To solve the system of equations:
- \( 6x + y = 20 \) \quad (Equation 1)
- \( -x - 4y = -11 \) \quad (Equation 2)
We can use the substitution or elimination method. Here, I'll use the substitution method.
First, solve Equation 1 for \( y \):
\[ y = 20 - 6x \]
Next, substitute \( y \) in Equation 2:
\[ -x - 4(20 - 6x) = -11 \]
Now, distribute the \(-4\):
\[ -x - 80 + 24x = -11 \]
Combine like terms:
\[ 23x - 80 = -11 \]
Now, isolate \( x \):
\[ 23x = -11 + 80 \] \[ 23x = 69 \] \[ x = 3 \]
Now that we have \( x \), plug it back into the equation for \( y \):
\[ y = 20 - 6(3) \] \[ y = 20 - 18 \] \[ y = 2 \]
The solution to the system of equations is:
\((x, y) = (3, 2)\)
Thus, the final answer is:
\[ \boxed{(3, 2)} \]