How do you solve -8x - 8y = 0 and -8x + 8y = -128?

Sep 3, 2017

See a solution process below:

Explanation:

Step 1) Solve the first equation for $x$:

$- 8 x - 8 y = 0$

$- 8 x - 8 y + + \textcolor{red}{8 y} = 0 + \textcolor{red}{8 y}$

$- 8 x - 0 = 8 y$

$- 8 x = 8 y$

$\frac{- 8 x}{\textcolor{red}{- 8}} = \frac{8 y}{\textcolor{red}{- 8}}$

$1 x = - 1 y$

$x = - y$

Step 2) Substitute $- y$ for $x$ in the second equation and solve for $y$:

$- 8 x + 8 y = - 128$ becomes:

$\left(- 8 \cdot - y\right) + 8 y = - 128$

$8 y + 8 y = - 128$

$16 y = - 128$

$\frac{16 y}{\textcolor{red}{16}} = - \frac{128}{\textcolor{red}{16}}$

$y = - 8$

Step 3) Substitute $- 8$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = - y$ becomes:

$x = - - 8$

$x = 8$

The Solution Is: $x = 8$ and $y = - 8$ or $\left(8 , - 8\right)$