# How do you solve the system of equations -4x + 6y = 4 and - x + 8y = 1?

Nov 28, 2016

$\left\{\begin{matrix}x = - 1 \\ y = 0\end{matrix}\right.$

#### Explanation:

$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ - x + 8 y = 1\end{matrix}\right.$

$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ \left(- x + 8 y = 1\right) \cdot \left(- 4\right)\end{matrix}\right.$

$\left\{\begin{matrix}- 4 x + 6 y = 4 \\ 4 x - 32 y = - 4\end{matrix}\right.$

$- 26 y = 0$

Therefore

$y = 0$

Plug $0$ in for $y$ in one of the equations like so:

$- x + \left(8 \cdot 0\right) = 1$

$- x = 1$

$x = - 1$

Check your answer by putting $0$ in for $y$ and $- 1$ in for $x$ in the other equation like so:

$\left(- 4 x - 1\right) + \left(6 \cdot 0\right) = 4$

$4 + 0 = 4$

$4 = 4$

:)