Answer
The solutions are $(-1, 0)$ and $(-2, 0)$.
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:
$-x^2 - 3x - 2 = x^2 + 3x + 2$
We want to move all terms to the left side of the equation.
First, we will subtract $x^2$ from both sides of the equation:
$-2x^2 - 3x - 2 = 3x + 2$
Subtract $3x$ from both sides of the equation:
$-2x^2 - 6x - 2 = 2$
Subtract $2$ from both sides of the equation:
$-2x^2 - 6x - 4 = 0$
Factor out common terms:
$-2(x^2 + 3x + 2) = 0$
Divide each side by $-2$:
$x^2 + 3x + 2 = 0$
Factor the quadratic equation. The factors $2$ and $1$ will work:
$(x + 2)(x + 1) = 0$
Set each factor equal to $0$.
First factor:
$x + 2 = 0$
Subtract $2$ from each side of the equation:
$x = -2$
Second factor:
$x + 1 = 0$
Subtract $1$ from each side of the equation:
$x = -1$
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:
$y = (-1)^2 + 3(-1) + 2$
Evaluate the exponent first:
$y = 1 + 3(-1) + 2$
Multiply next:
$y = 1 - 3 + 2$
Add or subtract from left to right:
$y = 0$
Let's solve for $y$ using the other solution for $x$:
$y = (-2)^2 + 3(-2) + 2$
Evaluate the exponent first:
$y = 4 + 3(-2) + 2$
Multiply next:
$y = 4 - 6 + 2$
Add or subtract from left to right:
$y = 0$
The solutions are $(-1, 0)$ and $(-2, 0)$.
