Chinese remainder theorem

Statement. If $m_1,\ldots,m_n$ are pairwise coprime, and $a_1,\ldots,a_n$ are arbitrary integers, the system $x\equiv a_i\pmod{m_i}$ has a unique solution modulo $M=m_1\cdots m_n$.

Solve. $x=2+5k$. Need $2+5k\equiv3\pmod7$, i.e. $5k\equiv1\pmod7$. Since $5^{-1}\equiv3\pmod7$ (as $5\cdot3=15\equiv1$), $k\equiv3\pmod7$, so $k=3+7m$. Then $x=2+5(3+7m)=17+35m$.

Conclusion. $x\equiv17\pmod{35}$. Check: $17=3\cdot5+2$, $17=2\cdot7+3$. Correct.