Solving Linear-Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}y=x^2-2x-6& \text{(I)}\\ y=4x+10& \text{(II)}\end{array}$$ The $y\text{-}$variable is isolated in Equation (I). This allows us to substitute its value, $x^2-2x-6,$ for $y$ in Equation (II). Notice that in Equation (II), we have a quadratic equation in terms of only the $x\text{-}$variable. $$\begin{array}{c}{x}^2-6x-16=0\iff 1x^2+(\text{-}6)x+(\text{-}16)=0\end{array}$$ We can substitute $a=1,$ $b=\text{-}6,$ and $c=\text{-}16$ into the Quadratic Formula. This result tells us that we have two solutions for $x.$ One of them will use the positive sign and the other will use the negative sign.

$x=\frac{6\pm 10}{2}$
$x_1=\frac{6+10}{2}$	$x_2=\frac{6-10}{2}$
$x_1=\frac{16}{2}$	$x_2=\frac{\text{-}4}{2}$
$x_1=8$	$x_2=\text{-}2$

Now, consider Equation (I). $$\begin{array}{c}y=x^2-2x-6\end{array}$$ We can substitute $x=8$ and $x=\text{-}2$ into the above equation to find the values for $y.$ Let's start with $x=8.$ We found that $y=42$ when $x=8.$ One solution of the system, which is a point of intersection of the parabola and the line, is $(8,42).$ To find the other solution, we will substitute $\text{-}2$ for $x$ in Equation (I) again. We found that $y=2$ when $x=\text{-}2.$ Therefore, our second solution, which is the other point of intersection of the parabola and the line, is $(\text{-}2,2).$