Better way to solve systems of linear congruences in general

[professionalbronco]

Note: this requires knowledge about modular inverses.
Suppose you have the system

$$x\equiv 3\pmod{4}$$ $$x\equiv 2\pmod{11}$$

which was used in the mini-question.
By CRT (Chinese Remainder Theorem), we know that there is a unique solution modulo $44$ that satisfies the system. From the first congruence, we see that $x$ can be expressed as $4a+3$ where $a$ is a positive integer.
Substituting this into the second congruence, we get

$$4a+3\equiv 2\pmod{11}\implies 4a\equiv -1\pmod{11}\implies 4^{-1}\cdot 4a\equiv -1\cdot 4^{-1}\pmod{11}\implies a\equiv 8\pmod{11}.$$

This means that $a=11b+8$, where $b$ is another positive integer. Substituting this into $x=4a+3$, we get $x=4(11b+8)+3=44b+35$. Therefore, $x\equiv 35\pmod{44}$. From this, you would get the answer of $0$ for the mini-question.

Anyways, I made this post to emphasize a different way to solve systems of linear congruences, which I found from AoPS.
(I usually use this method to solve modular congruences, but if numbers are small enough, I usually end up using guess and check. Also, mods, can you check this?)

[西瓜]

This is a very nice and interesting way to solve modular congruences!

What is this method called?

[professionalbronco]

@西瓜 There isn't a name. This method can be used to solve systems of modular congruences in general: bigger numbers, etc.

[西瓜]

Okay, thanks! @professionalbronco