Better way to solve systems of linear congruences in general

Note: this requires knowledge about modular inverses.
$$x\equiv 3\pmod{4} $$ $$x\equiv 2\pmod{11} $$
Suppose you have the systemwhich was used in the miniquestion.
$$4a+3\equiv 2\pmod{11}\implies 4a\equiv 1\pmod{11}\implies 4^{1}\cdot 4a\equiv 1\cdot 4^{1}\pmod{11}\implies a\equiv 8\pmod{11}. $$
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 getThis 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 miniquestion.
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?) 
This is a very nice and interesting way to solve modular congruences!
What is this method called?

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

Okay, thanks! @professionalbronco