Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 30

{$(-3,\sqrt 7),(-3,-\sqrt 7),(4,0)$}

Let us multiply the first equation by $-1$ .After adding the two equations, we get $x^2-x=12$ or, $(x-4)(x+3)=0$; or, $x=${$-3,4$} From first equation $x+y^2=4$ when $x=-3$, we have $y=\pm \sqrt 7$ From first equation $x+y^2=4$ when $x=4$, we have $y=0$ Thus, solution set is {$(-3,\sqrt 7),(-3,-\sqrt 7),(4,0)$}