Solution system $3x \equiv 6\,\textrm{mod}\,\, 12$, $2x \equiv 5\,\textrm{mod}\,\, 7$ , $3x \equiv 1\,\textrm{mod}\,\, 5$ Have solution the following congruence system?
$$\begin{array}{ccl}
      3x & \equiv & 6\,\textrm{mod}\,\, 12\\
      2x & \equiv & 5\,\textrm{mod}\,\, 7\\
      3x & \equiv & 1\,\textrm{mod}\,\, 5
    \end{array}$$
Point of Interest: This question requires some special handling due to the mixture of factors among the moduli. This is more than the run of the mill Chinese Remainder Theorem problem

Using $\#12$ of this,
$$2x\equiv5\pmod7\equiv5+7\iff x\equiv6\pmod7\ \ \ \ (1)$$
$$3x\equiv1\pmod5\equiv1+5\iff x\equiv2\pmod5\ \ \ \ (2)$$
$$3x=12k+6\iff x=4k+2\implies x\equiv2\pmod4\ \ \ \ (3)$$
$$(2),(3)\implies x\equiv2\pmod{\text{lcm}(5,4)}\implies x\equiv2\pmod{20}\ \ \ \ (4)$$
Now safely use CRT on $(1),(4)$
