How to solve the following congruence with fewer steps I have the following congruences
$$
\begin{align*}
2n&\equiv3\pmod5\\
3n&\equiv4\pmod7\\
4n&\equiv5\pmod9\\
5n&\equiv6\pmod{11}
\end{align*}
$$
I already know a way to solve it (multiply each congruence by the modular inverse of the coefficient of $n$, and then solve using Chinese Remainder Theorem), but the answer I get is 
$$n = 3464 \equiv -1 \pmod {5 \cdot 7 \cdot 9 \cdot 11}$$
I'm thinking that this probably isn't a coincidence, so I am wondering if there was a quicker way to solve this problem. Thanks!

Rewrite as
$$2n\equiv-2\!\!\!\pmod5,\\ 3n\equiv-3\!\!\!\pmod7,\\ 4n\equiv-4\!\!\!\pmod9,\\ 5n\equiv-5\!\!\!\pmod{11}.$$
Then there is an obvious answer, and the CRT proves that this answer is unique modulo $5\times7\times9\times11$.
