Solve $15\cdot x=27\pmod{18}$ EXERCISE:

Solve   $15\cdot x\equiv 27\pmod{18}$

SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equiv 9\pmod{6}$
So, from now and then I can't understand how to proceed!
The book continue with this:
"We have a unique solution of   $5\cdot x\equiv 9\pmod{6}$ which is":
$x=5\cdot 9 =45\equiv 3\pmod 6$ and $15\cdot 3\equiv 27\pmod{18}\equiv 9\pmod{18} $
The other solutions are:$(9,15)$. So all solutions are these: $(3,9,15)$
Can anyone explain me how the author of the book continue the exercise?I really can't understand how he ends up with these solutions!Is there another way to reach these solutions?
I would really appreciate a thorough  explanation, since I've just started working on these type of exercises and I have to clear my mind on them. 
Thanks for your time !

It is possible to solve this problem using algebraic symbols or augmented matrices. For example the following uses augmented matrices to solve the problem:
$$\begin{array}{c|c|c|}
 \hline
 \text{x} & \text{y} & \text{c} & \text{s} & \text{t} \\
\hline
\text{15} & \text{-18} & \text{27} \\
& \text{+3} & \text{+3} & \text{-15} \\
 & & & & \text{3} \\
\hline
\text{5} & \text{-6} & \text{9} \\
& \text{1} & \text{1} & \text{-5} \\
\hline
\text{1} & & \text{3} & \text{9} \\
& \text{1} & \text{1} & \text{-5}
\\
\hline
\end{array}$$
I have included animations for the construction of the augmented matrices in the PDF document "How I solved 15x≡27 (mod 18)" and a description of the "method" and its relation to the Euclidean Algorithm and the Extended Euclidean Algorithm.
