What are the methods of solving linear congruences? I have been trying very hard to solve this type of congruence equation:
$$ax = b \pmod n$$
and finally managed to actually solve a few by using properties like
$$\text{if}\quad
\begin{cases}
ax=b \pmod n \\ 
cx=d \pmod n 
\end{cases}
\quad\text{then}\quad acx = bd \pmod n$$
but still there are some simple  congruences which I am not able to solve  like $$25x=15\pmod{29}.$$
I tried to make use of both the above and transitive property of congruences, but that is not working here.
Now, I wanted to ask if there is any other method to solve congruences. I am asking this because I have very little knowledge of congruences. I have Burton's book of number theory and that helped me much better than Zuckerman's text did. But, still there are some topics that I am not able to do yet.
$$25x\equiv 15\pmod{29}\tag{1}$$

$29$ is a prime number, an one can show (I think Gauss showed) that that implies every number in $\{1,2,3,\ldots,28\}$ must have a mod-$29$ multiplicative inverse within that set.  So let's find the inverse $y$ of $25$, so that $25y\equiv1\pmod{29}$, and then multiply both sides of $(1)$ by $y$, getting
$$
x\equiv15y\pmod{29}.
$$
We want:
$$
\begin{align}
25y &amp; \equiv 1 \pmod {29} \\
25y &amp; = 1 - 29z \\
25y+29z &amp; = 1.
\end{align}
$$

If we divide $29$ by $25$, we get $1$, with remainder $4$:
$$
29-\{1\}25 = 4\tag{2}
$$
Now divide $25$ by $4$, getting $6$, with remainder $1$:
$$
25-\{6\}4 = 1\tag{3}
$$
Because $(2)$ is true, we can put $29-\{1\}25$ in place of $4$ in $(2)$:
$$
25-\{6\}(29-\{1\}25).
$$
Now collect the $29$s and $25$s:
$$
-\{6\}29 + \{7\}25 = 1.\tag{4}
$$
So we have $y=7$ and $z=-6$.

Thus $(4)$ tells us that
$$
7\cdot25\equiv 1\pmod{29}
$$

Thus multiply both sides of $(1)$ by $7$:
$$
x\equiv7\cdot15\equiv 18 \pmod{29}.
$$
See Calculating the Modular Multiplicative Inverse without all those strange looking symbols for the way to find the inverse of $322$ mod $701$.  It turns out to be $455$.
Solution using Euler's function:
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