Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 11

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

$3x^2 - y^2 = 11$ (1) $x^2 + 4y^2 = 8$ (2) 4(1) + (2) $13x^2 = 52$ $x^2 = 4$ x = -2, +2 Substitute x= -2 $(-2)^2 + 4y^2 = 8$ (2) $4 + 4y^2 = 8$ $y^2 = 1$ $y = +1, -1$ x=2 would yield the same result $(-2, 1) (-2, -1) (2, 1) (2,-1)$