To solve the system of equations given by:
- \( 8x - 3y = -22 \) (Equation 1)
- \( y = 10 + 4x \) (Equation 2)
We can substitute Equation 2 into Equation 1 to find the values of \( x \) and \( y \).
Substituting \( y \) in Equation 1:
\[ 8x - 3(10 + 4x) = -22 \]
Now distribute the \(-3\):
\[ 8x - 30 - 12x = -22 \]
Combine like terms:
\[ 8x - 12x - 30 = -22 \]
This simplifies to:
\[ -4x - 30 = -22 \]
Next, isolate \( -4x \) by adding 30 to both sides:
\[ -4x = -22 + 30 \] \[ -4x = 8 \]
Now, divide both sides by \(-4\):
\[ x = -2 \]
Now that we have the value of \( x \), we can substitute it back into Equation 2 to find \( y \):
\[ y = 10 + 4(-2) \] \[ y = 10 - 8 \] \[ y = 2 \]
So the solution to the system of equations is:
\[ x = -2, \quad y = 2 \]
In ordered pair form, the solution is:
\[ (-2, 2) \]