How do we use EEA and why do we multiply by 4 throughout? 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, $x \equiv 100 \pmod{30}$.
