# How do you solve 8x + 6y = 4 and 3x + 5y = 28?

Sep 8, 2015

$\left(x , y\right) = \left(- \frac{74}{11} , \frac{106}{11}\right) = \left(6 \frac{8}{11} , 9 \frac{7}{11}\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} 8 x + 6 y = 4$
[2]$\textcolor{w h i t e}{\text{XXX}} 3 x + 5 y = 28$

Multiply [1] by 5 and [2] by 6 so we have two equations with the same coefficient for $y$
[3]$\textcolor{w h i t e}{\text{XXX}} 40 x + 30 = 20$
[4]$\textcolor{w h i t e}{\text{XXX}} 18 x + 30 y = 168$

Subtract [4] from [3]
[5]$\textcolor{w h i t e}{\text{XXX}} 22 x = - 148$

Divide [5] by $22$
[6]$\textcolor{w h i t e}{\text{XXX}} x = - \frac{74}{11}$

Substituting $- \frac{74}{11}$ for $x$ in [2]
[7]$\textcolor{w h i t e}{\text{XXX}} 3 \cdot \left(- \frac{74}{11}\right) + 5 y = 24$

Do some (tedious) arithmetic
[8]$\textcolor{w h i t e}{\text{XXX}} y = \frac{106}{11}$