Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 21

{$(3,2),(3,-2),(-3,2),(-3,-2)$}

After adding the given two equations, we get $13x^2=117$ when we multiply the second equation by $4$. or, $x^2=9$; or $x=\pm 3$ From first equation $x^2-4y^2=-7$ when $x=3$, we have $y=\pm 2$ From first equation $x^2-4y^2=-7$ when $x=-3$, we have $y=\pm 2$ Thus, solution set is {$(3,2),(3,-2),(-3,2),(-3,-2)$}