Answer
The solution sets are $(5, -10)$ and $(-3, -2)$.
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:
$x^2 - 3x - 20 = -x - 5$
We want to move all terms to the left side of the equation.
$x^2 - 3x + x - 20 + 5 = 0$
Combine like terms:
$x^2 - 2x - 15 = 0$
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 sum up to $b$. In this exercise, $ac = -15$ and $b = -2$. The factors $-5$ and $3$ will work:
$(x - 5)(x + 3) = 0$
Set each factor equal to $0$.
First factor:
$x - 5 = 0$
Add $5$ to each side of the equation:
$x = 5$
Second factor:
$x + 3 = 0$
Subtract $3$ from each side of the equation:
$x = -3$
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:
$y = -x - 5$
Substitute the solution $5$ for $x$:
$y = -5 - 5$
Subtract to solve
$y = -10$
Let's solve for $y$ using the other solution, $x = -3$:
$y = -(-3) - 5$
Subtract to solve:
$y = -2$
The solution sets are $(5, -10)$ and $(-3, -2)$.
