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:
2y+7x=55y7x=12\begin{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}
We notice that the first equation has a 7x7x term and the second equation has a 7x-7x term. These terms will cancel if we add the equations together—that is, we'll eliminate the xx terms:
Solving for yy, we get:
7y+0=77y=7y=1\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:
2y+7x=521+7x=52+7x=57x=7x=1\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=1x=\blueD{-1}, y=1y=\goldD{1}.
We can check our solution by plugging these values back into the the original equations. Let's try the second equation:
5y7x=12517(1)=?125+7=12\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:
9y+4x20=07y+16x80=0\begin{aligned} -9y+4x - 20&=0\\\\ -7y+16x-80&=0 \end{aligned}
We can multiply the first equation by 4-4 to get an equivalent equation that has a 16x\purpleD{-16x} term. Our new (but equivalent!) system of equations looks like this:
36y16x+80=07y+16x80=0\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}
Adding the equations to eliminate the xx terms, we get:
Solving for yy, we get:
29y+00=029y=0y=0\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:
36y16x+80=036016x+80=016x+80=016x=80x=5\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=5x=\blueD{5}, y=0y=\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.
3x+8y=152x8y=10\begin{aligned} 3x+8y &= 15\\\\ 2x-8y &= 10 \end{aligned}
x=x=
y=y=
Want more practice? Check out these exercises:
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