Answer
The solutions of the system are $(9,6)$ and $(7,-2)$
Work Step by Step
$y=4x-30$
$y=x^2-12x+33$
Substitute $4x-30$ for y
$4x-30=x^2-12x+33$
$x^2-16x+63=0$
$(x-9)(x-7)=0$
$x-9=0$ or $x-7=0$
$x=9$ or $x=7$
Find corresponding y-values. Use either original equation:
$y=4x-30$
$y=4(9)-30$ or $y=4(7)-30$
$y=6$ or $y=-2$
The solutions of the system are $(9,6)$ and $(7,-2)$.
