I ran into a snag in attempting to solve the following problem.
$$x\equiv r\pmod p$$ $$x\equiv0\pmod q$$
I did the following
$$x=k_1q=k_2p+r$$ $$k_1q\equiv r\pmod p$$ $$k_1\equiv rq^{-1}\pmod p$$ $$x=k_1q=(k_3p+rq^{-1})q=$$ $$k_3pq+rq^{-1}q$$ $$x\equiv rq^{-1}q\pmod{pq}$$
I guess the problem occurs in the last $3$ lines. $q^{-1}$ makes sense when working $\pmod p$, but is there some notation to indicate the inverse is mod $p$ outside of the relavant equivalence? Secondly, is there any way to simplify that term or is that how it has to remain?