smc

Solve the second equation by taking the square root of both sides. $$y^2=9$$ $$y=\pm 3$$ Solve the first equation by substituting $y^2$ into the first equation then solving for $x$. $$x^2+9=9$$ $$x^2=0$$ $$x=0$$ The two solutions are $(0,3)$ and $(0,-3)$, so the answer is $\boxed{\text{two}}$.