Question

Solve the system of equations.
−6x+2y=−32

4x+3y=−9

Answer 1

To solve the system of equations

1. \(-6x + 2y = -32\) (Equation 1)
2. \(4x + 3y = -9\) (Equation 2)

we can use either substitution or elimination methods. Here, I'll use the elimination method.

Let's start by multiplying Equation 1 by 1 so that it stays the same, and we will multiply Equation 2 by 1 so that both equations are aligned for elimination.

1. \(-6x + 2y = -32\)
2. \(4x + 3y = -9\)

Next, to eliminate \(x\), we will adjust these equations. We can multiply Equation 1 by 2 to make the coefficient of \(x\) in both equations the same in terms of magnitude (this will give us \(12x\) for Equation 1 and \(12x\) for Equation 2 when both are made positive):

1. \(2(-6x + 2y) = 2(-32) \rightarrow -12x + 4y = -64\)
2. \(3(4x + 3y) = 3(-9) \rightarrow 12x + 9y = -27\)

Now we have the new system:

1. \(-12x + 4y = -64\)
2. \(12x + 9y = -27\)

Adding these two equations:

\[
(-12x + 4y) + (12x + 9y) = -64 - 27
\]

The \(x\) terms cancel out:

\[
0 + (4y + 9y) = -91
\]
\[
13y = -91
\]

Now, divide both sides by 13:

\[
y = -\frac{91}{13} \approx -7
\]

Next, we can substitute \(y = -7\) back into one of the original equations to find \(x\). We'll use Equation 1:

\[
-6x + 2(-7) = -32
\]
\[
-6x - 14 = -32
\]

Now, add 14 to both sides:

\[
-6x = -32 + 14
\]
\[
-6x = -18
\]

Now, divide by -6:

\[
x = 3
\]

Thus, the solution to the system of equations is:

\[
\boxed{(3, -7)}
\]