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# Elimination method review (systems of linear equations)

The elimination method is a technique for solving systems of linear equations. This article reviews the technique with examples and even gives you a chance to try the method yourself.

## What is the elimination method?

The elimination method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

### Example 1

We're asked to solve this system of equations:
$\begin{array}{rl}2y+7x& =-5\\ \\ 5y-7x& =12\end{array}$
We notice that the first equation has a $7x$ term and the second equation has a $-7x$ term. These terms will cancel if we add the equations together—that is, we'll eliminate the $x$ terms:
Solving for $y$, we get:
$\begin{array}{rl}7y+0& =7\\ \\ 7y& =7\\ \\ y& =1\end{array}$
Plugging this value back into our first equation, we solve for the other variable:
$\begin{array}{rl}2y+7x& =-5\\ \\ 2\cdot 1+7x& =-5\\ \\ 2+7x& =-5\\ \\ 7x& =-7\\ \\ x& =-1\end{array}$
The solution to the system is $x=-1$, $y=1$.
We can check our solution by plugging these values back into the original equations. Let's try the second equation:
$\begin{array}{rl}5y-7x& =12\\ \\ 5\cdot 1-7\left(-1\right)& \stackrel{?}{=}12\\ \\ 5+7& =12\end{array}$
Yes, the solution checks out.
If you feel uncertain why this process works, check out this intro video for an in-depth walkthrough.

### Example 2

We're asked to solve this system of equations:
$\begin{array}{rl}-9y+4x-20& =0\\ \\ -7y+16x-80& =0\end{array}$
We can multiply the first equation by $-4$ to get an equivalent equation that has a $-16x$ term. Our new (but equivalent!) system of equations looks like this:
$\begin{array}{rl}36y-16x+80& =0\\ \\ -7y+16x-80& =0\end{array}$
Adding the equations to eliminate the $x$ terms, we get:
Solving for $y$, we get:
$\begin{array}{rl}29y+0-0& =0\\ \\ 29y& =0\\ \\ y& =0\end{array}$
Plugging this value back into our first equation, we solve for the other variable:
$\begin{array}{rl}36y-16x+80& =0\\ \\ 36\cdot 0-16x+80& =0\\ \\ -16x+80& =0\\ \\ -16x& =-80\\ \\ x& =5\end{array}$
The solution to the system is $x=5$, $y=0$.
Want to see another example of solving a complicated problem with the elimination method? Check out this video.

## Practice

Problem 1
Solve the following system of equations.
$\begin{array}{rl}3x+8y& =15\\ \\ 2x-8y& =10\end{array}$
$x=$
$y=$

Want more practice? Check out these exercises:

## Want to join the conversation?

• what if you are using three variables and you have three different equations?
for example: (three variables,three equations)
x-y-2z=4
-x+2y+z=1
-x+y-3z=11
• First of all, the only way to solve a question with 3 variables is with 3 equations. Having 3 variables and only 2 equations wouldn't allow you to solve for it. To start, choose any two of the equations. Using elimination, cancel out a variable. Using the top 2 equations, add them together. That results in y-z=5. Now, look at the third equation and cancel out the same variable that you originally cancelled out. In this case, we canceled out x. Adding the first equation to the 3rd equation would get rid of his. Adding would give -5z=15. We got lucky because both the x and the y cancelled out. If they didn't both cancel out, you would just have t solve the two equations which you should know how to do. Back to the problem, -5z=15, so z=-3.Plug that into the equation y-z=5 to solve for y. y-(-3)=5, so y+3=5. That gives y=2. Plug both of those into any of the three original equations and solve for x. You get x=0. Your final solution is x=0, y=2, and z=-3 or (0,2,-3).
• What if the numbers before x and y can not make up?

4x-3y=8
5x-2y=-11
• I don't completely understand what you mean, but I have an idea of what you're asking. You can multiply both equations by a number to get one of the x or y absolute values the same, multiply the top equation by 2, to get 8x-6y=16, and the second equation by -3 to get -15x+6y=33, then add the equations to get -7x=49, so x is equal to -7. Plug in -7 for x to solve for y which would be -12
Hope this helps
• what about unsorted equations?
-4y-11x=36
20=-10x-10y
• Yeka,

Use algebraic manipulation to get the x and y terms on the same side of the =. For the second equation, add 10x to both sides, add 10y to both sides, and subtract 20 from both sides. Then, proceed with the elimination method.
• what if you have -14x + 9y = 46 and 14x - 9y = 102. Elimination will cancel both x and y out. What do you do?
• You are correct, both the X and Y cancel out leaving you with: 0=148. This is a false statement. It is telling you that the system has no solution. It also means that the 2 lines are parallel. Parallel lines have no points in common which is why the system has no solution.
Hope this helps.
• kesvi patel
are we supposed to do last divide step ?
• After you add/subtract the new equations, you eliminate one of the variables and divide. After solving one of them, plug your solved variable to one of the original problems.
This might help you understand more clearly:

12x + 2y = 90 ... (1)
6x + 4y = 90 ... (2)

(2)*2 12x + 8y = 180 ... (2)'

(1)-(2)' 12x + 2y = 90
- 12x + 8y = 180
-6y = -90
You eliminated x
and now you solve y by dividing.
y = 15
y = 15 plugged into (2). (Always pick the easier problem to solve your problem accurately.)
6x + 4(15) = 90
6x + 60 = 90
6x = 90 - 60
6x = 30
x = 5

(x,y) = (5,15)

Hope this helps!
• In an honors class in 8th grade, I am generally confused. Can someone maybe give me the steps in a simple worded form?
• Well, it depends on what you need. lets go with the equations:

10y-11x=-4
-2y+3x=4

For this, you want to first make the bottom equation have an equivalent value for one of the variables so that we can eliminate it. the bottom equation is perfect, because:

-2y(5)+3x(5)=4(5)
this makes -10y+15x=20

which in that case...

10y-11x=-4
-10y+15x=20
=
4x=16
4x/4=16/4
x=4

from there, substitute 4 for x to solve for y, and you have your coordinate variables.
• I don't understand how we can do anything to the equasions 10y - 11x = -4 and -2y + 3x = 4!!
and every time I do this, it marks me wrong for trying to multiply the 3 and -11. And the hint doesn't help either!
• You should multiply by 3 and +11 to create -33x and +33x. You need opposite signs to eliminate the x. By using -11, you are making the x terms have the same sign and when you add the 2 equations, no variable is eliminated.

Try that and see how it works out.
Comment back with your work if you still have issues.