# How do you solve 6y + 4x = 12 and -6x + y = -8?

Sep 5, 2015

I found:
$x = \frac{3}{2}$
$y = 1$

#### Explanation:

You can rearrange the two equations and write:
$\left\{\begin{matrix}4 x + 6 y = 12 \\ - 6 x + y = - 8\end{matrix}\right.$
we can now multiply the second equation by $- 6$ to get:
$\left\{\begin{matrix}4 x + 6 y = 12 \\ \textcolor{red}{36 x - 6 y = 48}\end{matrix}\right.$
we now add together (by column) the two equations:
$40 x + 0 = 60$ so that:
$x = \frac{60}{40} = \frac{6}{4} = \frac{3}{2}$
substitute this value into the first equation:
$4 \left(\frac{3}{2}\right) + 6 y = 12$
$6 + 6 y = 12$
$y = 1$

hope it helps!