 # Systems of Equations (Elimination)

Here you’ll learn the final method for solving a system of equations. Below you can see instructions on how to proceed, followed by an example of how to solve a system of equations with this method.

Rule

### TheEliminationMethod

1.
Choose one of the variables that you want to eliminate.
2.
Multiply the equations $I$ and $II$ with the numbers that make the variable you chose have the same coefficient, but with opposite signs.
3.
Write the new equations below the first two.
4.
Add equation $II$ to $I$ and write the answer below these equations. Now you’ve got one equation with one unknown. Solve it.
5.
Put the answer you find into equation $I$ and solve for the final variable.
6.
Write your answer with coordinates: 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.
I decide to get rid of $y$.
2.
Since you have $2y$ in $II$ and $y$ in $I$, you have to multiply $I$ with $-2$ to eliminate $y$. In this case you don’t have to multiply equation $II$ with anything: $\begin{array}{llll}\hfill I\phantom{\rule{2em}{0ex}}y+2x=1& \phantom{\rule{1em}{0ex}}|×\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill II\phantom{\rule{2em}{0ex}}\underline{2y-x=2}& \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
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. I choose equation $I$: $\begin{array}{llll}\hfill y+2x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+2×\left(0\right)& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+0& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
6.
Write the answer with coordinates: ANSWER: $\left(x,y\right)=\left(0,1\right)$