Answer
$(-3,- \sqrt 6)$, $(-3,\sqrt 6)$, $(3,- \sqrt 6)$, $(3,\sqrt 6)$
Work Step by Step
Re-arrange the given two equations and then use the substitution method to get
$x^2+y^2=15$ and $-x^2-y^2=-15$
Now, $(2x^2+2y^2)+(-x^2-y^2)=24 -15$
This gives: $x^2=9 \implies x=3,-3$
when $x=3$ then we have $(3)^2+y^2=15 \implies y =\sqrt 6, -\sqrt 6$
when $x=-3$ then we have $(-3)^2+y^2=15 \implies y =\sqrt 6, -\sqrt 6$
Hence, our answers are: $(-3,- \sqrt 6)$, $(-3,\sqrt 6)$, $(3,- \sqrt 6)$, $(3,\sqrt 6)$
