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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

- 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.
- Write your answer using coordinates:
ANSWER: $(x,y)=(a,b)$

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: $$\begin{array}{llll}\hfill I\phantom{\rule{2em}{0ex}}y+2x=1& \phantom{\rule{1em}{0ex}}|\cdot ({-2})\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill II\phantom{\rule{2em}{0ex}}\underline{2y-x=2}& \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
- 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: $$\begin{array}{llll}\hfill y+2x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+2\times ({0})& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y+0& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
- 6.
- Write the answer using coordinates:
ANSWER: $(x,y)=(0,1)$