### Solving a System Using Elimination

- To solve a system using elimination
- Multiply equations by a constant so that a pair of terms will cancel out when added.
- Add the equations together. (A pair of terms should cancel)
- Solve the new equation created
- Use back-substitution in one of the original equations
- Solve this equation for the other variable
- Check you solution by substituting both values into the system to see if it makes all equations true.
- When multiplying by a constant, you may have to multiply one, or both of the equations to create a pair of terms that will cancel when added.
- If the system has only one solution, the method will find the x and y values that work.
- If the system has no solutions, then the elimination method will create a false statement.
- If the system has an infinite amount of solutions, then the elimination method will create a true statement. All points on the line are considered a solution to the system.

### Solving a System Using Elimination

2x - 4y = 10

6x + 4y = 14

- 2x âˆ’ 4y = 10
__+____6____x____+____4____y____=____14__ - 2x âˆ’ 4y = 10
__+____6____x____+____4____y____=____14__8x = 24 - x = 3
- 2x âˆ’ 4y = 102(3) âˆ’ 4y = 10
- 6 âˆ’ 4y = 10
- âˆ’ 4y = 4

2x + 8y = - 12

16x - 8y = 30

- 2x + 8y = âˆ’ 12+
__(____16____x__âˆ’__8____y____=____30____)__ - 18x = 18
- x = 1
- Plug x back into the equation to find y
- 2(1) + 8y = âˆ’ 12
- 2 + 8y = âˆ’ 12
- 8y = âˆ’ 14
- y = [( âˆ’ 14)/8]

4x + 4y = 36

- 8x - 4y = 24
__+ 4x + 4y = 36__ - 12x = 60
- x = 5
- 8x - 4y = 24

8(5) - 4y = 24 - 40 - 4y = 24
- - 4y = - 16

11x - 3y = 33

- 12x + 3y = 48
__+ 11x - 3y = 33__ - 23x = 81

7x - 2y = 11

- 7x - 5y = 17
__- 7x - 2y = 11__

- 3y = 6 - y = - 2
- 7x - 5y = 17

7x - 5( - 2) = 17 - 7x + 10 = 17
- 7x = 7

5x - 2y = 11

- 5x - 6y = 19
__- 5x - 2y = 11__ - 4y = 8
- y = 2
- 5x - 6(2) = 19
- 5x - 12 = 19
- 5x = 31

2x - 7y = 6

- 2x - 3y = 18
__- 2x - 7y = 6__ - 10y = 12
- y = [12/10] = [6/5]
- y = 1[1/5]
- 2x âˆ’ 3( 1[1/5] ) = 18
- 2x âˆ’ 3[3/5] = 18
- 2x = 21[3/5]

7x - 2y = 11

- 7x - 5y = 17
__- 7x - 2y = 11__

- 3y = 6 - y = - 2
- 7x - 5y = 17

7x - 5( - 2) = 17 - 7x + 10 = 17
- 7x = 7

5x - 2y = 11

- 5x - 6y = 19
__- 5x - 2y = 11__ - 4y = 8
- y = 2
- 5x - 6(2) = 19
- 5x - 12 = 19
- 5x = 31

2x - 7y = 6

- 2x - 3y = 18
__- 2x - 7y = 6__ - 10y = 12
- y = [12/10] = [6/5]
- y = 1[1/5]
- 2x âˆ’ 3( 1[1/5] ) = 18
- 2x âˆ’ 3[3/5] = 18
- 2x = 21[3/5]

*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 a System Using Elimination

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 0:09
- Solving a System Using Elimination 0:27
- Elimination Method
- Elimination Example
- One Solution, No Solution, or Infinite Solutions
- Example 1 8:53
- Example 2 11:46
- Example 3 15:37
- Example 4 17:45

### Algebra 1 Online Course

### Transcription: Solving a System Using Elimination

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

*In this lesson we are going to work on solving a system of linear equations using the method of elimination.*0003

*We focus on how to use this elimination method or how it differs from substitution*0011

*and how we can recognize the many different types of solutions that could happen.*0017

*Do we get one solution, and infinite amount of solutions, or possibly no solution or whatsoever.*0021

