Chapter 10 - Systems of Equations and Inequalities - Section 10.6 Systems of Nonlinear Equations - 10.6 Assess Your Understanding - Page 794: 19

See graphs, $\left(-2,-2\right),\left(2,2\right)$

1. See graphs for $xy=4$ and $x^2+y^2=8$ 2. The 1st equation gives $y=\frac{4}{x}$, plug in the 2nd equation to get $x^2+\frac{16}{x^2}=8$ or $x^4-8x^2+16=0$ or $(x^2-4)^2=0$, thus $x=\pm2$ 3. Use the above with the 1st equation, we can find the intersect(s): $\left(-2,-2\right),\left(2,2\right)$