## Intermediate Algebra for College Students (7th Edition)

{$(-3,0),(-2,4),(2,20)$}
Re-write the first equation as: $y=4x+12$ Second equation yields: $4x+12=x^3+3x^2$ or, $x^3+3x^2-4x-12=0$ or,$(x+3)(x^2-4)=0$ or, $x=${$-3,-2,2$} From first equation $-4x+y=12$ when $x=-3$, we have $y=0$ From first equation $-4x+y=12$ when $x=-2$, we have $y=4$ From first equation $-4x+y=12$ when $x=2$, we have $y=20$ Thus, solution set is: {$(-3,0),(-2,4),(2,20)$}