*The goal with the elimination method is to combine the equations in such a way that we eliminate one of our variables.*0030

*In that way, we will only have one that we will need to solve for.*0037

*In order to do this, sometimes we will have to multiply one of the equations or even both of the equations by some sort of a constant.*0042

*Now what we are looking to do when we multiply is we want a pair of coefficients that will end up canceling each other out.*0050

*Do I have - 3x and 3x?*0059

*When we combine those they would end up canceling out and that is what I want.*0062

*Once I get a pair of coefficients that I know will cancel, we will go ahead and actually add the 2 equations together.*0067

*In fact, some people call the elimination method the addition method because we end up adding the equations.*0073

*We will get a new equation after doing that and we will only have one type of variable in it.*0081

*We will end up solving that new equation.*0086

*We will have half of our solution at this point so we will be able to take that and substitute it back into one of the originals.*0090

*In fact, if you look at all the rest of stuffs from here on out, these are the same as the substitution method.*0096

*We simply have half of our solution and we are back substituting so we can find the other value.*0104

*Once we are all the way done through this entire process, it is not a bad idea to check the solution to make sure that does satisfy both of the equations.*0110

*Let us walk through this method, so you get a better sense of how it works.*0123

*Here I have 2x â€“ 7y = 2 and 3x + y = -20.*0129

*What my goal is to either eliminate these x values here or I need to eliminate my y values.*0136

*If I was just going through and I decide that I was going to add them together as they are,*0145

*you will see that this would not be a good idea because nothing happens.*0150

* 2x + 3x would give me 5x and â€“ 7y + y would give me -6y.*0154

*2 and -20 is -18, so I definitely created a new equation, but it still has xâ€™s and it still has yâ€™s.*0163

*There is not much that I can do with it.*0171

*What I want to happen is for some things to cancel out, so I only have one type of variable to solve for.*0174

*Let us go ahead and a do a little work on this so that we can get something to cancel out.*0181

*The first thing is we get to choose what we want to cancel out.*0188

*Is it going to be the x's or yâ€™s?*0191

*I want to choose my y since one of them is already negative and the other one is positive.*0194

*In order to make these guys came flat, let us take everything in the second equation and multiply it by 7.*0200

*The first equation exactly the way it is, no changes, and everything in the second equation will get multiplied by 7.*0210

*3 Ã— 7 = 21x, I have 7y, -20 Ã— 7 would be -140.*0218

*I still have a system and I still have a pair but now I noticed what is going on here with the y.*0231

*Since 1 is -7 and one is 7, when I add those together, we will end up canceling each other out.*0236

*Let us do that, let us go ahead and get to the addition part of the elimination method.*0244

*Adding up our x's, we will get 23x.*0248

*When we add up the y, we will get 0y so that term is gone.*0254

*Over on the other side - 138 and in this new equation the only thing I have in here is simply an x.*0259

*I can solve this new equation for x and figure out what it needs to be which I can do by dividing both sides by 23.*0270

*How many times 23 is going to 138?*0292

*6 Ã— 3 = 18, looks like 6 times so x = -6.*0303

*That is half of our solution and now we need to work on back substitution.*0313

*We will take this value here for our x and plug it back into one of the original equations.*0318

*It does not matter which one you plug this into, just as long as you put it into one of them it should turn out okay.*0323

*I end up solving this one for y, 2 Ã— -6 = -12 and we are adding 12 to both sides would give me 14 and y divided by -7, I have -2.*0335

*I have both halves of the solution.*0360

*I have x = -6 and that y = -2.*0362

*In the elimination method, we are working to eliminate one of our variables.*0370

*You might be curious what would have happened if we would have chosen to get rid of those x's instead.*0377

*We could have done it, but you would have to multiply your equations by something different.*0383

*Iâ€™m not going all the way through this, but just to show you how you could have started.*0390

*For example, if you multiply the first one by 3, it would have given you 6x - 21y = 6.*0394

*If you multiply the second equation by -2, - 6x - 2y = 40.*0405

*You would see that using those multiplications, the x's would have cancel out when you add them.*0414

