## Algebra 2 (1st Edition)

$(-5,0)(0,\sqrt5),(0,-\sqrt5)$
Adding the first and negative $25$ times the second equation we get: $5x^2+25x-125+125=0\\5x^2+25x=0\\x^2+5x=0\\x(x+5)=0$ Thus $x=0$ or $x=-5$ $y^2=5+x$ thus if $x=0$, then $y=\pm\sqrt{5}$ and if $x=-5$ then $y=0$. Thus the solutions are: $(-5,0)(0,\sqrt5),(0,-\sqrt5)$