To solve the system of equations:
- \( x - 6y = -35 \) (Equation 1)
- \( -7x - 6y = -43 \) (Equation 2)
We'll use the substitution or elimination method. Let's start with substitution.
From Equation 1, we can solve for \( x \):
\[ x = 6y - 35 \]
Now, substitute \( x \) into Equation 2:
\[ -7(6y - 35) - 6y = -43 \]
Distributing the \(-7\):
\[ -42y + 245 - 6y = -43 \]
Combine like terms:
\[ -48y + 245 = -43 \]
Now, isolate \( y \):
\[ -48y = -43 - 245 \] \[ -48y = -288 \] \[ y = \frac{-288}{-48} \] \[ y = 6 \]
Now substitute \( y = 6 \) back into Equation 1 to find \( x \):
\[ x - 6(6) = -35 \] \[ x - 36 = -35 \] \[ x = -35 + 36 \] \[ x = 1 \]
The solution to the system of equations is:
\[ \boxed{(1, 6)} \]