We want to find [tex]y[/tex] such that
[tex]\begin{cases}y\equiv3\pmod5\\y\equiv3\pmod6\\y\equiv3\pmod9\end{cases}[/tex]
6 and 9 are not coprime, so we split the moduli according to [tex]6=2\cdot3[/tex] and [tex]9=3^2[/tex] to get the system
[tex]\begin{cases}y\equiv3\equiv1\pmod2\\y\equiv3\equiv0\pmod3\\y\equiv3\pmod5\end{cases}[/tex]
If we take
[tex]y=1\cdot3\cdot5+0\cdot2\cdot5+3\cdot2\cdot3[/tex]
we can see that
By the Chinese remainder theorem, we've found that any [tex]y[/tex] satisfying
[tex]y\equiv15+0+18\equiv33\pmod{2\cdot3\cdot5}\equiv3\pmod30[/tex]
will satisfy each congruence above, and that any solution of the form [tex]y=3+30n[/tex] for any integer [tex]n[/tex] will work.
The smallest possible value of these occurs for [tex]n=0[/tex], so that 3 is the least positive solution.
