Answer
$(3, -1)$ and $(-2, 4)$
Work Step by Step
First, we need to get the $y$ term by itself for both equations:
$y = -x + 2$
$y = -x^2 + 8$
We will use substitution to solve this system of equations. We substitute one of the expressions given for the $y$ term, which would mean that we are going to set the two equations equal to one another to solve for $x$ first:
$-x^2 + 8 = -x + 2$
We want to move all terms to the left side of the equation:
$-x^2 + x + 8 - 2 = 0$
Combine like terms:
$-x^2 + x + 6 = 0$
Divide both sides by $-1$ so that the $x^2$ term is positive:
$x^2 - x - 6 = 0$
Factor the quadratic equation. The quadratic equation takes the form $ax^2 + bx + c = 0$. We need to find factors whose product is $ac$ but that sum up to $b$. In this exercise, $ac = -6$ and $b = -1$. The factors $-3$ and $2$ will work:
$(x - 3)(x + 2) = 0$
Set each factor equal to $0$.
First factor:
$x - 3 = 0$
Add $3$ to each side of the equation:
$x = 3$
Second factor:
$x + 2 = 0$
Subtract $2$ from each side of the equation:
$x = -2$
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 first equation:
$x + y - 2 = 0$
Substitute the solution $3$ for $x$:
$3 + y - 2 = 0$
Combine like terms:
$y + 1 = 0$
Subtract $1$ from each side of the equation:
$y = -1$
Let's solve for $y$ using the other solution $x = -2$:
$-2 + y - 2 = 0$
Combine like terms:
$y - 4 = 0$
Add $4$ to both sides of the equation:
$y = 4$
The solutions are $(3, -1)$ and $(-2, 4)$.
