Solving Linear-Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}y=\text{-}{x}^2+7& \text{(I)}\\ y=2x+4& \text{(II)}\end{array}$$ The $y$-variable is isolated in Equation (II). This allows us to substitute its value $2x+4$ for $y$ in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the $x$-variable. $$\begin{array}{c}{x}^2+2x-3=0\iff 1x^2+2x+(\text{-}3)=0\end{array}$$ Now, recall the Quadratic Formula. $$\begin{array}{c}x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$ We can substitute $a=1,$ $b=2,$ and $c=\text{-}3$ 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.

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

Now, consider Equation (II). $$\begin{array}{c}y=2x+4\end{array}$$ We can substitute $x=1$ and $x=\text{-}3$ into the above equation to find the values for $y.$ Let's start with $x=1.$ We found that $y=6,$ when $x=1.$ One solution of the system, which is a point of intersection of the parabola and the line, is $(1,6).$ To find the other solution, we will substitute $\text{-}3$ for $x$ in Equation (II) again. We found that $y=\text{-}2,$ when $x=\text{-}3.$ Therefore, our second solution, which is the other point of intersection of the parabola and the line, is $(\text{-}3,\text{-}2).$