I have the following system
$ x \equiv 2 (mod 4)$
$ x \equiv 2 (mod 6) $
$ x \equiv 2 (mod 7) $
And I can't apply the Chinese remainder Theorem.
I tried applying the Chinese remainder Theorem to the last 2 congruences, which gave me that the set of solutions of that 2 congruences is $ 2 + 42 \cdot \beta $ with $\beta$ belonging to $\Bbb Z$.
Then I solved the set of solutions of the first congruence, which is $ 6 + 4 \cdot \alpha $ with $\alpha$ belonging to $\Bbb Z$.
A common solution would be the one that satisfies $ 2 + 42 \cdot \beta = 6 + 4 \cdot \alpha$, equivalently, the one that satisfies $ 42 \cdot \beta + 4 \cdot \alpha = 4$, and this (because of Bezout Identity) have solution only if $mcd(42,4)=4$ which is not true. So this would mean this system have no solution, which is incorrect (I think).
Then, what can I do?