 3; that is 3; 3 - 0--that is 3; 2 - 4--and that is just going to be...2 - 4 is going to give me -2.*0801

*0 - -8 (working down here) is going to be 0 + 8, which equals 8 in this position.*0818

*1 - 6 is -5; 4 - 0 is 4; and then, 3 - -1 gives me 3 + 1, which equals 4.*0831

*So again, with subtraction, just be really careful with the signs.*0845

*And I end up with a difference matrix right here that resulted from taking an element*0848

*in the first matrix and subtracting the corresponding element of the other matrix from that.*0856

*Example 3: Find the product--and it is the product of a scalar and a matrix, so this is scalar multiplication.*0864

*And here, the scalar is -2.*0873

*I am rewriting this down here.*0876

*Recall that, in scalar multiplication, you are going to multiply the scalar times each element in the matrix,*0884

*and the result is going to be a scalar product that has the same dimensions as the original.*0890

*My original here is a 3x3 matrix; so I am going to take -2 times 2 for the first position; -2 times -1;*0895

*-2 times 6; -2 times 3; and continue on, multiplying each one...-2 times 0; the third row:*0909

*-2 times 1; -2 times 4; and finally, -2 times 3.*0924

*Then, when I do my multiplication, I am going to end up with -4, 2, -12, -6, here is -4, 0, -2, -8, and then -6.*0932

*So again, in scalar multiplication, simply multiply each element in the matrix by the scalar*0956

*to get a scalar product matrix with the same dimensions as the original.*0963

*Example 4 is slightly more complicated: we are asked to find -3A + 6B.*0970

*OK, so what I need to do is multiply matrix A by -3; multiply matrix B by 6; and then add those together.*0978

*Just to make sure I can add them eventually, I verify the dimensions as 3 rows, 2 columns and 3 rows, 2 columns.*0989

*So, I will be able to add them.*1000

*OK, just to note, looking at the distributive property, it says that k, a scalar, times A + B (these two matrices) equals KA + KB.*1002

*So, there is actually another way I could do this: I could say, "All right, I am going to factor out the -3, and then I am going to end up with A - 2B."*1015

*This would be another way to do that; however, it is debatable which way is easier.*1038

*I am going to go ahead and just follow this original.*1044

*But you actually, if you felt like this way was easier, could have done it this way.*1046

*OK, so starting out, the first thing we are going to need to do is multiply the A by the scalar -3.*1053

*OK, and from that I can find 3A; I want to find 3A.*1082

*So, this is -3 and A, and I want to find -3A.*1089

*OK, so recall that all we are going to do is multiply each element of the matrix by the scalar.*1097

*And that will give me 3 here; -3 times 0 is 0; -3 times 2 is -6; -3 times -1 is 3; -3 times 3 is 9.*1105

*And let's see, then I have -3 times 4, which is -12; OK.*1122

*Now, what I also need to do is find 6B; so here, I have 3A--I need to find 6B.*1128

*So, I have B over here; 0, -3, -4, 6, 4, and 9; so, I have B, and I am going to multiply it by 6, and that is going to give me 6B.*1139

*So, figuring this out, it is: 6 times 0 is 0; 6 times -3 is -18; 6 times -4 is -24.*1157

*6 times 6 is 36; 6 times 4 is 24; and 6 times 9--that is 54.*1171

*All right, so here I have 3A; here I have 6B; now, I need to add those.*1178

*So, let me go ahead and copy 6B right over here, so I can add it.*1184

*OK, 0, -18, -24, 36, 24, and 54: now, I am erasing that; these two are no longer equal.*1200

*OK, all we need to do with matrix addition is to add the corresponding elements.*1215

*So, I am going to add 3 and 0; and this is going to give me 3.*1222

*I am going to add 0 and -18 to get -18; -6 and -24 is -30; 3 and 36 is 39;*1230

*9 and 24...or excuse me, this actually should be -9; correct that--that is -3 times 3; that is -9 + 24 is 15;*1241

