# How do solve the following linear system?:  -9x + 6y = -2 , 3x-y=7 ?

Mar 20, 2017

$\left(x , y\right) = \left(\frac{40}{9} , \frac{19}{3}\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} - 9 x + 6 y = - 2$
[2]$\textcolor{w h i t e}{\text{XXX}} 3 x - y = 7$

Multiplying [2] by $3$
[3]$\textcolor{w h i t e}{\text{XXX}} 9 x - 3 y = 21$

[4]$\textcolor{w h i t e}{\text{XXX}} 3 y = 19$

Dividing [4] by $3$
[5]$\textcolor{w h i t e}{\text{XXX}} y = \frac{19}{3}$

Substituting $\frac{19}{4}$ for $y$ in [2]
[6]$\textcolor{w h i t e}{\text{XXX}} 3 x - \frac{19}{3} = 7$

Adding $\frac{19}{4}$ to both sides of [6] and simplifying the right side
[7]$\textcolor{w h i t e}{\text{XXX}} 3 x = \frac{40}{3}$

Dividing [7] by $3$
[8]$\textcolor{w h i t e}{\text{XXX}} x = \frac{40}{9}$

You might want to simplify this as $\left(x , y\right) = \left(3 \frac{4}{9} , 6 \frac{1}{3}\right)$
however
leaving these values in their improper fraction forms will probably make the verification easier (...and you always do this; right?)