# How do you solve -2x -8y = 4 and x+ 16y = 22?

Sep 6, 2015

$\left\{\begin{matrix}x = - 10 \\ y = 2\end{matrix}\right.$

#### Explanation:

Your system of equations looks like this

$\left\{\begin{matrix}- 2 x - 8 y = 4 \\ x + 16 y = 22\end{matrix}\right.$

Multiply the second equation by $2$ to get

$\left\{\begin{matrix}- 2 x - 8 y = 4 \\ x + 16 y = 22 | \cdot \left(2\right)\end{matrix}\right.$

$\left\{\begin{matrix}- 2 x - 8 y = 4 \\ 2 x + 32 y = 44\end{matrix}\right.$

Add these two equations by taking the left-hand sides and the right-hand sides separately

$- \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} - 8 y + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} + 32 y = 4 + 44$

$24 y = 48 \implies y = \frac{48}{24} = 2$

Take this value of $y$ into one of the original equations and solve for $x$

$x + 16 \cdot \left(2\right) = 22$

$x + 32 = 22 \implies x = - 10$

The solution set for this system will be

$\left\{\begin{matrix}x = - 10 \\ y = 2\end{matrix}\right.$