How do you solve 8x - 8y = 16 and 4x + 4y = -72 ?

Work with one equation to get a term in the other, then substitute in, to get the system $x = - 8 , y = - 10$

Explanation:

We have a system of 2 equations:

$8 x - 8 y = 16$
$4 x + 4 y = - 72$

Since this is in the substitution method category, we'll use that method.

If we take the second equation and multiply both sides by 2, we can get the term $8 x$. We can take what $8 x$ will equal and substitute it into the first equation:

$\left(4 x + 4 y\right) \left(2\right) = \left(- 72\right) \left(2\right)$

$8 x + 8 y = - 144$

$8 x = - 144 - 8 y$

So let's now substitute. Starting with the first equation:

$8 x - 8 y = 16$

$- 144 - 8 y - 8 y = 16$

$- 16 y = 160$

$y = - 10$

We can now solve for x. I'm going to do it with both equations so that we can make sure we didn't make a mistake (if we get 2 different answers in checking the 2 equations, we did!)

$8 x - 8 y = 16$

$8 x - 8 \left(- 10\right) = 16$

$8 x + 80 = 16$

$8 x = - 64$

$x = - 8$

and now the other one:

$4 x + 4 y = - 72$

$4 x + 4 \left(- 10\right) = - 72$

$4 x - 40 = - 72$

$4 x = - 32$

$x = - 8$

$x = - 8 , y = - 10$