To solve the system of equations
\[ -4x + 6y = 38 \quad (1) \] \[ x - 5y = -20 \quad (2) \]
we can use substitution or elimination. Here, we will use substitution.
From equation (2), we can express \( x \) in terms of \( y \):
\[ x = 5y - 20 \]
Now, we substitute this expression for \( x \) into equation (1):
\[ -4(5y - 20) + 6y = 38 \]
Distributing \(-4\):
\[ -20y + 80 + 6y = 38 \]
Combine like terms:
\[ -20y + 6y + 80 = 38 \] \[ -14y + 80 = 38 \]
Now isolate \( y \):
\[ -14y = 38 - 80 \] \[ -14y = -42 \]
Dividing both sides by \(-14\):
\[ y = 3 \]
Now substitute \( y \) back into the expression we found for \( x \):
\[ x = 5(3) - 20 \] \[ x = 15 - 20 \] \[ x = -5 \]
Thus, the solution to the system is:
\[ (x, y) = (-5, 3) \]
The correct response is:
\[ \boxed{(-5, 3)} \]