To solve the system of equations:
- \(-3x - 4y = -23\) (Equation 1)
- \(3x - 5y = 5\) (Equation 2)
We can use either the substitution method or the elimination method. For this solution, let’s use the elimination method.
First, we can eliminate \(x\) by multiplying Equation 2 by 1 and adding it to Equation 1 after multiplying it by \( -1\):
Starting with:
- \( -3x - 4y = -23 \) (Equation 1)
Multiply Equation 2 by 1:
- \( 3x - 5y = 5 \) (Equation 2)
Now we multiply Equation 1 by -1:
-1 * (−3x - 4y = −23) → \( 3x + 4y = 23 \) (Equation 3)
Now we will add Equation 2 and Equation 3:
\[ (3x + 4y) + (3x - 5y) = 23 + 5 \] \[ 3x + 4y + 3x - 5y = 28 \] \[ 6x - y = 28 \]
Now let's isolate \(y\): \[ -y = 28 - 6x \] \[ y = 6x - 28 ; (Equation 4) \]
Now we can substitute \(y\) in Equation 2:
\[ 3x - 5(6x - 28) = 5 \] \[ 3x - 30x + 140 = 5 \] \[ -27x + 140 = 5 \] \[ -27x = 5 - 140 \] \[ -27x = -135 \] \[ x = 5 \]
Now we substitute \(x = 5\) back into Equation 4 to find \(y\): \[ y = 6(5) - 28 = 30 - 28 = 2 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (5, 2) \]
So the answer is \((5, 2)\).