Answer
{$(3, \sqrt 6), (3, -\sqrt 6),(-3, \sqrt 6), (-3, -\sqrt 6)$}
Work Step by Step
On subtracting the given two equations, we have $x^2=9$ or, $x=${$-3,3$}
From the second equation, we have
$x^2+y^2=15$
For $x=3$
Then $3^2+y^2=15$
or, $y^2=6$
Thus, $y=\pm \sqrt 6$
From the second equation, we have
$x^2+y^2=15$
For $x=-3$
Then $(-3)^2+y^2=15$
or, $y^2=6$
Thus, $y=\pm \sqrt 6$
Hence, the solution set: {$(3, \sqrt 6), (3, -\sqrt 6),(-3, \sqrt 6), (-3, -\sqrt 6)$}
