Find the integers $n$ such that 6 divides $n-4$ and 10 divides $n-8$.
We will, once again, use the Chinese Remainder Theorem. Note that $6=2 \times 3$ and that $10=2 \times 5$. Saying that $n \equiv 4(\bmod 6)$ is equivalent to saying that $n \equiv 4(\bmod 2)$ and $n \equiv 1(\bmod 3)$. Saying that $n \equiv 8(\bmod 10)$ is equivalent to saying that $n \equiv 8(\bmod 2)$ and $n \equiv 8(\bmod 5)$.

We are looking for integers $n$ such that $n \equiv 0(\bmod 2)$, $n \equiv 1(\bmod 3)$, and $n \equiv 3(\bmod 5)$. Note that $n=-2$ is such an integer. According to the Chinese Remainder Theorem, the integers we are looking for are therefore the integers $n \equiv -2(\bmod 30)$, that is, the integers of the form $30k-2$, where $k$ is an integer.