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## Practice Questions

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## Table of Contents

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## Related Books

### Solving Linear Equations in One Variable

- If only one value will make an equation true it is conditional. If any number will make the equation true it is an identity. If no number will make the equation true, it is a contradiction.
- To solve a linear equation we work to isolate the variable on one side of the equation. To help out we use the Addition and Multiplication property of equality.
- The addition and multiplication property of equality says we may add the same number to both sides of an equation, and we may multiply the same number on both sides of an equation (as long as it is not zero.)
- To help out with the solving process you can multiply all terms by a common denominator. This clears out fractions.
- It is a good idea to check the solution in the original problem to ensure that it is correct.

### Solving Linear Equations in One Variable

- 40 + x - 40 = - 22 - 40

- 10 - 54 = 54 - y - 54
- - 44 = - y

- r − [1/5] + [1/5] = [2/5] + [1/5]

- [5/8] − [2/8] = [2/8] − t − [2/8]
- [3/8] = − t

x = unknown number

- 35 = 6 + x
- 35 - 6 = 6 + x - 6

n = unknown number

- n - 12 = 66
- n - 12 + 12 = 66 + 12

n = unknown number

- 18 = n - 32
- 18 + 32 = n - 32 + 32

x = unknown number

- 6 + 8 + x = 30
- 14 + x = 30
- 14 + x - 14 = 30 - 14

x = unknown number

- 55 - x = 23
- 55 - x - 55 = 23 - 55
- - x = - 32

x = unknown number

- 21.8 + 60.1 + x = 111.5
- 81.9 + x = 111.5
- 81.9 + x - 81.9 = 111.5 - 81.9

- 12 ×( [y/12] ) = 3 ×12

- 9 ×( [g/9] ) = 17 ×9

- [22i/22] = [132/22]
- i = [132/22]

- [10/3]s = [2/4]
- ( [3/10] ) ×[10/3]s = [2/4] ×( [3/10] )
- s = [6/40] = [3/20]

**s**

**=**[(

**3**)/(

**20**)]

- [30/7]l = [3/5]
- ( [7/30] ) ×[30/7]l = [3/5] ×( [7/30] )
- l = [21/150] = [7/50]

**l**

**=**[(

**7**)/(

**50**)]

- [( − 12j)/( − 12)] = [480/12]

- 4[1/5]x = 2[3/8]
- [21/5]x = [19/8]
- ( [5/21] ) ×[21/5]x = [19/8]v ×( [5/21] )

**x**

**=**[(

**95**)/(

**168**)]

- [3/10]d = 3[7/8]
- [3/10]d = [31/8]
- ( [10/3] ) ×[3/10]d = [31/8] ×( [10/3] )
- d = [310/24] = [155/12]

**d**

**=**[(

**155**)/(

**12**)]

- [52/h] = 4
- h ×[52/h] = 4 ×h
- 52 = 4h

- [112/9]w = [11/13]
- ( [9/112] ) ×[112/9]w = [11/13] ×( [9/112] )

**w**

**=**[(

**99**)/(

**1456**)]

- 3y - 11 + 11 = 37 + 11
- 3y = 48

- 89 - 17 = 17 - 6x - 17
- 72 = - 6x
- 12 = - x

- 4( [(a − 7)/4] ) = 11(4)
- a - 7 = 44
- a - 7 + 7 = 44 + 7

- 3(16) = ( [(45 − t)/3] )3
- 48 = 45 - t
- 48 - 45 = 45 - t - 45
- 3 = - t

- 5(9) = ( [(17 + b)/5] )5
- 45 = 17 + b
- 45 - 17 = 17 + b - 17

- x, x + 1 ,x + 2 represent the three consecutive numbers
- x + (x + 1) + (x + 2) = 54
- 3x + 3 = 54
- 3x + 3 - 3 = 54 - 3
- 3x = 51

- x, x + 1, x + 2, x + 3, x + 4 represent the five consecutive numbers.
- x + (x + 1) + (x + 2) + (x + 3) + (x + 4)
- 5x + 10 = 100
- 5x + 10 - 10 = 100 - 10
- 5x = 90

