The Chinese remainder theorem (two moduli)

Substitute. $2+5k\equiv3\pmod7$, so $5k\equiv1\pmod7$.

Inverse. $5^{-1}\equiv3\pmod7$ (since $5\cdot3=15\equiv1$). So $k\equiv3\pmod7$, i.e. $k=3+7m$.

Solution. $x=2+5(3+7m)=17+35m$.

Conclusion. $x\equiv17\pmod{35}$.