To solve the system of equations using substitution, we have the following equations:
- \( -4x + 6y = 38 \)
- \( x - 5y = -20 \)
First, let's solve the second equation for \( x \):
\[ x = 5y - 20 \]
Now we can substitute this expression for \( x \) into the first equation:
\[ -4(5y - 20) + 6y = 38 \]
Distribute \(-4\):
\[ -20y + 80 + 6y = 38 \]
Combine like terms:
\[ -14y + 80 = 38 \]
Now, isolate \( y \):
\[ -14y = 38 - 80 \] \[ -14y = -42 \] \[ y = \frac{-42}{-14} = 3 \]
Now that we have \( y \), we can substitute back to find \( x \):
\[ x = 5(3) - 20 \] \[ x = 15 - 20 \] \[ x = -5 \]
So the solution to the system of equations is \( (x, y) = (-5, 3) \).
The correct response is:
(−5, 3).