# Elimination method review (systems of linear equations)

CCSS Math: HSA.REI.C.6
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 $7x$ 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{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 $x=\blueD{-1}$, $y=\goldD{1}$.
We can check our solution by plugging these values back into the 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 $-4$ to get an equivalent equation that has a $\purpleD{-16x}$ 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:
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 $x=\blueD{5}$, $y=\goldD{0}$.
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{aligned} 3x+8y &= 15\\\\ 2x-8y &= 10 \end{aligned}
$x=$
$y=$
Want more practice? Check out these exercises: