To solve the system of equations
- \( 5x - 4y = -44 \) (Equation 1)
- \( -x + 3y = 22 \) (Equation 2)
we can use either substitution or elimination. Here, I will use the elimination method.
First, we can multiply Equation 2 by 5 to align the coefficients of \(x\):
\[ 5(-x + 3y) = 5(22) \] \[ -5x + 15y = 110 \quad \text{(Equation 3)} \]
Now we have the modified system:
- \( 5x - 4y = -44 \) (Equation 1)
- \(-5x + 15y = 110 \) (Equation 3)
Next, we will add Equations 1 and 3:
\[ (5x - 4y) + (-5x + 15y) = -44 + 110 \] \[ (5x - 5x) + (-4y + 15y) = 66 \] \[ 0 + 11y = 66 \] \[ 11y = 66 \] \[ y = \frac{66}{11} = 6 \]
Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\). Let's use Equation 2:
\[ -x + 3(6) = 22 \] \[ -x + 18 = 22 \] \[ -x = 22 - 18 \] \[ -x = 4 \] \[ x = -4 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-4, 6) \]