Is it possible to solve this congruency system? 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.

$$x=6k+4\implies x=3(2k+1)+1$$

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.