Answer
{$(\dfrac{4 \sqrt 2}{3},\dfrac{2}{3}),(-\dfrac{4 \sqrt 2}{3},\dfrac{2}{3})$}
Work Step by Step
Let us multiply the second equation by $-1$.After adding the two equations, we get $y^2-(y-3)^2=-5$
or, $6y=4$;
or $y=\dfrac{2}{3}$
From first equation $x^2+y^2=4$ when $y=\dfrac{2}{3}$, we have $x=\pm \dfrac{4 \sqrt 2}{3}$
Thus, solution set is {$(\dfrac{4 \sqrt 2}{3},\dfrac{2}{3}),(-\dfrac{4 \sqrt 2}{3},\dfrac{2}{3})$}
