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## Solving Quadratic Systems 1.5 - Solution

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).
$\begin{cases}y=x^2-3x-3 & \, \text {(I)}\\ y=\text{-}2x^2-x+5 & \text {(II)}\end{cases}$
$\begin{cases}{\color{#0000FF}{\text{-}2x^2-x+5}}=x^2-3x-3 \\ y=\text{-}2x^2-x+5 \end{cases}$
${\color{#8C8C8C}{\text{(I): }}}$ Simplify
$\begin{cases}0=x^2-3x-3-(\text{-}2x^2-x+5) \\ y=\text{-}2x^2-x+5 \end{cases}$
$\begin{cases}0=x^2-3x-3+2x^2+x-5 \\ y=\text{-}2x^2-x+5 \end{cases}$
$\begin{cases}0=3x^2-2x-8 \\ y=\text{-}2x^2-x+5 \end{cases}$
$\begin{cases}3x^2-2x-8=0 \\ y=\text{-}2x^2-x+5 \end{cases}$
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.
$x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{\text{-} ({\color{#FF0000}{\text{-} 2}})\pm\sqrt{({\color{#FF0000}{\text{-} 2}})^2-4({\color{#0000FF}{3}})({\color{#009600}{\text{-} 8}})}}{2({\color{#0000FF}{3}})}$
Solve for $x$
$x=\dfrac{2\pm\sqrt{(\text{-}2)^2-4(3)(\text{-}8)}}{2(3)}$
$x=\dfrac{2\pm\sqrt{4-4(3)(\text{-}8)}}{2(3)}$
$x=\dfrac{2\pm\sqrt{4+96)}}{6}$
$x=\dfrac{2\pm\sqrt{100}}{6}$
$x=\dfrac{2 \pm 10}{6}$

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 $x=2.$
$y=\text{-}2 x^2-x+5$
$y=\text{-}2 ({\color{#0000FF}{2}})^2-{\color{#0000FF}{2}}+5$
Solve for $y$
$y=\text{-}2(4)-2+5$
$y=\text{-}8-2+5$
$y=\text{-}5$
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.
$y=\text{-}2x^2-x+5$
$y=\text{-}2\left( {\color{#0000FF}{\text{-} \dfrac{4}{3}}} \right)^2-\left({\color{#0000FF}{\text{-}\dfrac{4}{3}}}\right)+5$
Solve for $y$
$y=\text{-}2\left(\dfrac{16}{9}\right)-\left(\text{-}\dfrac{4}{3}\right)+5$
$y=\text{-}\dfrac{32}{9}-\left(\text{-}\dfrac{4}{3}\right)+5$
$y=\text{-}\dfrac{32}{9}+\dfrac{4}{3}+5$
$y=\text{-}\dfrac{32}{9}+\dfrac{12}{9}+5$
$y=\text{-}\dfrac{32}{9}+\dfrac{12}{9}+\dfrac{45}{9}$
$y=\dfrac{25}{9}$
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 $\left( \text{-} \frac{4}{3},\frac{25}{9}\right).$