Find the integers $n$ such that 5 divides $3 n-2$ and 7 divides $2 n+1$.
We are going to use the Chinese Remainder Theorem. First, by considering modulo 5, we observe that $3 n-2 \equiv 0(\bmod .5)$ if and only if $n \equiv 4$ (mod. 5$). Similarly, by considering modulo 7, we observe that $2 n+1 \equiv 0(\bmod .7)$ if and only if $n \equiv 3(\bmod .7)$.

Note that 5 and 7 are coprime: according to Bézout's theorem, there exist integers $u$ and $v$ such that $5 u+7 v=1$. For example, $(u, v)=(3,-1)$ works. According to the Chinese Remainder Theorem, the integers $n$ such that $n \equiv 4(\bmod .5)$ and $n \equiv 3(\bmod .7)$ are therefore the integers $n \equiv$ $4(7 v)+3(5 u) \equiv-11(\bmod .35)$.

The integers we are looking for are thus the integers of the form $35 k-11$, where $k$ is an integer.