How would I go about solving the following simultaneous congruence:

Question

How do I solve the following simultaneous congruence:

$25x \equiv 18 (\mod 48)$

$x \equiv 11 (\mod 35)$.

Answer 1

\begin{align} x \equiv 11 \pmod{35} &\implies x = 11 + 35t \quad \text{for some integer $t$} \\ 25x \equiv 18 \pmod{48} &\implies 25(11 + 35t) \equiv 18 \pmod{48} \\ &\implies 11t+35 \equiv 18 \pmod{48} \\ &\implies t = 29 \pmod{48}\\ &\implies x = 11 + 29*35\pmod{1680} \\ &\implies x = 1026 \pmod{1680} \end{align}

Answer 2

We have $25x-18=48k$ and $x-11=35l$ and hence $25*11-18=48k-25*35l=48k-875l$. We need to find a solution for $257=48k-875l$. Using the Euclidean algorithm we find $$1=48(237)-875(13)$$ and hence $$257 = 48(257*237)-875(257*13).$$Therefore setting $l=257*13$ we find $x=116946 \equiv 1026 \mod (35*48)$ is one of the solutions. Note that one can also find the inverse of $25 \mod 48$ i.e. use the Euclidean algorithm to find $$(25)25-13(48)=1$$ and hence replace the first equation by $x \equiv 18 \mod 48$. Then you can use the Chinese remainder theorem.