*In the elimination method, eliminate one of your variables.*0421

*Onto the next important thing with the elimination method.*0428

*We have many different things that could happen.*0432

*We could have one solution, no solution or an infinite amount of solutions.*0433

*Visually, we can see that the lines either cross, do not cross, or perhaps they are the same line.*0439

*The way you recognize this when using the elimination method is that you might go through that method and everything will work out just fine.*0457

*You actually be able to find your solution.*0464

*That is when you know it has one solution, everything is good.*0466

*If you go through the system and all your work looks good, but you create a false statement*0471

*then you will know that there is actually no solution to the system.*0476

*In fact, they are 2 parallel lines and they never cross.*0479

*I think I mentioned this earlier, but go ahead and check your work if it looks like you are getting a false statement*0483

*just to make sure that the false statement is created from them being parallel and not from you making a mistake.*0488

*If you go through and you create a true statement, then this is an indication they have an infinite amount of solution.*0496

*Again, check your work, but make sure that you know whether it has an infinite amount or not.*0503

*These are exactly the same criteria that using the substitution method, so they are nice and easy to keep track of.*0512

*False statement, no solution, true statement, infinite amount of solutions and everything works out normally, then you just have one solution.*0518

*Let us get into some examples and see this in action.*0526

*In this first example, we will look at solving 2x - 5y = 11 and 3x + y = 8.*0534

*Think about our goal with the elimination method.*0543

*We want to get rid of these x's or we want to get rid of the y.*0546

*You get to choose which one you want rid of just a matter how you will end up manipulating these to ensure that they do cancel each other out.*0550

*I think the better ones to go after are probably these y, since one is already positive and one is already negative.*0561

*The way we are going to do this is when I take the second equation and we are going to multiply everything there by 5.*0570

*Let us see the result this will have.*0579

*We have not touched the first equation and everything in the second equation by 5 would be 15x + 5y = 40.*0582

*I can see that when we add together our y values, they will cancel each other out.*0597

*2x + 15x that is 17x, -5y + 5y they would cancel each other and get 0y.*0605

*11 + 40 =51, so in this new equation I can see that we only have x to worry about and we can continue solving for x.*0617

*We will do that by dividing both sides by 17, x =3.*0625

*Now that I have half of my solution and I know what x is, let us take it and end up substituting it back into one of our original equations.*0636

*(2 Ã— 3) - 5y = 11.*0647

*Our goal here is to get that y all by itself.*0656

*We will multiply the 2 and 3 together and get 6, then we will subtract 6 from both sides, - 5y is equal to 5.*0660

*Now, dividing both sides by -5, we will have y = 8, -1.*0680

*We have both halves of our solutions and I can say that our solution is 3, -1.*0693

*Let us try another one.*0703

*For this next one we will try solving 3x + 3y = 0 and 4x + 2y = 3.*0709

*In this one, it looks like all of my terms on x and y are all positive.*0716

*It is not clear which one will be the easier one to get rid off, we just have to pick one and go with it.*0721

*Let us go ahead and give these x's a try over here and see if we can eliminate them.*0726

*Some things that you can do to help you eliminate some of these variables, is first just try and get them to be the exact same number.*0732

*What some of my students realizes is that if you just take this coefficient and multiply it up here,*0740

*then take the other coefficient and multiply it into your second equation.*0746

*That is usually enough to make them exactly the same .*0751

*Iâ€™m going to do that here.*0753

*Iâ€™m going to take the entire second equation and we will multiply it by 3.*0755

*We will take everything in the first equation and we will multiply that by 4.*0762

*Let us see, 4 Ã— 3 = 12x + 12y, 0 Ã— 4 = 0, so that will be my new first equation.*0768

*Everything in the second one by 3, 12x + 6y = 9 will be the second one.*0783

*We are off to a good start and I can see at least that the x's are now exactly the same.*0792

*We want them to cancel out, let us go ahead and take one of our equations and multiply it entirely through by a negative number.*0798

*Let us do it to the second one, everything through in the second one by -1.*0806

*It will give us -12x - 6y = -9, the first equation is exactly the same so we would not touch that one.*0810

