Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 28

{$(0,-4),(2\sqrt 3,2),(-2\sqrt 3,2)$}

Let us multiply the first equation by $-1$.After adding the two equations, we get $y^2+2y=8$ or, $y^2+2y-8=0$; or $(y+4)(y-2)=0$ or $y=${$-4,2$} From first equation when $y=-4$, we have $x=0$ From first equation when $y=2$, we have $x=\pm 2\sqrt {3}$ Thus, solution set is {$(0,-4),(2\sqrt 3,2),(-2\sqrt 3,2)$}