- [(3m + 10)/4] + 6 − 6 = 18 − 6
- [(3m + 10)/4] = 12
- 4( [(3m + 10)/4] ) = 12(4)
- 3m + 10 = 48
- 3m + 10 - 10 = 48 - 10
- 3m = 38

- − 8 − 10 = [(5 − c)/2] + 10 − 10
- − 18 = [(5 − c)/2]
- 2( − 18) = ( [(5 − c)/2] )2
- - 36 = 5 - c
- - 36 - 5 = 5 - c - 5
- - 41 = - c

- [(6y − 3)/10] − 12 + 12 = 36 + 12
- [(6y − 3)/10] = 48
- 10( [(6y − 3)/10] ) = 48(10)
- 6y - 3 = 480
- 6y - 3 + 3 = 480 + 3
- 6y = 483

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Linear Equations in One Variable

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:08
- Solving Linear Equations in One Variable 0:34
- Conditional Cases
- Identity Cases
- Contradiction Cases
- Solving Linear Equations in One Variable Cont. 2:00
- Addition Property of Equality
- Multiplication Property of Equality
- Steps to Solve Linear Equations
- Example 1 4:22
- Example 2 8:21
- Example 3 12:32
- Example 4 14:19
- Example 5 17:25
- Example 6 22:17

### Algebra 1 Online Course

### Transcription: Solving Linear Equations in One Variable

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

*In this lesson we will look at more of the nuts and bolts of solving linear equations.*0002

*Specifically some of the things that we will actually do is we will look at types of solutions.*0010

*What does it mean to actually have a solution versus when it is consistent or inconsistent?*0015

*The things that will make this handy is how you can also deal with fractions when they are in the solving process.*0023

*It is something that my students always ask me.*0031

*What a solution means is a value that when you substitute it in for the variable, it makes the equation true.*0039

*We are looking for that magic value that would make the whole thing true.*0046

*In some cases, only one value would end up making the whole thing true.*0051

*If you only have one value that will do that or sometimes only two or three, we call this conditional.*0057

*The condition is that this variable must be that value.*0064

*There are other cases where we actually end up where any value will do.*0069

*It does not really matter what the value is as soon as you substitute it in for the variable it will make it a true statement.*0075

*If any value will do we say that the equation is an identity.*0083

*Lastly, there are very few types of equations that no matter what you try and substitute in for that variable, nothing seems to work.*0089

*No matter what that value is if nothing works then we say that that equation is a contradiction.*0100

*I have hidden a couple of those into the examples later on so watch for the one that is an identity and one that is a contradiction*0106

*and how we actually pick that out of just one that actually have a solution.*0116

*How do you exactly solve a linear equation?*0123

*You will use a few properties to help you out in the solving process.*0126

*One of them is your addition property of equality.*0130

*It is a handy property but what it says is that you can add the same amount to both sides of the equation.*0133

*In fact, many people like to look at this as almost like a balance scale.*0140

*Whatever you add to one side of your equation then you better be sure to add it to the other side as well to make sure that it all balances out.*0146

*If you are going to add + 4 over here, make sure you also add + 4 to the other side.*0155

*There is also the multiplication property of equality.*0163

*For this one you are allowed to multiply both sides by the same value and it will keep the equation the same.*0168

*You do have to be a little bit more careful with that property because it works as long as you do not multiply both sides by 0.*0175

*But you can use any other number you want.*0184

*You can multiply both sides by 3, multiply both sides by (x) as long as (x) is not 0, it will keep that equation the same.*0186

*One thing that I like to do when solving equation is to think of this process,*0197

*it often help with fractions as well as to get the variable that we are looking for all alone.*0203

*The very first thing that I like to do is scan it over and see if it does contain any fractions*0209

*and immediately clear them out if I can using some sort of common denominator.*0213

*After I have done that, I like to simplify each side of the equation as much as possible before shifting things to one side or the next.*0217

*We will simplify each side as much as possible first.*0225

*After we simplify both sides, we will try and isolate the variable on one side of the equation.*0230

