How do solve the following linear system?:  9 x-6y =3 , -3x -8y = -9 ?

Feb 3, 2016

$\left(x , y\right) = \left(\frac{13}{15} , \frac{4}{5}\right)$

Explanation:

Given
[equation 1]$\textcolor{w h i t e}{\text{XXX}} 9 x - 6 y = 3$
[equation 2]$\textcolor{w h i t e}{\text{XXX}} - 3 x - 8 y = - 9$

If we multiply [equation 2] by $3$ then the coefficient of $x$ will be of the same magnitude as in [equation 1]
[equation 3]$\textcolor{w h i t e}{\text{XXX}} - 9 x - 24 y = - 27$

Adding [equation 1] and [equation 3]
[equation 4]$\textcolor{w h i t e}{\text{XXX}} - 30 y = - 24$

Dividing [equation 4] by $\left(- 30\right)$ gives
[equation 5]$\textcolor{w h i t e}{\text{XXX}} y = \frac{24}{30} = \frac{4}{5}$

Substituting $\frac{4}{5}$ for $y$ in [equation 1]
[equation 6]$\textcolor{w h i t e}{\text{XXX}} 9 x - 6 \times \frac{4}{5} = 3$

[equation 7]$\textcolor{w h i t e}{\text{XXX}} 9 x = 3 + \frac{24}{5} = \frac{39}{5}$

Dividing [equation 7] by $9$ gives
[equation 8]$\textcolor{w h i t e}{\text{XXX}} x = \frac{39}{45} = \frac{13}{15}$