Question

Solve the system of equations 4x-8y=-36 and -6x+y=32

Answer 1

To solve the system of equations

1) \( 4x - 8y = -36 \)
2) \( -6x + y = 32 \)

we can use the method of substitution or elimination. Here, we'll use substitution.

First, let's solve the second equation for \( y \):

\[
y = 6x + 32
\]

Now we can substitute this expression for \( y \) into the first equation:

\[
4x - 8(6x + 32) = -36
\]

Distributing the \(-8\):

\[
4x - 48x - 256 = -36
\]

Combining like terms:

\[
-44x - 256 = -36
\]

Now, we can isolate \( x \) by adding 256 to both sides:

\[
-44x = -36 + 256
\]
\[
-44x = 220
\]

Dividing both sides by -44:

\[
x = \frac{220}{-44} = -5
\]

Now that we have \( x \), we can substitute it back into the equation we found for \( y \):

\[
y = 6(-5) + 32
\]
\[
y = -30 + 32
\]
\[
y = 2
\]

Thus, the solution to the system of equations is

\[
(x, y) = (-5, 2)
\]

To verify, we can substitute \( x = -5 \) and \( y = 2 \) back into the original equations:

1) \( 4(-5) - 8(2) = -20 - 16 = -36 \) (True)
2) \( -6(-5) + 2 = 30 + 2 = 32 \) (True)

Both equations are satisfied, confirming our solution is correct.

The final answer is:

\[
(x, y) = (-5, 2)
\]