Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 22

$(-1,-2)$, $(-1,2)$, $(1,-2)$, $(1,2)$

Here, we have $-2(2x^2-y^2)=(-2)(-2) \implies -4x^2+2y^2=4$ Now, $(3x^2-2y^2) +(-4x^2+2y^2) =-5+4$ This gives: $-x^2=-1 \implies x^2=1$ when $x=1$ then we have $2(1)^2-y^2=2 \implies y^2=4$ This gives: $y=2, -2$ when $x=1$ then we have $2(-1)^2-y^2=2 \implies y^2=4$ This gives: $y=2, -2$ Hence, our answers are: $(-1,-2)$, $(-1,2)$, $(1,-2)$, $(1,2)$