Solving Quadratic Systems

We want to solve the given system of equations using the Substitution Method. $\begin{cases}y=x^2-3x-3 & \, \text {(I)}\\ y=\text{-}2x^2-x+5 & \text {(II)}\end{cases}$ The $y\text{-}$variable is isolated in Equation (II). This allows us to substitute its value, $\text{-}2x^2-x+5,$ for $y$ in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the $x\text{-}$variable. $\begin{gathered} 3x^2-2x-8=0 \quad \Leftrightarrow \quad {\color{#0000FF}{3}}x^2+({\color{#FF0000}{\text{-} 2}})x+({\color{#009600}{\text{-} 8}})=0 \end{gathered}$ Now, recall the Quadratic Formula. $\begin{gathered} x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}$ We can substitute ${\color{#0000FF}{a}}={\color{#0000FF}{3}},$ ${\color{#FF0000}{b}}={\color{#FF0000}{\text{-} 2}},$ and ${\color{#009600}{c}}={\color{#009600}{\text{-} 8}}$ into this formula to solve the quadratic equation.

This result tells us that we have two solutions for $x.$ One of them will use the positive sign and the other one will use the negative sign.

$x=\dfrac{2 \pm 10}{6}$
$x_1=\dfrac{2+10}{6}$	$x_2=\dfrac{2-10}{6}$
$x_1=\dfrac{12}{6}$	$x_2=\dfrac{\text{-} 8}{6}$
$x_1=2$	$x_2=\text{-} \dfrac{4}{3}$

Now, consider Equation (II). $\begin{gathered} y=\text{-}2 x^2 - x + 5 \end{gathered}$ We can substitute $x=2$ and $x=\text{-} \frac{4}{3}$ into the above equation to find the values for $y.$ Let's start with $x=2.$ We found that $y=\text{-} 5,$ when $x=2.$ One solution of the system, which is a point of intersection of the two parabolas, is $(2,\text{-} 5).$ To find the other solution, we will substitute $\text{-} \frac{4}{3}$ for $x$ in Equation (II) again. We found that $y=\frac{25}{9},$ when $x=\text{-} \frac{4}{3}.$ Therefore, our second solution, which is the other point of intersection of the two parabolas, is $\left( \text{-} \frac{4}{3},\frac{25}{9}\right).$