Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 6

{$-3,0$} and {$2,3$}

Given: $x^2+y=9$ and $x-y+3=0$ This implies that $y=x+3$ Then, we have $x^2+x+3=9$ or, $x^2+x-6 =0$ This gives: $(x+3)(x-2) \implies x=-3,2$ a) At $x= -3$, we have $y=x+3=-3+3=0$ b) At $x=2$, we have $y=x+3=2+3=5$ Thus, the solutions set are: {$-3,0$} and {$2,3$}