7. [6] Find all ordered pairs $(x, y)$ such that
$$
(x-2 y)^{2}+(y-1)^{2}=0 .
$$
Answer: $(2,1)$ The square of a real number is always at least 0 , so to have equality we must have $(x-2 y)^{2}=0$ and $(y-1)^{2}=0$. Then $y=1$ and $x=2 y=2$.