We are given a system of two linear equations with two variables, $x$ and $y$. We need to find the values of $x$ and $y$ that satisfy both equations. The system of equations is: $6x - 3y = -39$ $-3x + 8y = 39$

1. Problem Description

We are given a system of two linear equations with two variables, $x$ and $y$. We need to find the values of $x$ and $y$ that satisfy both equations. The system of equations is:

$6x - 3y = -39$

$-3x + 8y = 39$

2. Solution Steps

We can use the substitution or elimination method to solve this system. Let's use the elimination method.

First, we can multiply the second equation by 2 to make the coefficient of $x$ in the second equation equal to the negative of the coefficient of $x$ in the first equation.

$2(-3x + 8y) = 2(39)$

$-6x + 16y = 78$

Now, we have the following system of equations:

$6x - 3y = -39$

$-6x + 16y = 78$

Next, we can add the two equations to eliminate $x$.

$(6x - 3y) + (-6x + 16y) = -39 + 78$

$13y = 39$

Now we can solve for $y$:

$y = \frac{39}{13}$

$y = 3$

Now that we have the value of $y$, we can substitute it into either of the original equations to solve for $x$. Let's substitute it into the first equation:

$6x - 3(3) = -39$

$6x - 9 = -39$

$6x = -39 + 9$

$6x = -30$

$x = \frac{-30}{6}$

$x = -5$

Thus, the solution to the system of equations is $x = -5$ and $y = 3$.

3. Final Answer

$x = -5, y = 3$