Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 23

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

Let us multiply the first equation by $-2$ and multiply the second equation by $3$.After adding the two equations, we get $-17y^2+17=0$ or, $y^2=1$; or $y=\pm 1$ From first equation $2x^2-3y^2-5=0$ when $y=1$, we have $x=\pm 2$ From first equation $2x^2-3y^2-5=0$ when $y=-1$, we have $x=\pm 2$ Thus, solution set is {$(2,1),(-2,1),(2,-1),(-2,-1)$}