To solve the system of equations:
- \( 5x + 2y = 14 \) (Equation 1)
- \( x - 6y = 22 \) (Equation 2)
We can solve for one variable and substitute it into the other equation. Let's solve Equation 2 for \( x \):
\[ x = 6y + 22 \]
Now, substitute \( x \) into Equation 1:
\[ 5(6y + 22) + 2y = 14 \]
Distributing the 5:
\[ 30y + 110 + 2y = 14 \]
Combine like terms:
\[ 32y + 110 = 14 \]
Now, subtract 110 from both sides:
\[ 32y = 14 - 110 \] \[ 32y = -96 \]
Now, divide by 32:
\[ y = \frac{-96}{32} = -3 \]
Now, substitute \( y = -3 \) back into the expression we derived for \( x \):
\[ x = 6(-3) + 22 \] \[ x = -18 + 22 \] \[ x = 4 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (4, -3) \]