Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 36

{$(-2,8),(0,0)$}

After adding the two equations, we get $x^3+2x^2=0$ or, $x^2(x+2)=0$; or $x=${$-2,0$} From first equation $x^3+y=0$ when $x=-2$, we have $y=8$ From first equation $x^3+y=0$ when $x=0$, we have $y=0$ Thus, solution set is: {$(-2,8),(0,0)$}