Answer
{$(-3,-1),(3,1),(-1,-3),(1,3)$}
Work Step by Step
Re-write the first equation as: $y=\dfrac{3}{x}$
First equation yields: $x^2+(\dfrac{3}{x})^2=10$ or, $x^4-10x^2+9=0$
or,$(x^2-9)(x^2-1)=0$
or, $x=${$-3,-1,1,3$}
From first equation when $x=-1$, we have $y=-3$
From first equation when $x=-3$, we have $y=-1$
From first equation when $x=1$, we have $y=3$
From first equation when $x=3$, we have $y=1$
Thus, solution set is {$(-3,-1),(3,1),(-1,-3),(1,3)$}
