Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 8

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

Given: $x^2-y=1$ and $2x^2+3y=17$ This implies that $x^2-y =1 \implies x^2=1+y$ Then, we have $2(1+y)+3y=17$ or, $2+2y+3y=17 \implies 5y=15$ This gives: $y=3$ At $y= 3$, we have $x^2=1+3 \implies x=\pm \sqrt 4 =\pm 2$ Thus, the solutions set are: {$2,3$} and {$-2,3$}