Chapter 5, Systems of Equations and Inequalities - Section 5.4 - Systems of Nonlinear Equations - 5.4 Exercises - Page 465: 11

$(\pm2,\pm1)$

If we add $4$ times the first equation to the second one, we get: $13x^2=52\\x^2=4\\x=\pm2$ Plugging this back into the second one, we get: $(\pm2)^2+4y^2=8\\4+4y^2=8\\4y^2=4\\y^2=1\\y=\pm1$ Hence, the solutions are: $(\pm2,\pm1)$