Answer
The solutions are $(7, -6)$ and $(-2, 12)$.
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:
$x^2 - 7x - 6 = 8 - 2x$
We want to move all terms to the left side of the equation.
$x^2 - 7x + 2x - 6 - 8 = 0$
Combine like terms:
$x^2 - 5x - 14 = 0$
Factor the quadratic trinomial. The quadratic equation takes the form $x^2 + bx + c$. We need to find factors of $c$ whose sum is equal to $b$.
In this exercise, $c = -14$ and $b = -5$ so look for factors of $-14$ whose sum is $-5$. These factors are $-7$ and $2$. This means that the factored form of the trinomial is $(x-7)(x+2)$. Thus,
$(x - 7)(x + 2) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation.
First factor:
$x - 7 = 0$
$x = 7$
Second factor:
$x + 2 = 0$
$x = -2$
Now that we have the two possible values for $x$, we can plug them into one of the original equations to find the corresponding $y$ values. Let's use the second equation:
$y = 8 - 2x$
Substitute the solution $7$ for $x$:
$y = 8 - 2(7)$
Multiply first to simplify:
$y = 8 - 14$
Subtract to solve:
$y = -6$
Let's solve for $y$ using the other solution, $x = -2$:
$y = 8 - 2(-2)$
Multiply first to simplify:
$y = 8 + 4$
Subtract to solve:
$y = 12$
The solutionts are $(7, -6)$ and $(-2, 12)$.
