To solve the system of equations:
- \(-6x + 2y = -32\) (Equation 1)
- \(4x + 3y = -9\) (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 \((3, -7)\).