To solve the system of equations
\[ -4x + 6y = 38 \quad \text{(1)} \] \[ x - 5y = -20 \quad \text{(2)}, \]
we can use the substitution or elimination method. Here, we will use the substitution method.
First, let's solve equation (2) for \(x\):
\[ x = 5y - 20. \]
Now, we will substitute this expression for \(x\) into equation (1):
\[ -4(5y - 20) + 6y = 38. \]
Expanding this gives:
\[ -20y + 80 + 6y = 38. \]
Combining like terms yields:
\[ -14y + 80 = 38. \]
Now, subtract 80 from both sides:
\[ -14y = 38 - 80, \] \[ -14y = -42. \]
Next, divide both sides by -14:
\[ y = \frac{-42}{-14} = 3. \]
Now that we have \(y\), we will substitute it back into the expression for \(x\):
\[ x = 5(3) - 20, \] \[ x = 15 - 20, \] \[ x = -5. \]
Thus, the solution to the system of equations is
\[ \boxed{(-5, 3)}. \]
To verify, we can substitute \(x = -5\) and \(y = 3\) back into both original equations:
- For the first equation:
\[ -4(-5) + 6(3) = 20 + 18 = 38 \quad \text{(correct)}. \]
- For the second equation:
\[ -5 - 5(3) = -5 - 15 = -20 \quad \text{(correct)}. \]
Both equations are satisfied, confirming the solution is indeed \( \boxed{(-5, 3)} \).