I am trying to understand the following example and solution, but I am confused by why we use EEA and why we need to multiply by $4$?
Example. Find a solution to $$\begin{align}x &\equiv 88 \phantom{1}\mod 6 \\x &\equiv 100 \mod 15 \end{align}$$
Solution 1:
From the first equation we know we want $x − 88 = 6k$ for some integer $k,$ so $x$ is of the form $88+6k$.
From the second equation, we also want $88+6k \equiv 100 \mod 15$, so we want $6k\equiv 12\mod 15$.
Use the extended Euclidean Algorithm to find that $15(1) + 6(−2) = 3.$ Multiply through by $4$ to get $15(4) + 6(−8) = 12.$ Thus $6(−8) \equiv 12 \mod 15$, and we can take $k = −8$. This results in $x = 88+6(−8) = 40$.
This is how I would solve it.
$x \equiv 88 \pmod{6}$ implies $x = 88 + 6s$ for some integer $s$.
So $x \equiv 100 \pmod {15}$ becomes.
Eliminating $x$, we get \begin{align} x &\equiv 100 \pmod {15}\\ 88 + 6s &\equiv 100 \pmod {15} \\ 6s &\equiv 12 \pmod{15} &(\text{$3$ divides $6,12,$ and $15$})\\ 2s &\equiv 4 \pmod 5 & (2^{-1}\equiv 3 \pmod 5) \\ s &\equiv 2 \pmod 5 &(\text{After mult. through by 3}) \end{align}
So $s =2+5t$ for some integer $t$.
Hence \begin{align} x &= 88 + 6s \\ x &= 88 +6(2+5t) \\ x &= 100 + 30t \end{align}
In other words, $x \equiv 100 \pmod{30}$.