*It may be difficult to do especially if there is more than one copy of that variable but if we can get them together usually we can isolate it just fine.*0236

*Lastly, it is always a good idea to check the solution just to make sure it actually works in the original problem.*0244

*That is always a good thing to do in case we make a mistake in one of these earlier steps.*0251

*We will definitely be able to find that out by the four steps when we check it by substituting it back into the equation.*0256

*Let us get into the examples and see how the solving process actually works.*0264

*I want solve (2 × x) +3 = 4x – 8, our goal is to figure out what the value for x is that would make this entire thing true.*0270

*In order to help this process out, I can see that it does not have any fractions, I’m going to try and simplify both sides and make it as compact as possible.*0281

*This means I'm going to use my distributive property on the left so they do not have my x inside parentheses 2x + 6.*0291

*Looking at both sides of the equation now, there is not much I can simplify if I was looking at just left or if I was looking at just the right.*0310

*What I want to do is to try and isolate my variable, in other words try and get it all alone.*0320

*Since I have the next on the left and on the right, I have to work on getting these together first.*0326

*In order to get them together, I will use my addition property of equality to add the same thing to both sides.*0333

*What I will add is -2x, we will put that on both sides and see the results.*0340

*2x and -2x, those two would cancel each other out and get rid of each other and just be left with 6.*0351

*On the right side, I have 4x - 2x so only 2x left – 8.*0358

*That is good, I worked out that way because now I only have one x to deal with and I can work on isolating that and getting it all by itself.*0368

*How am I going to do that? I better move this into the other side by adding 8 to both sides.*0377

*6 + 8 that will give me 14 and -8 + 8 will cancel each other out and be gone.*0384

*I have 14 = 2x, we just divide both sides by 2, I have 7 = x.*0400

*It looks like we have found a solution something that will make our equation true.*0415

*My question is does it actually work or not?*0420

*On to that last step, the one we actually see if this is the solution and we do that by substituting it back into the original problem.*0424

*I’m going to write the original problem, I’m going to write it all except for those x’s.*0431

*I’m going to take that value and put it in wherever I saw an x, let us see.*0443

*I had an x + 3 put 7 there and 4 × x now 4 × 7.*0451

*I'm doing here is simplify both sides of our equation now, let us see if it balances out.*0458

*2 × 7 + 3 is 10, 2 ×10 that would be 20, I’m not sure if these are equal, if we continue we find out.*0465

*4 × 7 is 28, 28 - 8 is 20, sure enough it looks like the value of 7 does make this equation true.*0479

*I know that 7 = x is my solution.*0494

*Let us try another this one, 3/4x – 1 = 7/5.*0498

*This one contains fractions so remember how I suggested taking care of them and you do not have to worry about quite as much.*0507

*We are going to try and find a common denominator that we can just multiply it by and get rid of our fractions.*0515

*Looking at our denominator, I have 4 and 5.*0522

*A common denominator in this case would be 20.*0529

*I’m going to multiply both sides by that 20, 20 and 20.*0541

*Let us write down the rest of our equation as well, 3/4x -1 and 7/5.*0555

*Sometimes it is no fun to deal with numbers like 20 but watch what it does when I start distributing it on the left side here,*0567

*and we will multiply the 20 and 7/5 together.*0573

*20 × 3/4 would be a 15x and 20 × 1 being – 20.*0578

*We do not have to deal with fractions on the outside of the equation anymore.*0591

*Perfect, I like it.*0593

*20 × 7/5 if we want to cancel out 5 from there, when you are looking at 4 × 7 = 28.*0595

*I just have to solve 15x – 20=28, let us work on getting an x all by itself by isolating it.*0608

*I will add 20 to both sides giving us 15x equal to 48 and then we will divide both sides by 15.*0622

*This will give us that x is equal to 48/15 but if we want we can even reduce that a little bit further.*0650

*I think 3 goes into the top and into the bottom.*0657

*This will use 16/5, not bad.*0665

*You have x equals 16/5, now get in front of that last step.*0672

*Does it equal to 16/5? Does it not? Let us find out by playing it back into the original equation.*0676

