To solve the system of equations:
- \(-4x + 6y = 38\) (Equation 1)
- \(x - 5y = -20\) (Equation 2)
We'll use the substitution or elimination method. Let's solve Equation 2 for \(x\):
From Equation 2: \[ x = 5y - 20 \]
Now, substitute this expression for \(x\) in Equation 1:
\[ -4(5y - 20) + 6y = 38 \]
Distribute \(-4\): \[ -20y + 80 + 6y = 38 \]
Combine like terms: \[ -20y + 6y + 80 = 38 \] \[ -14y + 80 = 38 \]
Now, isolate \(y\): \[ -14y = 38 - 80 \] \[ -14y = -42 \] \[ y = \frac{-42}{-14} = 3 \]
Now that we have \(y\), 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 from the options given is: \[ \text{(−5, 3)} \]