Solving simultaneous congruences with the Chinese remainder theorem Solve the following system of congruences using the Chinese remainder theorem:
$$\begin{align*}
2x &\equiv 3 \pmod{7} \\
x &\equiv 4 \pmod{6} \\
5x &\equiv 50 \pmod{55}
\end{align*}
$$
I was a little confused how to reduce the congruences into a form where the Chinese remainder theorem is applicable.

The congruence $2x\equiv 3\pmod{7}$ holds if and only if $x\equiv 5\pmod{7}$. And $5x\equiv 50\pmod{55}$ if and only if $x\equiv 10\pmod{11}$. Now the problem is in "standard" form. 
Remark: To get the first result, we can multiply both sides of $2x\equiv 3\pmod{7}$ by the inverse of $2$ modulo $7$. Note that $4$ is the inverse to $2$. Or more simply we replace the $3$ by $10$, and divide by $2$.
The second result comes from the general fact that if $k\ne 0$, then $ka\equiv kb\pmod{km}$ if and only if $a\equiv b\pmod{m}$. 