*and then, I have -12 and 54, to give me 42.*1251

*So, this here is -3A + 6B: so right here is my solution.*1256

*And what I did is took A; I multiplied it by the scalar -3 to get this -3A.*1264

*I took 6, and I multiplied it by the second matrix, B, to get 6B.*1272

*Then, I added -3A + 6B to get -3A + 6B as my solution.*1279

*That concludes this lesson on matrix operations at Educator.com; I will see you next lesson.*1291

*Welcome to Educator.com.*0000

*In today's lesson, we will be covering matrix multiplication.*0002

*In a previous lesson, we discussed scalar multiplication, which is multiplying a constant by a matrix.*0005

*This time, we will be talking about multiplying one matrix by another.*0012

*Before you proceed with matrix multiplication, you need to verify that the dimension requirement has been met.*0018

*So, suppose that matrix A is m by n, and matrix B has dimensions of p by q.*0024

*The product of these two matrices can be obtained only if n = p.*0031

*Now, what is that saying? What that is saying is that the number of columns of the first matrix must equal the number of rows of the second matrix.*0036

*So, in order to find AB, to find that product, to be allowed to do that kind of multiplication,*0045

*the number of columns of the first matrix must equal the number of rows of the second matrix.*0052

*OK, once you find that product, the resulting product matrix, AB, will have dimensions of m times q.*0099

*So, the product matrix dimensions will have the number of rows of the first matrix and the same number of columns as the second matrix.*0109

*To make this more concrete, let's look at an example.*0115

*If matrix A equals 2, -1, 4, 0, and matrix B equals 3, 4, 2, 1, 0, -3, 0, -2; my first question is, "Can I multiply them--can I find AB?"*0118

*Well, this has one row and four columns; it is a 1x4 matrix.*0137

*This has four rows and two columns; it is a 4x2 matrix; so this gives me 1x4 and 4x2.*0145

*So, all I have to do is see, "OK, does this second number equal this first number?" and yes, the second number does equal the first number.*0152

*Therefore, I can multiply those.*0162

*Now, what are going to be the dimensions of the product matrix?*0164

*Well, the dimensions of the product matrix...this is a 1x4, multiplied by a 4x2 matrix; and AB is going to have*0168

*the same number of rows as this first matrix (which is 1), and the same number of columns as the second matrix (which is 2).*0175

*So, the product matrix is going to be a 1x2 matrix.*0184

*So, the important thing is: before you multiply, make sure that you verify that the dimension requirement is met--*0188

*that the second number of the first matrix is equal to the first number of the second matrix.*0193

*And to predict the dimensions of the product matrix, you take this first number and this second number; and that will give you the product matrix dimensions, 1x2.*0200

*OK, multiplying by matrices is not exactly what you would expect.*0217

*It is not like addition and subtraction of matrices, where (in addition and subtraction) we just took the first matrix*0224

*and added the corresponding element of that to the corresponding element of the second matrix (and the same in subtraction).*0231

*You might think, "OK, I am just going to multiply the corresponding elements by each other."*0238

*But it is actually more complicated than that; and you need to take it step-by-step.*0243

*So, the best way to understand this is to go through an example: so let's look at a pair of matrices and multiply them out.*0247

*OK, this is my first matrix; I am going to call it A; and then, I want to multiply it by another matrix, which I am going to call B.*0264

*So, before I multiply, I have to make sure that they meet the dimension requirements.*0277

*And A has two rows, and it has four columns; B has four rows and two columns.*0284

*So, I am allowed to multiply these, because this second number equals the first number.*0292

*My product matrix, AB, is going to have two rows and two columns; it is going to be a square matrix with dimensions 2x2.*0299

*Now, as you read up here, it says that the element in row I and column J of the product...*0311

*so the element in a certain row and column of the product of the matrices A and B...*0317

*is obtained by forming the sum of the products of the corresponding elements in row I*0323

