Answer
{$(1,4),(-1,-4),(2\sqrt 2,\sqrt 2),(-2\sqrt 2,-\sqrt 2)$}
Work Step by Step
Re-write the second equation as: $x=\dfrac{4}{y}$
First equation yields: $2(\dfrac{4}{y})^2+y^2=18$ or, $y^4-18y^2+32=0$
or,$(y^2-16)(y^2-2)=0$
or, $y=${$4,-4,\sqrt 2,- \sqrt 2$}
From first equation when $y=4$, we have $x=1$
From first equation when $y=-4$, we have $x=-1$
From first equation when $y=\sqrt 2$, we have $x=2 \sqrt 2$
From first equation when $y=-\sqrt 2$, we have $x=-2 \sqrt 2$
Thus, solution set is: {$(1,4),(-1,-4),(2\sqrt 2,\sqrt 2),(-2\sqrt 2,-\sqrt 2)$}
