I've the following congruence system:
\begin{align*} I \quad 2x \equiv 0\mod 7 \\ II \quad x \equiv 1 \mod 5\\ III \quad x \equiv 3 \mod 4 \end{align*}
Now I tried to solve it:
\begin{align*} II \quad &x \equiv 1 \mod 5 \Rightarrow x=5x_1+1\\ \stackrel{I}{\Rightarrow} 2(5x_1+1) &\equiv 0 \mod 7 \\ \Leftrightarrow 10x_1+2 &\equiv 0 \mod \\ \Leftrightarrow 10x_1 &\equiv -2 \mod 7\\ \Leftrightarrow 10x_1 &\equiv 12 \mod 7\\ \Rightarrow 5x_1 &\equiv 6 \mod 7 \Rightarrow 5x_1=7x_2+6 \end{align*}
and now
$$x=5x_1+1=7x_2+6+1=7x_2+7 \Leftrightarrow x \equiv 0 \mod 7$$
This result is obviously no solution. If I try to solve it with euclidean algorithm, I'll get the correct result. Now I try to unterstand why the first idea is wrong. In general I understood the way of solving congruence systems, but never thought about why it works.
Any help is appreciated.