<s>[INST] Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5.
x % 3 = 2
x % 5 = 3
We break down the problem to:
x % 3 = 1
x % 5 = 0
Therefore,
5k % 3 = 1
2k % 3 = 1
k = 2
10, when remainder = 1
20, when remainder = 2
Now x % 3 = 0
and x % 5 = 1
3k % 5 = 1
k = 2
6, when remainder = 1
18, when remainder = 3
Therefore, final solution is 20 + 18 = 38.
38 is a solution
LCM of 3,5 = 15. 
38 - 15 - 15 = 8.
8 is the least number, that is the solution.
But now if I have 
x % 7 = 3
x % 4 = 2
How do I solve the question ?
 [/INST] It has nothing to see with primeness of the moduli, but to their coprimeness.
The systematic method consists in finding Bézout's coefficients for $7$  and $3$: $\enspace-1\cdot 7+2\cdot 4=1$
(For bigger numbers you can use the Extended Euclidean Algorithm).
The solution to the system of congruences \begin{equation*}\begin{cases}x\equiv 3 \pmod 7\\x\equiv 2 \pmod 4
\end{cases}
\end{equation*}
are then simply:$$x\equiv -2\cdot 7+3\cdot 2\cdot 4=10 \pmod{\operatorname{lcm(7,4)}}.$$
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