Solving system of congruences where $m$'s are not coprime I'm really struggling with how to find the solution to a system of congruences where the m_i's are not relatively prime. For example:
$x ≡ 3 (mod4)$ 
$x ≡ 1 (mod6)$.
I know this has a unique solution mod 12 but I'm stuck here. I read through some similar questions but I'm still confused. Since I can't use the Chinese Remainder Theorem, how do I find the solution?
 A: Write $$4k+3 =x=6l+1$$ so we have
$$2k+1=3l \Longrightarrow 2\mid 3l-1 \Longrightarrow 2\mid l-1 \Longrightarrow l-1=2n$$
Thus $l=2n+1$ and $k=3n+1$ and finally $x= 12n+7$. Or if you want $x\equiv 7 \pmod {12}$.
A: For small values, like your problem, you can use the method of adding the modulus:
$\pmod{6}: x\equiv 1\equiv 7 $.  
Then noting that also $7\equiv 3 \pmod{4}$, you have your solution:
$x\equiv 7 \pmod{12}$
(Sometimes you may have to add the modulus more than once.)
A: You use the CRT for each modulus, then combine the results.  For your system,
$x \cong 3 \pmod{4}$ just gives you that congruence (because $4$ is a power of a prime) and then $x \cong 1 \pmod{6}$ gives \begin{align*}
    x &\cong 1 \pmod{2}  \\
    x &\cong 1 \pmod{3}  \text{.}
\end{align*}
The first of these is redundant with the previous equation ("$x$ is odd" is compatible with but tells you less than $x \cong 3 \pmod{4}$.)  So your three equations reduce to \begin{align*}
    x &\cong 3 \pmod{4}  \\
    x &\cong 1 \pmod{3}  \text{.}
\end{align*}
Then reassemble using the CRT: $x \cong 7 \pmod{12}$.
