We are given a system of two linear equations: $-12x = -12y + 72$ $3y - 6x - 36 = 0$ We need to find the solution to this system.

1. Problem Description

We are given a system of two linear equations:

$-12x = -12y + 72$

$3y - 6x - 36 = 0$

We need to find the solution to this system.

2. Solution Steps

First, simplify the first equation by dividing by -12:

$x = y - 6$

Next, simplify the second equation by dividing by 3:

$y - 2x - 12 = 0$

Now, solve for $y$ in the second equation:

$y = 2x + 12$

Substitute the expression for $x$ from the first equation into the equation for $y$:

$y = 2(y - 6) + 12$

$y = 2y - 12 + 12$

$y = 2y$

$y - 2y = 0$

$-y = 0$

$y = 0$

Now, substitute $y = 0$ back into the equation $x = y - 6$:

$x = 0 - 6$

$x = -6$

So the solution is $x = -6$ and $y = 0$.

Check the solution in the original equations:

$-12x = -12y + 72$

$-12(-6) = -12(0) + 72$

$72 = 72$

$3y - 6x - 36 = 0$

$3(0) - 6(-6) - 36 = 0$

$0 + 36 - 36 = 0$

$0 = 0$

The solution is correct.

3. Final Answer

$x = -6, y = 0$