<s>[INST] Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
 [/INST] Using the reduced equations , we get
$$ \color{ red} {x = 3k+2 \quad \quad {\text(1)}}$$
Substituting it into the next gives us :
$$\color {#F0A}{3k+2\equiv 3\mod 4 \implies  k \equiv 3 \mod 4 \implies k = 4j +3 \quad \quad \text { (2)}}$$
From $(1)$ and $(2)$ we get :
$$\color{ blue}{x = 12j + 11}$$
Substituting it into the next gives us :
$$\color {#50F}{12j +11 \equiv 4 \mod 5 \implies j\equiv 4 \mod 5 \implies j = 5l +4 \quad \quad \text { (3)}}$$
From $(1)$ and $(3)$ we get :
$$\color{ green}{x = 60l +59 \quad \quad \text{ (4)}}$$
And finally from the last equation , we get :
$$\color {orange}{60l +59 \equiv 0 \mod 7 \implies l \equiv1 \mod 7 \implies l = 7y+1 \quad \quad \text{ (5)} }$$ 
And we get :
$$\boxed{\color{navy}{x = 60 (7y+1) +59 \implies x = 420y + 119}}$$
Hence all values of form $420y + 119$ is the solution starting from $119,539...$
</s>