Answer
{$(-1,1), (1,-1)$}
Work Step by Step
Since, we have first equation: $x^2+y^2=2$ and second equation: $x+y=0$
From the first equation, we have
$x^2+(-x)^2=2$
or, $x^2+x^2=2$
or, $2x^2=2$
Thus, $x=${$-1,1$}
From the second equation, we have
$-1+y=0$
or, $y=1$
Now, $1+y=0 \implies y=-1$
Hence, the solution set: {$(-1,1), (1,-1)$}
