Chapter 10 - Systems of Equations and Inequalities - Section 10.6 Systems of Nonlinear Equations - 10.6 Assess Your Understanding - Page 795: 48

$(-2,2),(3,-3)$

1. Factor the 1st equation to get $(x+y)(x-2y)=0$ 2. For $y=-x$, use the 2nd equation to get $-x^2+x+6=0$ or $(x+2)(x-3)=0$, thus $x=-2, 3$ and $y=2, -3$ 3. For $y=x/2$, use the 2nd equation to get $x^2/2+x+6=0$ or $x^2+2x+12=0$, and there is no real solution for this case. 4. Combine the above results to get the solutions: $(-2,2),(3,-3)$