Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 27

{$(-1,\sqrt 3),(-1,-\sqrt 3),(0,2),(0,-2)$}

Let us multiply the second equation by $-1$.After adding the two equations, we get $x^2+x=0$ or, $x(x+1)=0$; or $x=${$-1,0$} From first equation $x-y^2=-4$ when $x=-1$, we have $y=\pm \sqrt 3$ From first equation $x-y^2=-4$ when $x=0$, we have $y=\pm 2$ Thus, solution set is {$(-1,\sqrt 3),(-1,-\sqrt 3),(0,2),(0,-2)$}