*By multiplying it just that way, we are doing a little bit of prep work, now I can be assured that my x's will definitely cancel out.*0823

*Let us add these 2 equations together and end up solving for y.*0831

*12x - 12y that will give us 0x, that is gone.*0836

*12y - 6y will be 6y and 0 + -9 = -9.*0841

*y = -9/6 which we might as well just reduce that call it to say â€“ 3/2 and there is half of our solution.*0852

*Let us go ahead and take this, plug it back into the original and see if we can find our other half.*0866

*Iâ€™m just going to choose this one right here, 4x + 2 =- 3/2 equals 3.*0873

*In this equation, the only thing I need to solve for is that x, let us go ahead and do that.*0886

*2 Ã— -3/2 = - 3, adding 3 to both sides here I will be left with 4x = 6 and dividing both sides by 4, I will have that x = 3/2.*0895

*My final solution here, x = 3/2 and y = -3/2.*0917

*As long as you do proper prep work, you should be able to get your final solution just fine.*0926

*Remember that it has 2 parts, the x value and y value.*0931

*Let us see if we can solve this one using elimination.*0939

*I have my x's that are both positive and my yâ€™s are both negative, but they are almost the same.*0947

*Let us start off by multiplying the second one by 5 and see if it gets a little bit closer to canceling.*0955

*5x - 5y, 5 Ã— 12 = 60 we multiply the entire second equation by 5.*0965

*It still looks like nothing will cancel out, so I will also multiply the second equation by -1.*0977

*-5x + 5y = - 60 looking pretty good here.*0983

*I will add these together and I can eliminate my x's but you will notice that it looks like the y will also eliminate.*0998

*If the x's and yâ€™s are gone, the only thing on the left side is just 0.*1007

*What is on the other side? 3 - 60 would be a -57 and that is a bit of a problem because the 0 does not equal -57.*1013

*Let us say that this is a false statement.*1026

*This is one of these situations and it is a good idea to go back through your work, check it and make sure it is all correct.*1031

*All of our steps do look good here, we multiplied it by 5, we multiplied by - 1.*1038

*We do not actually have a whole lot of spots where we could make mistakes.*1042

*What this false statement is telling us here is that there is no solution to the system.*1046

*These 2 lines are actually parallel and they are not crossing whatsoever.*1057

*Watch out for those special cases.*1061

*Let us try one last example, x â€“ 4y =2 and 4x â€“ 16y = 8.*1068

*I want to try and maybe eliminate, let us do the x's.*1077

*We will do this by multiplying the first equation by -4.*1081

* -4x, -4 Ã— -4 =16y equals -8.*1094

*In the second equation I do not need to manipulate that one, I will leave it exactly the same.*1105

*This is a lot like our previous example.*1113

*The x's are going to cancel out when they are added together, but then again, so are the y values.*1117

*Both of them are going to cancel out.*1122

*I have 0 on one side of my equal sign.*1124

*This one actually is not quite so bad because if I look at my -8 and 8, they will cancel each other out as well.*1128

*I will get 0 = 0, which happens to be a true statement.*1137

*Now, one statement is true but what is your x and y?*1141

*In indicates that the lines are exactly the same and there are on top of one another.*1145

*We have an infinite number of solutions.*1151

*Using one of these algebraic methods is a great way to figure out what the solution is.*1164

*Keep in the back of your mind what the graphs of these look like*1169

*so we can interpret what it means to have one solution, no solution, or an infinite amount of solutions.*1172

*Thank you for watching www.educator.com.*1178

1 answer

Last reply by: Professor Eric Smith

Mon Jun 16, 2014 3:00 PM

Post by David Saver on June 10, 2014

Great Teaching!

2 answers

Last reply by: Professor Eric Smith

Mon Feb 10, 2014 4:48 PM

Post by Henry Sith on February 10, 2014

I believe in one of the "Practice questions" there is an error, the stated question is: 12x + 3y = 4811x Ã¢Ë†â€™ 3y = 33

(BTW, Educator should look to make the questions easier to read)

Here is the error:

Step 1. 12x + 3y = 48

+11x Ã¢Ë†â€™ 3y = 33

Step 2. 22x = 81