Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 22

{$(1,2),(1,-2),(-1,2),(-1,-2)$}

After adding the given two equations, we get $-x^2=-1$ when we multiply the second equation by $-2$. or, $x^2=1$; or $x=\pm 1$ From first equation $3x^2-2y^2=-5$ when $x=1$, we have $y=\pm 2$ From first equation $3x^2-2y^2=-5$ when $x=-1$, we have $y=\pm 2$ Thus, solution set is {$(1,2),(1,-2),(-1,2),(-1,-2)$}