*I’m just testing it out and see if it holds true.*0682

*The original equation is ¾ and I have that x, that is where we will put our number -1 and we will see if it is equals 7/5.*0686

*Let us see our number was 16/5, check and see what happens.*0700

*I can take out a 4 from the 16 so this would be 12/5 – 1, it will give me a common denominator of 5.*0707

*I would have 12/5 – 5/5 and how do we know? That is equal to 7/5.*0734

*This one definitely checks out, I do know that 16/5 is my solution, it looks good.*0743

*We will have something like this one, more fractions here and that is good.*0755

*4 + 5x = (5 x 3) + x and see what we can do this one.*0758

*Let us simplify both sides of the equation first and then go from there.*0765

*I'm going to distribute my 5 into the parentheses.*0769

*4 + 5x = 15 + 5x, it looks pretty good.*0775

*I want to get my x’s together and hopefully isolate it.*0786

*We will subtract 5x from both sides, 4 = 15.*0791

*Look at what happened there, when I subtract 5x from both sides, I lost all of my x’s completely.*0805

*Even worse than that than I thought I have left over this 4 -15 that is not true.*0812

*That does not make any sense, 4 does not equal 15.*0819

*What I was left here is known as a false statement.*0823

*What this is telling me is that since I did not make any mistakes that you do not know the value for x is going to work.*0828

*This is an example of one of those situations where we have a contradiction.*0837

*No matter what value for x you are trying use, it simply not going to work in this equation.*0849

*In fact, no solution exists for it.*0854

*Let us look at another one and see if we have any better luck.*0860

*In this one is -x + 3 = (1 + 4) - 2x/2.*0863

*We have a fraction, I'm going to try and get rid of those fractions first by multiplying everything through by a 2.*0869

*I will multiply it on the left side and I will multiply it on the right side, both by 2.*0877

*Just to keep things nice and balanced.*0886

*-x + 3, 1 + 4 - 2x / 2.*0890

*We multiplied both sides by 2, we can go ahead and distribute it.*0901

*Let us see what the result will be.*0907

*-2x + 6 equals, I have 2 + and when it distributes on this second part here, it is going to get rid of that 2 in the bottom, 4 - 2x.*0911

*Looking good, let us continue trying to combine our terms specifically those x’s and the numbers.*0933

*I will add 2x to both sides and you will get 6 = 2 + 4.*0943

*You will notice in this one that we lost our x’s but this one is a little different.*0958

*Instead of being left with other nonsense, this one we are left with 6 = 6 which is actually something that is true.*0963

*Before you get too worried, of course check all the steps and make sure they are all correct, which they are.*0977

*What this is trying to indicate here is that we have an identity.*0982

*It does not matter the value of x, you can use any value and the equation is still going to ring true.*0985

*Let us mark this one as an identity.*0992

*In short, that is a great way that we can identify both of those situations.*0999

*If it has an actual solution, it is a conditional equation.*1003

*You usually go through the solving process and you will be able to figure out what that value is and able to test it and see if it actually works.*1007

*If it is a contradiction, so nothing works, then you will go through the solving process and usually your variable drops away entirely,*1014

*and what you are left with is a false statement, it does not make sense.*1022

*That is even after all of your steps are completely correct.*1026

*If you have an identity like this one then you will go through that solving process and you will be left with a true statement*1029

*which indicates that you could use any value want for x.*1036

*Let us try some more examples.*1042

*In this next one I have 1/3x – 5/12 = ¾ +1/2 x.*1046

*There is a lot of fractions in here, let us see if we can take care of them all by getting some sort of common denominator.*1053

*What would 3/12, 4, 2 go into?*1062

*I think our common denominator would have to be a 12.*1067

*We are going to multiply both sides by 12.*1072

*Let us write everything we got here, 1/3x – 5/12 and I have ¾ + 1/2x.*1084

*Then we are going to distribute through on both sides and see what we get.*1100

*1/3 of 12 would be 4x, 12 × 5/12 would give us 5, then I will have 12 × 3/4.*1108

