Answer
The solutions are $(5, -27)$ and $(-2, 8)$.
Work Step by Step
We will use substitution to solve this system of equations. We substitute one of the expressions for $y$, which would mean that we are going to set the two equations equal to one another to solve for $x$ first:
$-x^2 - 2x + 8 = x^2 - 8x - 12$
We want to move all terms to the left side of the equation.
$-x^2 - x^2 - 2x + 8x + 8 + 12 = 0$
Combine like terms:
$-2x^2 + 6x + 20 = 0$
Divide all terms by their greatest common factor, $-1$:
$x^2 - 3x - 10 = 0$
This equation is now in standard form, which is given by the formula:
$ax^2 + bx + c = 0$, where $a$ is the coefficient of the squared term, $b$ is the coefficient of the linear term, and $c$ is the constant term.
Factor the equation. Find factors that when multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $a$ is $1$, $b$ is $-3$, and $c$ is $-10$; therefore, $ac$ is $-10$. By looking at $ac$ and $b$, we see that one factor must be negative and one must be positive, but the negative factor should have the greater absolute value. The possibilities are:
$-5$ and $2$
$-10$ and $1$
The first option will work.
Rewrite the equation in factor form:
$(x - 5)(x + 2) = 0$
Using the Zero Product Property, set each factor equal to $0$, and solve for $x$:
First factor:
$x - 5 = 0$
$x = 5$
Second factor:
$x + 2 = 0$
$x = -2$
Substitute the values of $x$ into one of the original equations to solve for $y$:
First $x$ value:
$y = 5^2 - 8(5) - 12$
$y = 25 - 8(5) - 12$
$y = 25 - 40 - 12$
$y = -27$
Second $x$ value:
$y = (-2)^2 - 8(-2) - 12$
$y = 4 - 8(-2) - 12$
$y = 4 + 16 - 12$
$y = 8$
The solutions are $(5, -27)$ and $(-2, 8)$.
