To solve the system of equations:
- Equation 1: \(-4x + 6y = 38\)
- Equation 2: \(x - 5y = -20\)
We can use the substitution or elimination method. Here, I will use the substitution method.
First, we can express \(x\) in terms of \(y\) from Equation 2:
\[ x = 5y - 20 \]
Next, we can substitute this expression for \(x\) into Equation 1:
\[ -4(5y - 20) + 6y = 38 \]
Distributing \(-4\):
\[ -20y + 80 + 6y = 38 \]
Combine like terms:
\[ -14y + 80 = 38 \]
Now, isolate \(y\):
\[ -14y = 38 - 80 \] \[ -14y = -42 \] \[ y = \frac{-42}{-14} = 3 \]
Now that we have \(y\), substitute it back into the expression for \(x\):
\[ x = 5(3) - 20 \] \[ x = 15 - 20 \] \[ x = -5 \]
The solution to the system of equations is:
\[ x = -5, \quad y = 3 \]
So, the solution is \((-5, 3)\).