Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 37

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

Let us multiply the second equation by $-1$.After adding the two equations, we get $(y-2)^2+2y=0$ or, $y(y-2)=0$; or $y=${$0,2$} From first equation $x^2-2y=0$ when $y=0$, we have $x=0$ From first equation $x^2-2y=0$ when $y=2$, we have $x=\pm 2$ Thus, solution set is: {$(0,0),(2,2),(-2,2)$}