Answer
See graphs, $(1,1)$
Work Step by Step
1. See graphs
2. Use $y=\sqrt {x}$ in $y=2-x$ to get $2-x=\sqrt x$ or $x+\sqrt x-2=0$
3. Let $z=\sqrt x$, we have $z^2+z-2=0$ or $(z+2)(z-1)=0$, thus $z=-2, 1$
4. For $\sqrt x=-2$, there is no real solution.
5. For $\sqrt x=1$, we have $x=1$
6. Use the 2nd equation again to get $y=1$
7. We can find the intersect(s) $(1,1)$
