# Systems of Equations (Substitution)

Rule

### TheSubstitutionMethod

1.
Choose one equation and solve for one of the variables.
2.
Put the expression you got in the first step into the equation you haven’t used yet. Solve that equation.
3.
Take this solution and put it back into the first equation. Solve it.
4.

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

Example 1

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

using the substitution method

1.
Choose the equation you think looks easier and solve for one of the variables: $\begin{array}{llll}\hfill I\phantom{\rule{1em}{0ex}}& y+2x=1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & y=1-2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
2.
Insert the expression you just found into the equation you haven’t used yet: $\begin{array}{llll}\hfill y-x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(1-2x\right)-x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1-2x-x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -3x& =-2-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{−3}x}{\text{−3}}& =\frac{\text{−3}}{\text{−3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
3.
Then put this value back into the first expression you found: $\begin{array}{llll}\hfill & y=1-2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & y=1-2\cdot \left(1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & y=1-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & y=-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
4.