Question

−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 the substitution or elimination method. Here, I'll use the elimination method.

First, we can simplify Equation 1 by dividing all terms by 2:

\[
-3x + y = -16 \quad \text{(Equation 1 simplified)}
\]

Now, we can isolate \(y\) in Equation 1:

\[
y = 3x - 16 \quad \text{(Equation 3)}
\]

Next, we'll substitute Equation 3 into Equation 2:

\[
4x + 3(3x - 16) = -9
\]

Distributing the \(3\):

\[
4x + 9x - 48 = -9
\]

Combine like terms:

\[
13x - 48 = -9
\]

Now, add \(48\) to both sides:

\[
13x = 39
\]

Then, divide by \(13\):

\[
x = 3
\]

Now that we have \(x\), we will substitute \(x = 3\) back into Equation 3 to find \(y\):

\[
y = 3(3) - 16
\]
\[
y = 9 - 16
\]
\[
y = -7
\]

Thus, the solution to the system of equations is:

\[
x = 3, \quad y = -7
\]

In ordered pair form, the solution is \((3, -7)\).