We are given a system of two linear equations: $-8x + 4y = 0$ $x + 6y = -13$ We need to solve for $x$ and $y$ using the substitution method.

1. Problem Description

We are given a system of two linear equations:

$-8x + 4y = 0$

$x + 6y = -13$

We need to solve for $x$ and $y$ using the substitution method.

2. Solution Steps

First, solve the second equation for $x$ in terms of $y$:

$x = -6y - 13$

Now, substitute this expression for $x$ into the first equation:

$-8(-6y - 13) + 4y = 0$

$48y + 104 + 4y = 0$

$52y + 104 = 0$

$52y = -104$

$y = -104/52$

$y = -2$

Now that we have the value of $y$, we can substitute it back into the equation $x = -6y - 13$ to find $x$:

$x = -6(-2) - 13$

$x = 12 - 13$

$x = -1$

3. Final Answer

$x = -1$

$y = -2$

The solution is $(-1, -2)$.