Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 22

${(-2,0), (2,0)}$

$x^2+2y^2=4$ $x^2-y^2=4$ $x^2-y^2=4$ $2*(x^2-y^2)=2*4$ $2x^2-2y^2=8$ $x^2+2y^2=4$ $2x^2-2y^2=8$ $x^2+2y^2+2x^2-2y^2=4+8$ $3x^2=12$ $3x^2/3=12/3$ $x^2=4$ $\sqrt {x^2}=\sqrt 4$ $x= ±2$ $x^2-y^2=4$ $4-y^2=4$ $4-y^2+y^2-4=4+y^2-4$ $0=y^2$ $\sqrt 0 =\sqrt {y^2}$ $0=y$