Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 14

{$(-2,-2),(2,2)$}

Re-write the first equation as: $y=\dfrac{4}{x}$ First equation yields: $x^2+(\dfrac{4}{x})^2=8$ or, $x^4-8x^2+16=0$ or,$(x^2-4)(x^2-4)=0$ or, $x=${$-2,2$} From first equation when $x=-2$, we have $y=-2$ From first equation when $x=2$, we have $y=2$ Thus, solution set is {$(-2,-2),(2,2)$}