Solving Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}y+16x-22=4x^2& \text{(I)}\\ 4x^2-24x+26+y=0& \text{(II)}\end{array}$$ It looks like it will be the easiest to isolate the $y$-term in the first equation. The $y$-variable is isolated in Equation (I). This allows us to substitute its value, $4x^2-16x+22,$ for $y$ in Equation (II). Notice that in Equation (I) we have a quadratic equation in terms of only the $x$-variable. $$\begin{array}{c}{x}^2-5x+6=0\iff 1x^2+(\text{-}5)x+6=0\end{array}$$ Now recall the Quadratic Formula. $$\begin{array}{c}x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$ We can substitute $a=1,$ $b=\text{-}5,$ and $c=6$ into this formula to solve the quadratic equation. This result tells us that we have two solutions for $x.$

$x=\frac{5\pm 1}{2}$
$x_1=\frac{5+1}{2}$	$x_2=\frac{5-1}{2}$
$x_1=\frac{6}{2}$	$x_2=\frac{4}{2}$
$x_1=3$	$x_2=2$

Now, consider Equation (I). $$\begin{array}{c}y+16x-22=4x^2\end{array}$$ We can substitute $x=3$ and $x=2$ into the above equation to find the values for $y.$ Let's start with $x=3.$ We found that $y=10$ when $x=3.$ One solution of the system, which is a point of intersection of the two parabolas, is $(3,10).$ To find the other solution, we will substitute $2$ for $x$ in Equation (II) again. We found that $y=6$ when $x=2.$ Therefore, our second solution, which is the other point of intersection of the two parabolas, is $(2,6).$