*That goes in there 9 times, then (12 × ½) + 6x.*1124

*Notice that common denominator, we clear out all those fractions then you do not have to worry about them.*1134

*What we have left here is 4x - 5 = 9 + 6x.*1139

*If you want to continue trying to simplify this by getting the x’s together and getting them isolated.*1144

*Our x over here and x over here let us get them together by subtracting 4x from both sides, -5 = 9.*1153

*6x – 4x that would be 2x, that looks good.*1170

*I only have one x to deal with.*1177

*Let us go ahead and subtract 9 from both sides, -9 -9 -14 = 2x.*1180

*There is only one last thing to do, divide both sides by 2.*1194

*It looks like our solution is -7 or we are not too sure until we check it.*1200

*Let us substitute it back into the equation and see what happens.*1211

*I have 1/3 - 5/12 = ¾ + ½ and into all of those blank spots where we are used to have the x, let us go ahead and put that -7 in there.*1215

*Let us see if these things are true or not, also -7 × 1/3, that is -7/3 – 5/12, is that equal to ¾ - 87/2?*1237

*I do not know, we got to do a lot more simplifying before we can figure that out.*1259

*We will use our common denominator to help us out, the common denominator of 12.*1263

*Let us see if I need a common denominator over there that would be -28/12.*1270

*A common denominator of 12 on the other side would be 9/12 minus, top and bottom by 640, 2/12.*1279

*Let us see what we can do, -28 - 5 would be -33/12.*1294

*What is going on the other side? 9/12 – 42.*1306

*It looks like I have another -33/12.*1310

*They are exactly the same and what that means is that x does equal in -7.*1314

*You can see when you go through that process of checking it,*1322

*it can be a lot of work especially if you have lots of fractions and stuff that you are dealing with.*1324

*But it is a good idea just to make sure that the solution you come up with is the solution.*1329

*Let us do one more.*1338

*Here I have (3x -15)/4 + (x + 47)/7 = 11.*1339

*This one involves fractions, let us take care of them out of the gate by multiplying it by a common denominator, let us use 28.*1347

*A lot to keep track of in here, (3x -15)/4 + (x + 47)/7 = 28 × 11.*1364

*We will take that 28, we will actually distribute it to both parts on this left side.*1379

*We will take it to this giant fraction here and I will take it to this giant fraction over here.*1384

*What this is going to do is, it is going to allow us to cancel out those fractions in the bottom.*1388

*28/4 would give us 7, that is still going to be multiplied by that 3x -15 part.*1394

*When it distributes on the second piece, 28/7 reduces and that goes in there four times.*1404

*We still have a 4 multiplied by x + 47.*1411

*Over on the side we have 28 × 11 that is 308, some fairly big numbers.*1417

*Let us continue distributing and see what we can do.*1426

*Let us take this 3 × 7, we will also take the 15 × 7 and do the same thing with 4 on this side.*1429

*Hopefully free of those x’s and get them out of parentheses.*1440

*21x - (7 × 15) = 105 + 4x.*1444

*Now I have 4 × 47 = 188, again, some big numbers but just have to push on through.*1461

*I have more than one copy of x here let us get those guys together.*1476

*I also have some things that do not have an x, let us get those together as well.*1480

*21 + 4x would be 25x - 105 and 188, I think that would give us 83.*1487

*I only have a single x to deal with so I will simply work on getting that isolated.*1508

*Let us subtract 83 from both sides so that the 25x is the only thing on the left side.*1516

*308 – 83 will give us 225, almost done.*1529

*Let us go ahead and divide both sides by 25 now.*1539

*This will give us x = 9.*1549

*This one is a lengthy process to get all the way down to just x = 9.*1554

*Even when you get that far, go ahead and double check it just to make sure that it actually works out.*1558

*I have my (3 × x -15)/4 on one side and both in those blank spots, let us go ahead and put that 9.*1571

*Let us start to simplify and see what we will get.*1589

*3 × 9 = 27/4, I see a 9 + 47 = 56.*1592

*We are hoping that this will equal 11.*1610

*27 - 15 that is 12/4 that is 56/7, we are getting closer.*1614

*12/4 is 3, 56/7 let us see that goes in their 8 ×, 3 + 8 =11.*1625

*Sure enough this one checks out.*1639

*I know that x does equal 9.*1642

*When going through the solving process, you do have a lot to keep track of.*1648

