Answer
$(2,±2\sqrt{2})$ and $(-4,±4i)$
Work Step by Step
Add the two equations using the elimination method. This will eliminate the $y^2$ term.
$(4x-y^2)+(2x^2+y^2)=0+16$
$2x^2+4x=16$
Divide both sides by 2 and rearrange:
$x^2+2x=8$
$x^2+2x-8=0$
Factor:
$(x-2)(x+4)=0$
Therefore $x=2$ and $x=-4$
Rearrange the first equation:
$y=\sqrt{4x}$
And use this to solve for $y$.
When $x=2$:
$y=\sqrt{4(2)}=\sqrt{8}=±2\sqrt{2}$
When $x=-4$:
$y=\sqrt{4(-4)}=\sqrt{-16}=±4i$
Therefore the solutions are $(2,±2\sqrt{2})$ and $(-4,±4i)$.
