# Systems of Equations (Elimination)

Here you’ll learn the final method for solving a system of equations, the elimination method. Below are the instructions for how to use this method, followed by an example of solving a system of equations with this method.

Rule

### TheEliminationMethod

1.
Choose a variable that will be eliminated.
2.
Multiply the equations $I$ and $II$ by the numbers that would make the variable you chose have the same coefficient, but with opposite signs. This will allow you to cancel them.
3.
Write the new equations below the first two.
4.
Add equation $II$ to equation $I$, and write the answer below these equations. Now, you’re left with one equation, with one unknown. Solve it.
5.
Put the answer you find into equation $I$, and solve for the final variable.
6.

ANSWER: $\left(x,y\right)=\left(a,b\right)$

Example 1

Solve the system of equations $\begin{array}{llllllll}\hfill & I\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \hfill & y+2x=1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & II\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \hfill & 2y-x=2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

1.
Let’s decide to get rid of $y$.
2.
Since you have $2y$ in equation $II$ and $y$ in equation $I$, you have to multiply equation $I$ by $-2$ to eliminate $y$. In this case, you don’t have to multiply equation $II$ by anything: $\begin{array}{llll}\hfill I\phantom{\rule{2em}{0ex}}y+2x=1& \phantom{\rule{1em}{0ex}}|\cdot \left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill II\phantom{\rule{2em}{0ex}}\underline{2y-x=2}& \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
3.
Write the two equations again after applying the changes: $\begin{array}{llll}\hfill -2y-4x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \underline{\phantom{-}2y-\phantom{4}x}& \underline{=\phantom{-}2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
4.
Add the two equations together such that $y$ is eliminated, and solve for $x$: $\begin{array}{llll}\hfill -2y+2y-4x-x& =-2+2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0y-5x& =\phantom{-}0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\phantom{-}0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
5.
Put the answer into either equation $I$ or $II$. Here, you can choose whichever you like. Let’s pick equation $I$ in this case: $\begin{array}{llll}\hfill y+2x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+2×\left(0\right)& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+0& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
6.
ANSWER: $\left(x,y\right)=\left(0,1\right)$