Consider this system of congruence equations. \begin{cases} 
4x \equiv 14 \pmod m \\ 
3x \equiv 2 \pmod 5  
\end{cases}
I want to prove that for $m \in 4\mathbb{Z}$ there are no solutions(1). Moreover, I want to determine all m for which I have solutions(2). 
First of all, the second equation is equivalent to $ x \equiv 4$ (mod 5).
If $m$ and $5$ are coprime, the Chinese remainder theorem states that I have solution, so I pick $m$ and $5$ not coprime. In this case, $m = 5t$ for some $t \in \mathbb{Z}$. 
I can write:
$$ 4x = 14 + (5t)c$$
$$ 4x = 4 + 5(2 + tc)$$
$$ 4x = 4$$
$$ x \equiv 1 mod 5$$
However, I also have that $ x \equiv 4  \pmod 5$, so there are no solution. In the proof I did not use the fact that $m$ is a multiple of $4$, so I think the answer for (2) is that we have solution only for $(m,a) = 1$. Is that right?

If $m=4n$, then
\begin{align}
   4x \equiv 14 \pmod m
   &\implies 4x \equiv 14 \pmod{4n} \\
   &\implies 4n \mid 14-4x \\
   &\implies 2 \mid 7-2x \\
   &\implies 2 \mid 7
\end{align}
which is false.

$3x \equiv 2 \pmod 5 \implies x \equiv 4 \pmod 5 \implies x = 4 + 5u$ for some integer, $u$.
\begin{align}
   4x \equiv 14 \pmod m
   &\implies 4(4 + 5u) \equiv 14 \pmod m \\
   &\implies 16 + 20u \equiv 14 \pmod m \\
   &\implies 20u \equiv -2 \pmod m \\
   &\implies m \mid 2(10u+1)
\end{align}
If $u=0$, we get $x=4$ and $m \mid 2$
If $u=1$ we get $x=9$ and $m \mid 22$
$\dots$
So $x=4+5u$ and $m \mid 2(10u+1)$
Check
$3x = 12 + 15u \equiv 2 \pmod 5$
$4x - 14 = 16 + 20u - 14 = 20u + 2 = 2(10u+1)$
So $m \mid 2(10u+1) \implies m \mid 4x-14 \implies 4x \equiv 14 \pmod m$.
