Answer
{$(2,2),(-2,-2),(4,1),(-4,-1)$}
Work Step by Step
Re-write the second equation as: $x=\dfrac{4}{y}$
First equation yields: $(\dfrac{4}{y})^2+4y^2=20$ or, $y^4-5y^2+4=0$
or,$(y^2-4)(y^2-1)=0$
or, $y=${$2,-2,1,-1$}
From first equation when $y=2$, we have $x=2$
From first equation when $y=-2$, we have $x=-2$
From first equation when $y=1$, we have $x=4$
From first equation when $y=-1$, we have $x=-4$
Thus, solution set is: {$(2,2),(-2,-2),(4,1),(-4,-1)$}
