If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}
We notice that the first equation has a 7, x term and the second equation has a minus, 7, x term. These terms will cancel if we add the equations together—that is, we'll eliminate the x terms:
\begin{aligned} 2y+\redD{7x} &= -5 \\ +~5y\redD{-7x}&=12\\ \hline\\ 7y+0 &=7 \end{aligned}
Solving for y, we get:
\begin{aligned} 7y+0 &=7\\\\ 7y &=7\\\\ y &=\goldD{1} \end{aligned}
Plugging this value back into our first equation, we solve for the other variable:
\begin{aligned} 2y+7x &= -5\\\\ 2\cdot \goldD{1}+7x &= -5\\\\ 2+7x&=-5\\\\ 7x&=-7\\\\ x&=\blueD{-1} \end{aligned}
The solution to the system is x, equals, start color #11accd, minus, 1, end color #11accd, y, equals, start color #e07d10, 1, end color #e07d10.
We can check our solution by plugging these values back into the original equations. Let's try the second equation:
\begin{aligned} 5y-7x &= 12\\\\ 5\cdot\goldD{1}-7(\blueD{-1}) &\stackrel ?= 12\\\\ 5+7 &= 12 \end{aligned}
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{aligned} -9y+4x - 20&=0\\\\ -7y+16x-80&=0 \end{aligned}
We can multiply the first equation by minus, 4 to get an equivalent equation that has a start color #7854ab, minus, 16, x, end color #7854ab term. Our new (but equivalent!) system of equations looks like this:
\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}
Adding the equations to eliminate the x terms, we get:
\begin{aligned} 36y-\redD{16x} +80&=0 \\ {+}~-7y+\redD{16x}-80&=0\\ \hline\\ 29y+0 -0&=0 \end{aligned}
Solving for y, we get:
\begin{aligned} 29y+0 -0&=0 \\\\ 29y&=0 \\\\ y&=\goldD 0 \end{aligned}
Plugging this value back into our first equation, we solve for the other variable:
\begin{aligned} 36y-16x+80&=0\\\\ 36\cdot 0-16x+80&=0\\\\ -16x+80&=0\\\\ -16x&=-80\\\\ x&=\blueD{5} \end{aligned}
The solution to the system is x, equals, start color #11accd, 5, end color #11accd, y, equals, start color #e07d10, 0, end color #e07d10.
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{aligned} 3x+8y &= 15\\\\ 2x-8y &= 10 \end{aligned}
x, equals
y, equals

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).
-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.
• 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.

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!
• when plugging in... do i plug in to the original equation or the (new) equivalent equation?
• The new equivalent equation would be plugged in because then you will be able to solve for one of the variables. Then, you can plug that variable into the equation, and then solve for the last one using algebraic manipulation.
• Do you always have to eliminate the x term?
• No, you can pick the one that looks like it is the easiest to eliminate.
• Do you always have to eliminate the x term? Or could you do it with y, too?
• You can eliminate whichever term you want.
• what if my equations are like this:
4y=2x-8
5x-4y=20
where my equations are not lined up. how do I solve this?
• You need to change the first equation to standard form. Subtract 2x from both sides to get:
-2x+4y=-8
5x-4y=20

Now the variables line up. Hope this helps.