Chapter 10 - Review Exercises - Page 817: 47

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

Two equations, we have $xy-4=0$ and $y=x$ From the second equation, we have $x(x)-4=0$ or, $x^2=4$ Thus, $x=\pm 2$ From the second equation, we have $y-(-2)=0$ (when x=-2) or, $y=-2$ and From the second equation, we have $y-2=0$ (when x=2) or, $y=2$ Hence, the solution set: {$(-2,2),(2,2)$}