Applications of the modular inverse

Euclidean. $31=1\cdot23+8$, $23=2\cdot8+7$, $8=1\cdot7+1$.

Back. $1=8-7=8-(23-2\cdot8)=3\cdot8-23=3(31-23)-23=3\cdot31-4\cdot23$.

Inverse. $23^{-1}\equiv-4\equiv27\pmod{31}$.

Solution. $x\equiv27\cdot17=459$. $459=14\cdot31+25$, so $x\equiv25\pmod{31}$.

Check. $23\cdot25=575=18\cdot31+17$. Correct.

Conclusion. $x\equiv25\pmod{31}$.