Find the solution of $7x \equiv 29\pmod{31}$.
I have this solution but not understand it: Since $9$ is the inverse of $7 \bmod 31$, multiply both sides by $9$. Then $9\cdot 7 x \equiv 9\cdot 29 \pmod{31}$. And therefore
\begin{eqnarray*} x &\equiv& 261\pmod{31}\\ &\equiv& 13\pmod{31}. \end{eqnarray*}
The point is (as pointed out in the comments above) that $9\cdot 7 \equiv 1\pmod {31}$. That is, $9$ is the inverse of $7$ in the group $\mathbb{Z}_{31}^{\times}$.
You are trying to find an $x$ such that $7x \equiv 29 \pmod{31}$, that is, such that $7x = 29$ in the group $\mathbb{Z}_{31}^{\times}$. You are allowed to multiply by inverses, so is you do that $$\begin{align} 7x &\equiv 29 \pmod{31} & & \Rightarrow\\ 7^{-1}7x &\equiv 7^{-1}29 \pmod{31} & & \Rightarrow\\ 9\cdot7x &\equiv 9\cdot 29 \pmod{31} & & \Rightarrow\\ 63x &\equiv 261 \pmod{31} & & \end{align} $$ And modulo $31$ you have $63$ is $1$ and $261$ is $13$. So $$ x \equiv 13 \pmod{31} $$
