Answer
The solutions for this system of equations are $(-1, -8)$ and $(-3 , 0)$.
Work Step by Step
To solve this system of equations, we set the two equations equal to one another because they are both equal to the same number, $y$:
$3x^2 + 8x - 3 = x^2 - 9$
Subtract $x^2$ from both sides to gather this term on the left side of the equation:
$3x^2 - x^2 + 8x - 3 = - 9$
Add $9$ to both sides to move constants to the left side of the equation:
$3x^2 - x^2 + 8x - 3 + 9 = 0$
Simplify:
$2x^2 + 8x + 6 = 0$
We can now use the quadratic equation to solve for $x$:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a=2$, $b=8$, and $=6c$.
Let us plug in the numbers into the formula:
$x = \dfrac{-8 \pm \sqrt {8^2 - 4(2)(6)}}{2(2)}$
$x = \dfrac{-8 \pm \sqrt {64 - 48}}{4}$
$x = \dfrac{-8 \pm \sqrt {16}}{4}$
$x = \dfrac{-8 \pm 4}{4}$
Let us split this into two equations to solve for $x$. The first one is:
$x = \dfrac{-8 + 4}{4}$
$x = \dfrac{-4}{4}$
$x = -1$
Let us look at the other equation:
$x = \dfrac{-8 - 4}{4}$
$x = \dfrac{-12}{4}$
$x = -3$
Now that we have the values for $x$, we can plug these values into either of the original equations to find the corresponding values of $y$. Let us use the second equation because we can deal with fewer terms:
$y = (-1)^2 - 9$
$y = 1 - 9$
$y = -8$
Let us plug in the other value of $x$ into the second equation once again:
$y = (-3)^2 - 9$
$y = 9 - 9$
$y = 0$
The solutions for this system of equations are $(-1, -8)$ and $(-3 , 0)$.
