Answer
The solutions are $(-\frac{3}{2},-4)$ and $(2,10)$.
Work Step by Step
The given system is
$y=2x^2+3x-4$ ...... (1)
$y-4x=2$ ...... (2)
Substitute $2x^2+3x-4$ for $y$ in equation (2).
$\Rightarrow 2x^2+3x-4-4x=2$
Subtract $2$ from each side.
$\Rightarrow 2x^2+3x-4-4x-2=2-2$
Simplify.
$\Rightarrow 2x^2-x-6=0$
Factor the polynomial.
$\Rightarrow (2x+3)(x-2)=0$
Use zero product property.
$\Rightarrow 2x+3=0$ or $x-2=0$
Solve for $x$.
$\Rightarrow x=-\frac{3}{2}$ or $x=2$
Substitute $-\frac{3}{2}$ for $x$ in equation (2).
$\Rightarrow y-4(-\frac{3}{2})=2$
Simplify.
$\Rightarrow y+6=2$
Subtract $6$ from each side.
$\Rightarrow y+6-6=2-6$
Simplify.
$\Rightarrow y=-4$
Substitute $2$ for $x$ in equation (1).
$\Rightarrow y-4(2)=2$
Simplify.
$\Rightarrow y-8=2$
Add $8$ to each side.
$\Rightarrow y-8+8=2+8$
Simplify.
$\Rightarrow y=10$
Hence, the solutions are $(-\frac{3}{2},-4)$ and $(2,10)$.
