<s>[INST] Solving a linear congruence system How can this linear congruence system be solved by using Chinese remainder theorem?
\begin{align}
12x&\equiv -7 \pmod {13}\\
4x&\equiv 7 \pmod {9}\\
2x&\equiv -3 \pmod {11}
\end{align}
As far as I understand, we have to create some M numbers as it follows:
\begin{align}
M = 13 * 9 * 11 = 1287\\
M_1 = 9 * 11 = 99\\
M_2 = 13 * 11 = 143\\
M_3 = 13 * 9 = 117
\end{align}
But from here forward it becomes confusing for me. What should be done next?
Thank you in advance.
 [/INST] Note that CRT guarantees that an unique solution exist mod $9\cdot 11 \cdot 13$ since those are relatively primes but doesn’t give any particular method to solve the problem others that those used for the proof that you can find here.
To find the solution here we can proceed as follow


*

*$12x \equiv -7 \iff x\equiv 7 \pmod {13}$

*$4x\equiv 7 \iff x\equiv 4 \pmod {9}$

*$2x\equiv -3 \iff x\equiv 4 \pmod {11}$


this is the standard form for the system , then


*

*$x\equiv 7 \pmod {13}\implies x=7+13k$

*$x\equiv 4\implies 7+4k\equiv 4\implies 4k\equiv 6 \implies k\equiv 6  \pmod {9}\implies x=7+13(6+9h)=85+9\cdot13h$

*$x\equiv 4 \implies 8+7h\equiv4 \implies h\equiv 1 \pmod {11} \implies x=85+9\cdot13(1+11s)=202+9\cdot11\cdot 13s$


that is $$x=202 \pmod{9\cdot11\cdot 13}$$
</s>