Answer
$(4, 16)$ and $(-3, 9)$
Work Step by Step
We will use the substitution method to solve this system of equations because one of the equations already has one variable, $y$, isolated. We will use this expression to substitute into the second equation:
$y = x + 12$
$x^{2} = x + 12$
Bring all terms to the left-hand side to set the equation equal to $0$ to turn it into a quadratic equation:
$x^{2} - x - 12 = 0$
$(x - 4)(x + 3) = 0$
Now, set each factor equal to $0$ to solve for $x$:
$x - 4 = 0$ or $x + 3 = 0$
$x = 4$ or $ x = -3$
Now that we have the values for $x$, we can then plug these values into one of the equations to solve for $y$. Let's use the second equation to make it easier:
$y = x + 12$
$y = (4) + 12$
$y = 16$
$y = x + 12$
$y = (-3) + 12$
$y = 9$
The points where these two equations (graphs) intersect are $(4, 16)$ and
$(-3, 9)$.
