To solve the system of equations:
- \(-4x + 6y = 38\) (Equation 1)
- \(x - 5y = -20\) (Equation 2)
Let's solve these equations step by step.
First, we can solve Equation 2 for \(x\):
\[ x = 5y - 20 \]
Now we can substitute this expression for \(x\) back into Equation 1:
\[ -4(5y - 20) + 6y = 38 \]
Expanding this:
\[ -20y + 80 + 6y = 38 \]
Combining like terms:
\[ -14y + 80 = 38 \]
Now, isolate \(y\):
\[ -14y = 38 - 80 \] \[ -14y = -42 \]
Dividing by \(-14\):
\[ y = \frac{-42}{-14} = 3 \]
Now that we have \(y\), substitute it back into the expression for \(x\):
\[ x = 5(3) - 20 \] \[ x = 15 - 20 \] \[ x = -5 \]
Thus, the solution to the system of equations is:
\((x, y) = (-5, 3)\)
So the correct response is:
\((−5, 3)\)