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:
2y+7x=55y7x=12\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:
Solving for y, 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, equals, start color blueD, minus, 1, end color blueD, y, equals, start color goldD, 1, end color goldD.
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 minus, 4 to get an equivalent equation that has a start color purpleD, minus, 16, x, end color purpleD 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 x terms, we get:
Solving for y, 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, equals, start color blueD, 5, end color blueD, y, equals, start color goldD, 0, end color goldD.
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, equals
  • Svaret bør være
  • et heltall, som 6
  • en forkortet ekte brøk, som 3, slash, 5
  • en forkortet uekte brøk, som 7, slash, 4
  • et blandet tall, som 1, space, 3, slash, 4
  • et presist desimaltall, som 0, comma, 75
  • et multiplum av pi, som 12, space, p, i eller 2, slash, 3, space, p, i
y, equals
  • Svaret bør være
  • et heltall, som 6
  • en forkortet ekte brøk, som 3, slash, 5
  • en forkortet uekte brøk, som 7, slash, 4
  • et blandet tall, som 1, space, 3, slash, 4
  • et presist desimaltall, som 0, comma, 75
  • et multiplum av pi, som 12, space, p, i eller 2, slash, 3, space, p, i

Want more practice? Check out these exercises: