To solve the system of equations:
- \(-6x + 2y = -32\) (Equation 1)
- \(4x + 3y = -9\) (Equation 2)
We will use the method of substitution or elimination. Let's use the elimination method.
First, let's manipulate Equation 1 to make it easier to work with. Divide the entire equation by 2:
\[ -3x + y = -16 \quad \text{(Equation 3)} \]
Now we have the modified system of equations:
- \(-3x + y = -16\) (Equation 3)
- \(4x + 3y = -9\) (Equation 2)
Next, we can express \(y\) from Equation 3:
\[ y = 3x - 16 \]
Now, we can substitute this expression for \(y\) into Equation 2:
\[ 4x + 3(3x - 16) = -9 \]
Simplifying this:
\[ 4x + 9x - 48 = -9 \]
Combining like terms:
\[ 13x - 48 = -9 \]
Now, add 48 to both sides:
\[ 13x = 39 \]
Dividing by 13:
\[ x = 3 \]
Now that we have \(x\), we can substitute \(x = 3\) back into the expression for \(y\):
\[ y = 3(3) - 16 \] \[ y = 9 - 16 \] \[ y = -7 \]
Thus, the solution to the system of equations is:
\[ \boxed{(3, -7)} \]