Solution system $3x \equiv 6\,\textrm{mod}\,\, 12$, $2x \equiv 5\,\textrm{mod}\,\, 7$ , $3x \equiv 1\,\textrm{mod}\,\, 5$

Question

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

Answer 1

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 $(1),(4)$

Answer 2

The congruences simplify to $$\left\{ \begin{array}{ccc} x&\equiv& 2\pmod{4} \\ x&\equiv& 6 \pmod{7}\\ x&\equiv& 2 \pmod{5} \end{array} \right.$$

The first and third congruences combine to $x\equiv 2 \pmod{20}$.

Then adding the modulus, we have

$\pmod{20}$: $x\equiv 2\equiv 22\equiv 42\equiv 62$

And $62$ satisfies the remaining congruence. So the solution is $x\equiv 62\pmod{140}$