### Question:

Solve $12x\equiv9\pmod{15}$

### My try:

$\gcd(12,15)=3$ so it has at least $3$ solutions.

Now
$15=12\times1+3\\
3=15-12\times 1\\
3=15+2\times(-1)\\
\implies9=15\times3+12\times(-3)\\
\implies12\times(-3)\equiv9\pmod{15}$

So $x\equiv-3\pmod{15}$

Am I correct?