Solving Linear-Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}y=2x^2+3x-4& \text{(I)}\\ y-4x=2& \text{(II)}\end{array}$$ The $y$-variable is isolated in Equation (I). This allows us to substitute its value $2x^2+3x-4$ for $y$ in Equation (II). Notice that in Equation (II), we have a quadratic equation in terms of only the $x$-variable. $$\begin{array}{c}2x^2-x-6=0\iff 2x^2+(\text{-}1)x+(\text{-}6)=0\end{array}$$ Now, recall the Quadratic Formula. $$\begin{array}{c}x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$ We can substitute $a=2,$ $b=\text{-}1,$ and $c=\text{-}6$ 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.

$x=\frac{1\pm 7}{4}$
$x_1=\frac{1+7}{4}$	$x_2=\frac{1-7}{4}$
$x_1=\frac{8}{4}$	$x_2=\frac{\text{-}6}{4}$
$x_1=2$	$x_2=\text{-}\frac{3}{2}$

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