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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: