Sign In | Subscribe
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 1
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (5)

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 − 3y = 33

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

Here is the error:
Step 1. 12x + 3y = 48
+11x − 3y = 33
Step 2. 22x = 81

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

Solve the system:
2x - 4y = 10
6x + 4y = 14
  • 2x − 4y = 10 + 6x + 4y = 14
  • 2x − 4y = 10 + 6x + 4y = 14 8x = 24
  • x = 3
  • 2x − 4y = 102(3) − 4y = 10
  • 6 − 4y = 10
  • − 4y = 4
y = - 1
Solve the system:
2x + 8y = - 12
16x - 8y = 30
  • 2x + 8y = − 12+ (16x8y = 30)
  • 18x = 18
  • x = 1
  • Plug x back into the equation to find y
  • 2(1) + 8y = − 12
  • 2 + 8y = − 12
  • 8y = − 14
  • y = [( − 14)/8]
y = - [7/4]
8x - 4y = 24
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
y = 4
12x + 3y = 48
11x - 3y = 33
  • 12x + 3y = 48
    + 11x - 3y = 33
  • 23x = 81
x = 3[12/23]
7x - 5y = 17
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
x = 1
5x - 6y = 19
5x - 2y = 11
  • 5x - 6y = 19
    - 5x - 2y = 11
  • 4y = 8
  • y = 2
  • 5x - 6(2) = 19
  • 5x - 12 = 19
  • 5x = 31
x = 6[1/5]
2x - 3y = 18
2x - 7y = 6
  • 2x - 3y = 18
    - 2x - 7y = 6
  • 10y = 12
  • y = [12/10] = [6/5]
  • y = 1[1/5]
  • 2x − 3( 1[1/5] ) = 18
  • 2x − 3[3/5] = 18
  • 2x = 21[3/5]
x = 10[4/5]
x - 5y = 17
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
x = 1
5x - 6y = 19
5x - 2y = 11
  • 5x - 6y = 19
    - 5x - 2y = 11
  • 4y = 8
  • y = 2
  • 5x - 6(2) = 19
  • 5x - 12 = 19
  • 5x = 31
x = 6[1/5]
2x - 3y = 18
2x - 7y = 6
  • 2x - 3y = 18
    - 2x - 7y = 6
  • 10y = 12
  • y = [12/10] = [6/5]
  • y = 1[1/5]
  • 2x − 3( 1[1/5] ) = 18
  • 2x − 3[3/5] = 18
  • 2x = 21[3/5]
x = 10[4/5]

*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.


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

Transcription: Solving a System Using Elimination

Welcome back to

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

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

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

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

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

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

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

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

Do I have - 3x and 3x?0059

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

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

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

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

We will end up solving that new equation.0086

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

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

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

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

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

Here I have 2x – 7y = 2 and 3x + y = -20.0129

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

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

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

2x + 3x would give me 5x and – 7y + y would give me -6y.0154

2 and -20 is -18, so I definitely created a new equation, but it still has x’s and it still has y’s.0163

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

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

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

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

Is it going to be the x's or y’s?0191

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

In order to make these guys came flat, let us take everything in the second equation and multiply it by 7.0200

The first equation exactly the way it is, no changes, and everything in the second equation will get multiplied by 7.0210

3 × 7 = 21x, I have 7y, -20 × 7 would be -140.0218

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

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

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

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

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

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

I can solve this new equation for x and figure out what it needs to be which I can do by dividing both sides by 23.0270

How many times 23 is going to 138?0292

6 × 3 = 18, looks like 6 times so x = -6.0303

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

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

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

I end up solving this one for y, 2 × -6 = -12 and we are adding 12 to both sides would give me 14 and y divided by -7, I have -2.0335

I have both halves of the solution.0360

I have x = -6 and that y = -2.0362

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

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

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

I’m not going all the way through this, but just to show you how you could have started.0390

For example, if you multiply the first one by 3, it would have given you 6x - 21y = 6.0394

If you multiply the second equation by -2, - 6x - 2y = 40.0405

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

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

Onto the next important thing with the elimination method.0428

We have many different things that could happen. 0432

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

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

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

You actually be able to find your solution.0464

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

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

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

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

I think I mentioned this earlier, but go ahead and check your work if it looks like you are getting a false statement0483

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

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

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

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

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

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

In this first example, we will look at solving 2x - 5y = 11 and 3x + y = 8.0534

Think about our goal with the elimination method.0543

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

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

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

The way we are going to do this is when I take the second equation and we are going to multiply everything there by 5.0570

Let us see the result this will have.0579

We have not touched the first equation and everything in the second equation by 5 would be 15x + 5y = 40.0582

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

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

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

We will do that by dividing both sides by 17, x =3.0625

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

(2 × 3) - 5y = 11.0647

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

We will multiply the 2 and 3 together and get 6, then we will subtract 6 from both sides, - 5y is equal to 5.0660

Now, dividing both sides by -5, we will have y = 8, -1.0680

We have both halves of our solutions and I can say that our solution is 3, -1.0693

Let us try another one.0703

For this next one we will try solving 3x + 3y = 0 and 4x + 2y = 3.0709

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

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

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

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

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

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

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

I’m going to do that here. 0753

I’m going to take the entire second equation and we will multiply it by 3.0755

We will take everything in the first equation and we will multiply that by 4.0762

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

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

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

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

Let us do it to the second one, everything through in the second one by -1.0806

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

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

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

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

12y - 6y will be 6y and 0 + -9 = -9.0841

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

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

I’m just going to choose this one right here, 4x + 2 =- 3/2 equals 3.0873

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

2 × -3/2 = - 3, adding 3 to both sides here I will be left with 4x = 6 and dividing both sides by 4, I will have that x = 3/2.0895

My final solution here, x = 3/2 and y = -3/2.0917

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

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

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

I have my x's that are both positive and my y’s are both negative, but they are almost the same.0947

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

5x - 5y, 5 × 12 = 60 we multiply the entire second equation by 5.0965

It still looks like nothing will cancel out, so I will also multiply the second equation by -1.0977

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

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

If the x's and y’s are gone, the only thing on the left side is just 0.1007

What is on the other side? 3 - 60 would be a -57 and that is a bit of a problem because the 0 does not equal -57.1013

Let us say that this is a false statement.1026

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

All of our steps do look good here, we multiplied it by 5, we multiplied by - 1.1038

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

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

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

Watch out for those special cases.1061

Let us try one last example, x – 4y =2 and 4x – 16y = 8.1068

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

We will do this by multiplying the first equation by -4.1081

-4x, -4 × -4 =16y equals -8.1094

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

This is a lot like our previous example.1113

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

Both of them are going to cancel out.1122

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

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

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

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

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

We have an infinite number of solutions.1151

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

Keep in the back of your mind what the graphs of these look like1169

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

Thank you for watching