Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 14

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

Here, we have $y=\dfrac{4}{x}$ Now, $x^2+(\dfrac{4}{x})^2=8$ This gives: $x^4-8x^2+16=0$ This yields : $(x^2-4) (x^2-4)=0$ when $x=-2$ then we have $y=\dfrac{4}{-2}=-2$ when $x=2$ then we have $y=\dfrac{4}{2}=2$ Hence, our answers are: $(-2,-2)$, $(2,2)$