Answer
The solution sets are and $(-\frac{4}{3}, \frac{25}{9})$ and $(2, -5)$.
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:
$x^2 - 3x - 3 = -2x^2 - x + 5$
We want to move all terms to the left side of the equation.
$x^2 + 2x^2 - 3x + x - 3 - 5 = 0$
Combine like terms:
$3x^2 - 2x - 8 = 0$
Let's factor this quadratic equation by splitting the middle term. We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $ac$ is $-24$ and $b$ is $-2$. Let's look at the possibilities:
$-6$ and $4$
$-12$ and $2$
$-24$ and $1$
The first option will work.
Let's rewrite the equation and split the middle term using these two factors:
$3x^2 - 6x + 4x - 8 = 0$
Group the first two and last two terms:
$(3x^2 - 6x) + (4x - 8) = 0$
Factor common terms out:
$3x(x - 2) + 4(x - 2) = 0$
Group the factors:
$(3x + 4)(x - 2) = 0$
Set each factor equal to $0$.
First factor:
$3x + 4 = 0$
Subtract $4$ from each side of the equation:
$3x = -4$
Divide each side of the equation by $3$:
$x = -\frac{4}{3}$
Second factor:
$x - 2 = 0$
Add $2$ to 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:
Let's substitute in $x = -\frac{4}{3}$:
$y = (-\frac{4}{3})^2 - 3(-\frac{4}{3}) - 3$
Evaluate the exponent first:
$y = \frac{16}{9} - 3(-\frac{4}{3}) - 3$
Multiply next:
$y = \frac{16}{9} + \frac{12}{3}) - 3$
Find equivalent fractions with common denominators to add the fractions together:
$y = \frac{16}{9} + \frac{36}{9} - \frac{27}{9}$
Add or subtract from left to right:
$y = \frac{25}{9}$
Let's substitute $x = 2$:
$y = (2)^2 - 3(2) - 3$
Evaluate the exponent first:
$y = 4 - 3(2) - 3$
Add or subtract from left to right:
$y = 4 - 6 - 3$
Add or subtract from left to right:
$y = -5$
The solution sets are and $(-\frac{4}{3}, \frac{25}{9})$ and $(2, -5)$.