*(in a certain row of matrix A) and the corresponding column, J, of matrix B.*0329

*What does this mean? Well, instead of just saying row I and column 1, let's start with row 1, column 1.*0335

*I want to find the element that goes right here, in row 1, column 1.*0345

*And the way I am going to do that is: I am going to go over here to row 1, and I am going to go here in column 1;*0350

*and I am going to multiply 3 by 4 and find that product, and then I am going to add that to the next product, 2 by 0.*0361

*Then, I am going to add that to the next product, 0 by 1, to 0 by 3, and so on.*0369

*So, row 1, column 1; then I am going to go to the corresponding row in A and the corresponding column in B.*0375

*And working this out, this gives me 3 times 4, plus 2 times 0, plus 0 times 3, plus 1 times -2.*0380

*Working that out, that is going to give me 12 + 0 + 0 - 2; 12 - 2 is 10.*0399

*Therefore, my row 1, column 1 element is 10.*0410

*OK, next let's work on row 1, column 2; that means I am going to go to row 1 here in A, and column 2 here, in B, and do the same operations.*0414

*So, row 1, column 2: that is 3 times -1, plus 2 times 2, plus 0 times 1, plus 1 times 0.*0429

*OK, working that out, this gives me -3 + 4 + 0 + 0; that is 4 and -3, so that is 1.*0446

*Therefore, my row 1, column 2 element for the product is 1.*0461

*All right, now I need to work with this second row; and I am going to think about what position this is right here.*0468

*This is row 2, column 1; that means I am going to go to row 2 here and column 1 in B and do the same thing.*0472

*OK, row 2, column 1: -1 times 4, plus 0 times 0, plus 4 times 3 (so the third element here and the third element here),*0487

*plus the fourth element here and the fourth element there.*0507

*All right, figuring this out, this is going to give me -4 + 0 + 12 + -2, or - 2; look at it either way.*0512

*This is going to give me -4 and 12, which is 8, minus 2...actually, let's make this clearer...plus 12, minus 2...it is going to give me...*0526

*let's see...that is -4, 0, 4 and 3, 2 and -2; OK, so this is going to give me 12 - 6, or 6.*0546

*Now, the next row and column position: I have row 2, column 2.*0562

*Here is row 2; here is column 2; it is going to give me -1 times -1, plus 0 times 2, plus 4 times 1, plus 2 times 0.*0577

*OK, so that is going to come out to...-1 times -1 is 1, plus 0, plus 4, plus 0; 4 + 1 is 5, so I am going to get 5 right here.*0601

*So again, if I just looked here, and I said, "OK, that is row 1, column 2," then I would get that*0618

*by going to row 1 here and column 2 here and multiplying each of those, and then finding the sum of their products.*0626

*So, even with a bigger matrix, you can always find a particular position by using this method.*0638

*OK, there are some properties that govern matrix multiplication that you need to be familiar with.*0646

*If A, B, and C are matrices for which products are defined, and k is any scalar, then these properties hold.*0653

*The first property you will recognize as the associative property for multiplication.*0660

*And what it says is that I can either multiply A times B, find that product, and then multiply by C;*0667

*or I can multiply B times C, find that product, and then multiply A; and I will get the same result.*0679

*OK, next I see the same thing here (it is the associative property), except this time, it also involves multiplying a scalar.*0687

*So, here I have matrix multiplication, and then I also have a scalar.*0695

*And I can either multiply the two matrices and then multiply the scalar times that product;*0698

*or I can multiply the scalar times the first matrix, then multiply that product by the second matrix;*0704

*or the scalar and the second matrix, and the product by the first matrix.*0712

*So, it doesn't matter which order I do the multiplication in, in this situation; these two first, then the other two.*0716

*Next, you are going to recognize the distributive property.*0726

*And the distributive property, as usual, says that, if I have these two matrices, B + C,*0732

*and I am multiplying A by them, another approach that would give the equal result*0739

*is to first multiply A and B, and then add that product to the product of A and C--the distributive property.*0743

