I need help to determine if this congruency system can be solved and if it can be solved how do I do it: $$\begin{cases}x\equiv2\text{ (mod $3$)}\\ x\equiv4\text{ (mod $6$)}\\ \end{cases}$$
I do know that from the system I obtain the following:
$$\begin{align} x\equiv2\text{ (mod $3$)}\\ x\equiv4\text{ (mod $2$)}\\ x\equiv4\text{ (mod $3$)}\\ \end{align}$$
I do not know what to conclude from here. I think this system doesn't have solution, but if it is so how do I prove it.
On
Doesn't have solution, since $$\begin{cases} x\equiv 2 \pmod{3} \\ x\equiv 4 \pmod{3} \end{cases}$$ they implie $2\equiv 4 \pmod{3}$, but this is false.
On
$\begin{align} {\bf Hint}\qquad\ \ \ 2+3j &= x = 4 + 6k\,\Rightarrow\ 3\,j-6\,k = 2\,\Rightarrow\, 3\mid 2,\,\ \rm contradiction.\\[.5em] {\rm generally}\ \ \ a+mj &= x = b + nk\,\Rightarrow\, \color{#c00}mj-\color{#c00}nk = b-a\,\Rightarrow\, \bbox[6px,border:1px solid #c00]{\color{#c00}{\gcd(m,n)}\mid b-a }\end{align}$
i.e. $ $ they're consistent $\!\iff\!$ they become equivalent when reduced mod the $\rm\color{#c00}{gcd}$ of the moduli.
The above necessary condition for a solution to exist is also a sufficient condition. Furthermore, a system of congruences is solvable $\iff$ pairwise solvable as above.
$$x=6k+4\implies x=3(2k+1)+1$$
Thus if $$x\equiv 4\pmod {6} $$ then $$x\equiv 1\pmod 3$$
That is the system
$$\begin{cases} x\equiv 2 \pmod{3} \\ x\equiv 4 \pmod{6} \end{cases}$$
is not consistent.