# Affine open subset of an affine variety that is not the complement of a hypersurface?

Can it happen that an affine open subset of an affine variety is not the complement of a hypersurface? I think this must be able to happen because if not, I imagine I would have encountered a proposition to that effect in one of the books I am studying. However, I so far don't have an example: in all the toy examples I am thinking of, the minute I construct an open subset so that its complement isn't a hypersurface, it also stops being affine.

By the way, I am interested in this question due to a homework assignment (hence the homework tag) but the question itself is not the one I was assigned. I am not including the assigned question out of fear somebody might answer it or inadvertently give me a bigger hint than I want.

@QiaochuYuan - in reference to Hailong Dao's example at the MathOverflow link, why does $Yf+Xg=1$ imply that $U$ is affine? –  Ben Blum-Smith Nov 2 '12 at 15:57