What does dense mean in algebraic geometry?

Question

• If $f, g$ are regular functions on a variety X, then the set of points where $f-g=0$ is closed and dense, hence equal to $X. $ Why is it dense and why does the closure of such a set satisfy the equation?

• An open subset of a variety is dense (which means that the closure of an open subset is the whole variety). Why is it dense? Many thanks for your comment.