*The most important thing is truly working getting those x’s all by themselves and isolate it so that they are the only thing on one side of the equation.*1652

*If you have to deal with fractions, use a common denominator to clear them all out.*1660

*And also definitely go back and check your solution to make sure it is a solution.*1664

*Thank you for watching www.educator.com*1671

0 answers

Post by Genevieve Carisse on October 21 at 12:08:18 PM

Professor, what is the order to use?

For example, 2x+6=4x-8. Instead of -2x first, why not do -6 or + 8 first? Thank you!

1 answer

Last reply by: Professor Eric Smith

Tue Jun 6, 2017 11:22 AM

Post by Carl Kellogg on June 6 at 10:04:56 AM

Why did you pursue the path of teaching mathematics?

0 answers

Post by Karen Johnson on March 17 at 09:47:57 PM

Example 4, you distributed the 2 to the 1, but not to 4 - 2x?

1 answer

Last reply by: Professor Eric Smith

Thu Mar 30, 2017 2:04 PM

Post by Karen Johnson on March 17 at 09:30:03 PM

Eric Smith,

In example 2, you changed the -1 to - 5/5? how did you do this?

1 answer

Last reply by: Professor Eric Smith

Fri Aug 26, 2016 7:07 PM

Post by Raymond Hayden on February 23, 2016

Some of your questions have incorrect answers. For example:

40+x=-22 is x=-62,

you have x=-64 as the answer.

There are other errs as well, I just can't call them to mind right now. This is confusing and should be corrected. Thank you.

1 answer

Last reply by: Professor Eric Smith

Tue Sep 22, 2015 4:16 PM

Post by Oscar Prado on September 22, 2015

Hi there on your second example how did you have up with 20 times 3/4 is 15?

1 answer

Last reply by: Professor Eric Smith

Sat Mar 28, 2015 5:29 PM

Post by antonio cooper on March 26, 2015

Sir, I would like to inform you that on your practice questions (#1)40+X=-22.

Your answer is X=-64, however unless I am mistaken -40+-22 is -62. Thank you for your lessons.

1 answer

Last reply by: Professor Eric Smith

Thu Nov 13, 2014 10:31 AM

Post by TAHA sakor on November 13, 2014

Why do we have to multiplication the equation twenty

1 answer

Last reply by: Professor Eric Smith

Thu Nov 13, 2014 10:29 AM

Post by TAHA sakor on November 13, 2014

what is consistent

1 answer

Last reply by: Professor Eric Smith

Sun Jul 6, 2014 2:27 PM

Post by ajaa trevino on July 2, 2014

In Example 4 why did you multiply everything by 2?

2 answers

Last reply by: Professor Eric Smith

Mon Jun 23, 2014 1:25 PM

Post by dzung tran on June 16, 2014

What if someone thought, on Example 1,instead of adding a negative 2x,they divided a negative 2x?The problem would look like x+6=2x-8.

If I continued to solve it, it would turn to 6=x-8, and then to 14=x. I'm not saying that I did this, but I followed the rules and whatever I did to one side, I did to the other. Why is it different when I do a different operation?

1 answer

Last reply by: Professor Eric Smith

Thu Jan 9, 2014 3:59 PM

Post by Mohamed Elnaklawi on December 30, 2013

In example one, how did you simplify x+3, and get 6?

1 answer

Last reply by: Professor Eric Smith

Thu Jan 9, 2014 4:14 PM

Post by Araksya Fernandes on December 7, 2013

in Example V i got -7 as well, but from 4x-5=9+6x

I said 4x-6x=5+9

-2x=14

x= 14:(-2)

x=-7

If i do it this way, am I assured that I will get the answer right?

2 answers

Last reply by: Professor Eric Smith

Tue Sep 3, 2013 6:39 PM

Post by steven schwartzle on August 31, 2013

Do you still use PEMDAS for all Linear Equations??