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?
 A: \begin{cases}
x\equiv 0\;(mod\;2)\\
\hline
x\equiv 1\;(mod\;5)\\
x\equiv 2\;(mod\;5)\\
\hline
x\equiv 6\;(mod\;11)\\
\end{cases}
So you want a solution modulo $2 \cdot 5 \cdot 11 = 110$.
This is how I would solve it.
\begin{array}{r|rrr}
   & 2 & 5 & 11 \\
\hline
55 & 1 & 0 & 0\\
22 & 0 & 2 & 0\\
10 & 0 & 0 & -1 \\
\hline
\end{array}
Note that the top row is the three prime moduli that we are using.
The left column is 
   $\dfrac{2 \cdot 5 \cdot 11}{2} = 55 \quad $, 
   $\dfrac{2 \cdot 5 \cdot 11}{5} = 22 \quad$, and
   $\quad \dfrac{2 \cdot 5 \cdot 11}{11} = 10$.
The remaining entries show $55, 22, 10$ modulo $2, 5, 11$.
The goal is to multiply $55, 22$ and $10$ by the appropriate integers so that the three diagonal elements are all $1$.


*

*$55$ already gives us a diagonal element of $1$.

*Since $2 \cdot 3 \equiv 1 \pmod 5$, we change $22$ to $22 \cdot 3 = 66$.

*Since $-1 \cdot -1 \equiv 1 \pmod{11}$, we change $10$ to $10 \cdot (-1) = -10$.


\begin{array}{r|rrr}
   & 2 & 5 & 11 \\
\hline
55 & 1 & 0 & 0\\
22 & 0 & 2 & 0\\
10 & 0 & 0 & -1 \\
\hline
55 & 1 & 0 & 0\\
66 & 0 & 1 & 0\\
-10 & 0 & 0 & 1 \\
\hline
\end{array}
We use those numbers, $55, 66, -10$ as follows
$\left. \begin{align}
   x &\equiv 0 \pmod 2 \\
   x &\equiv 1 \pmod 5 \\
   x &\equiv 6 \pmod{11}
\end{align} \right\}
\iff x \equiv 0(55) + 1(66) + 6(-10) \equiv 6 \pmod{110}$
$\left. \begin{align}
   x &\equiv 0 \pmod 2 \\
   x &\equiv 2 \pmod 5 \\
   x &\equiv 6 \pmod{11}
\end{align} \right\}
\iff x \equiv 0(55) + 2(66) + 6(-10) \equiv 72 \pmod{110}$
A: Though in this case it is simpler to substitute the root of the linear polynomial into the quadratic (as in Integrand's answer), it is instructive to explain the general method of solution that you attempted in your question. The stumbling block there concerns how to split & recombine the systems via CRT, so let's closely examine this logic to see how it works generally. 
Suppose that $\,p,q,\bar q$ are pair-coprime integers and $\,f,g\,$ are integer coefficient polynomials.
$f(x)\equiv 0\pmod{\!pq}\!\!\overset{\small \rm CRT}\iff \begin{align} f(x)&\equiv 0\pmod{\!p}\iff x\in {\rm r}_p(f) := {\rm roots\ of} \,f\bmod p\\ f(x)&\equiv 0\pmod{\!q}\iff x\in {\rm r}_q(f)\end{align}$ 
$g(x)\equiv 0\pmod{\!p\bar q}\!\!\overset{\small \rm CRT}\iff \begin{align} g(x)&\equiv 0\pmod{\!p}\iff x\in {\rm r}_p(g)\\ g(x)&\equiv 0\pmod{\!\bar q}\iff x\in {\rm r}_{\bar q}(g)\end{align}$
Therefore $\,x\,$ is a root of both polynomial congruences iff $\,x\,$ satisfies
$$\begin{align}&\bmod p\!:\,\ x\in {\rm r}_p(f)\ \ \& \ \,x\in {\rm r}_p(g)\iff x\in {\rm r}_p(f)\cap {\rm r}_p(g)\\
&\bmod q\!:\,\ x \in {\rm r}_q(f)\\
&\bmod \bar q\!:\ x\in {\rm r}_{\bar q}(g)\end{align}\qquad\qquad\ \ \ $$
By CRT, each choice of a root for each modulus corresponds to a unique root $\!\bmod pq\bar q,\,$ i.e.
$$\begin{align}
x&\equiv r_i\in {\rm r}_p(f)\cap {\rm r}_p(g)\!\!\pmod{\!p}\\
x&\equiv s_j \in  {\rm r}_q(f)\ \ \ \,\qquad \pmod{\!q}\\
x&\equiv t_k \in {\rm r}_{\bar q}(g)\qquad\ \ \ \pmod{\!\bar q}\end{align} \iff x\equiv x_{i,j,k}\!\!\!\pmod{\!pq\bar q}\qquad$$
so the number of roots $\!\bmod pq\bar q\,$ is $\,|{\rm r}_p(f)\cap {\rm r}_p(g)|\cdot |{\rm r}_q(f)| \cdot  |{\rm r}_{\bar q}(g)|$. 

In the OP we have $\,p,q,\bar q = 2,5,11,\,$ so applying the above:
$\!\!\!\begin{align}\bmod 2\!:\ &0\equiv f = x^2\!+2x+2\equiv x^2\!\!\iff\! x\equiv 0\\
&0\equiv g = \,7\,x\,-\,20\,\equiv\, x\,\iff\: x\equiv 0,\ {\rm thus\ we\ have}\ \ {\rm r}_2(f)\cap {\rm r}_2(g) \equiv \{0\}\end{align}$
$\!\!\bmod 5\!:\ 0\equiv f = x^2+2x+2\equiv (x\!-\!1)(x\!-\!2)\!\iff\! x\equiv \color{#0a0}{1,2 =: r},\ \,{\rm so}\,\ {\rm r}_5(f) \equiv \{1,2\}$
$\!\!\bmod 11\!:\ 0\equiv g\equiv 7x-20\iff x\equiv \frac{20}7\equiv \frac{-2}{-4}\equiv \frac{1}2\equiv \frac{12}2\equiv 6,\ \,{\rm thus}\ \ {\rm r}_{11}(g) \equiv \{6\}$
So, as above, by CRT the common roots are precisely the solutions of 
$$\begin{align}
x&\equiv 0\!\!\pmod{\!2}\\
x&\equiv r\!\!\pmod{\!5},\ \color{#0a0}{r\in \{1,2\}}\\
x&\equiv 6\!\!\pmod{\!11}\end{align}\qquad$$
Now $\,x\equiv 0\equiv 6\pmod{\!\!2},\,x\equiv 6\pmod{\!\!11}\iff x\equiv 6\pmod{\!\!22}\,$ by CCRT
so $\,x = 6\!+\!22k.\,$ CRT combining this with the remaining congruence mod $5$ yields
$\!\!\bmod\color{#c00} 5\!:\,\ \color{#0a0}r\equiv x\equiv 6\!+\!22k\equiv 1\!+\!2k\iff 2k\equiv r\!-\!1\smash{\overset{\times 3\!}\iff} \color{#c00}{k\equiv 3r\!-\!3}$
So we obtain $\ x = 6+22\color{#c00}k = 6\!+\!22(\color{#c00}{3r\!-\!3+5n}) \equiv 6,72\pmod{\!\!110},\,$ for $\,\color{#0a0}{r = 1,2}$.
A: 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 $x=\{6,72\}\bmod 110$.
