System of congruences where $\gcd(m, n)\ne1$ I have to solve this system of congruences:
$$
\begin{cases}
x^2+2x+2\equiv 0\pmod{10}\\
7x\equiv 20\pmod{22}
\end{cases}
$$
after some calculations
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
since $x\equiv 6\pmod{2}$ and $x\equiv 0\pmod{2}\\$ are equal, we get:
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 6\pmod{11}\\
\end{cases}\implies x\equiv 46\pmod{55}
$$
$$
\begin{cases}
x\equiv 0\pmod{2}\\
x\equiv 2\pmod{5}\\
\end{cases}\implies x\equiv 2\pmod{10}
$$
but, $\gcd(55,10)\ne1$, so I cannot apply the Chinese theorem. What have I done wrong?

Solving the linear congruence gives $x\equiv 6\bmod 22$. Write $x=22n+6$ and substitute into the quadratic congruence:
$$
(22n+6)^2 +2(22n+6)+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+4n+6+4n+2+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+8n \equiv 0\bmod 10
$$Everything is even, so let's cancel 2:
$$
\Rightarrow 2n^2+4n\equiv 0 \bmod 5
$$
$$
\Rightarrow 2n^2\equiv n \bmod 5
$$
$$
\Rightarrow n^2\equiv 3n \bmod 5
$$So, $n$ is either $0$ or $3$ mod $5$. Together, these give the solutions $x=\{6,72\}\bmod 110$.