*And again, the distributive property would work if I placed these as follows, added A and B, and then multiplied C.*0751

*The same thing: I could say the product of A and C, plus the product of B and C.*0763

*What is very important to realize is that the order that you multiply matrices in matters a lot.*0768

*So, if I say, "Oh, I am going to do AB; does this equal BA?" no, it does not always equal that.*0775

*Sometimes it does; but very often it does not--these are two different things.*0783

*Therefore, matrix multiplication is not commutative; it does not follow the commutative property,*0788

*and so you can't simply say, "A times B is the same as B times A"; you can't just change the order of those two.*0797

*The first example: suppose A is 3x4; B is a matrix that is 4x3; C is 3x3; and D is 4x4.*0812

*What are possible defined products of the four matrices?*0821

*Recall that, for a product to be defined, the second number of the first matrix has to equal the first number of the other matrix.*0824

*So, let's first look at AA; is that defined?*0834

*Well, if I have 3x4, and I am trying to multiply it by 3x4; these two are not equal, so this is not defined.*0840

*OK, and let's make a list here of the ones that are defined, because that is what they are asking me for--where the product is defined.*0852

*If I take A times B, well, the second number of A is equal to the first number of B; so that is defined.*0863

*Now, let's do it in the opposite order, BA: if I have 4x3 and 3x4, that is also defined.*0871

*OK, now AC: 3x4 and 3x3--that is not defined.*0882

*Now, if I put C first--if I say CA--it is 3x3 and 3x4, and these two are equal, so CA would have a defined product.*0888

*OK, now 3x4 and 4x4; that is AD; that is defined, because of the 4 and the 4.*0903

*However, if I put D first--if I say I am going to do DA--that is going to give me the D first, which is 4x4, and then A, 3x4; that is not defined.*0910

*OK, moving on to B: B times B (BB) is 4x3 and 4x3; that is not defined.*0923

*BC is 4x3 and 3x3; these two are equal, so BC is defined.*0934

*CB is 3x3 and 4x3--not defined; OK, BD--4x3 and 4x4--not defined.*0940

*But if I put the D first, then I would get 4x4 and 4x3; that is defined, so DB is defined.*0953

*OK, now I am up to C; 3x3 and 3x3--CC--that is a square matrix, so that is defined; the multiplication is defined for that.*0965

*OK, now, CD--3x3 and 4x4--is not defined; then I try the D first--it is not going to matter: 4x4 and 3x3 is still not defined.*0975

*OK, let's go to D: I can multiply DD, because I have 4x4 and 4x4, and that means that, of course,*0988

*since this is a square matrix, this second number and the first number there are going to be equal.*1000

*So, these are my defined products; I have 3, 6, 8 defined products.*1005

*So, these are all the ones where the second number of the first matrix--*1010

*the number of columns--was equal to the first number of the second matrix--the number of rows.*1016

*And I can perform multiplication of these sets of matrices.*1021

*OK, now doing some matrix multiplication: before I proceed, I am going to make sure that the product is defined--that I can do it.*1029

*So, I have a 2x2 matrix here, and I have a 2x2 matrix here; therefore, multiplication is allowed.*1038

*I am rewriting this down here, since this second number is equal to this first number (so multiplication is allowed).*1052

*Using the method we discussed: my product matrix is going to be 2x2 also--the first number here and the second number there.*1066

*I am expecting a 2x2 matrix; so let's first look for this position--row 1, column 1.*1076

*I am going to go to row 1 here and column 1 here and work with those two.*1086

*I am going to say 3 times 0 (that product), and I am going to add it to the product of 2 times 4.*1091

*That is going to give me 0 + 8, which is 8; so the element in row 1, column 1 is 8.*1099

*Now, row 1, column 2--I am going to go to row 1 here--row 1 of this first matrix--and column 2 of the second matrix.*1111

*And that is going to give me...row 1 is 3, times 6; so row 1, column 2 *