A CRT application

System. $x\equiv2\pmod3$, $x\equiv3\pmod5$, $x\equiv2\pmod7$.

First two. $x\equiv8\pmod{15}$ (from $8\equiv2\pmod3$, $8\equiv3\pmod5$).

Combine with third. $8+15k\equiv2\pmod7$, so $1+k\equiv2\pmod7$ (since $8\equiv1$ and $15\equiv1$), $k\equiv1\pmod7$. Thus $x=8+15=23$.

Conclusion. $x=23$ is the smallest positive solution.

Check. $23\equiv2\pmod3$, $23\equiv3\pmod5$, $23\equiv2\pmod7$. Correct.