Answer
{$(-3,-1),(1,3)$}
Work Step by Step
Two equations, we have $x^2+y^2=10$ and $y=x+2$
From the second equation, we have
$x^2+(x+2)^2=10$ or, $x^2+2x-3=0$
or, $(x+3)(x-1)=0$
Thus, $x=${$-3,1$}
From the second equation when $x=-3$, we have
$y=-3+2$
or, $y=-1$
and
From the second equation when $x=3$, we have
$y=1+2$
or, $y=3$
Hence, the solution set: {$(-3,-1),(1,3)$}
