# How do you solve the system -2x+3y=-9 and -x+-3y=-3?

Jun 24, 2015

Solve the two possibilities separately:
$\left\{\begin{matrix}- 2 x + 3 y = - 9 \\ - x + 3 y = - 3\end{matrix}\right. \mathmr{and} \left\{\begin{matrix}- 2 x + 3 y = - 9 \\ - x - 3 y = - 3\end{matrix}\right.$
giving $\left(x , y\right) = \left(6 , 1\right)$ and $\left(x , y\right) = \left(4 , - \frac{1}{3}\right)$

#### Explanation:

Case 1:
$\left\{\left(\begin{matrix}1 \textcolor{w h i t e}{\text{XXX")-2x+3y=-9) \\ (2color(white)("XXX}} - x + 3 y = - 3\end{matrix}\right)\right.$
subtracting  from 
$\left[3\right] \textcolor{w h i t e}{\text{XXXX}}$$- x = - 6$
$\left[4\right] \textcolor{w h i t e}{\text{XXXX}}$$x = 6$
substituting $x = 6$ in 
$\left[5\right] \textcolor{w h i t e}{\text{XXXX}}$$- 6 + 3 y = - 3$
$\left[6\right] \textcolor{w h i t e}{\text{XXXX}}$$y = 1$

Case 2:
$\left\{\left(\begin{matrix}7 \textcolor{w h i t e}{\text{XXX")-2x+3y=-9) \\ (8color(white)("XXX}} - x - 3 y = - 3\end{matrix}\right)\right.$

$\left[9\right] \textcolor{w h i t e}{\text{XXXX}} - 3 x = - 12$
$\left[10\right] \textcolor{w h i t e}{\text{XXXX}} x = = 4$

$\left[11\right] \textcolor{w h i t e}{\text{XXXX}} - 4 - 3 y = - 3$
$\left[12\right] \textcolor{w h i t e}{\text{XXXX}} y = - \frac{1}{3}$