how to solve $min\{x \in \mathbb N_0 \quad |x \cdot 714\quad mod \quad 1972 \quad = \quad 1292 \quad mod \quad 1972 \} $ (modulo equation) Question: How can I solve: $min\{x \in \mathbb N_0 \quad |x \cdot 714 \equiv 1292 \mod 1972 \} $ ?
I only know about:  
$x \cdot a \equiv _m b \Rightarrow m|x \cdot a - b$
different way of notation:
$min\{x \in \mathbb N_0 \quad | x \cdot 714 \equiv_{1972} 1292\}$ 
How to go on?
I appreciate every hint. 

As $714=34\cdot21$, $1292=34\cdot 38$, $1972=34\cdot 58$, this is equivalent to solving $21 x\equiv 38\mod 58$.
We have to find the inverse of $21$ modulo $58$. The tool for this is the Extended Euclidean algorithm:
$$\begin{array}{rrrl}
\hline
r_i&u_i&v_i&q_i\\
\hline
58&0&1\\
21&1&0&2\\
\hline
16&-2&1&1\\
5&3&-1&3\\1&-11&4\\
\hline
\end{array}$$
Thus the inverse of $21$ mod $58$ is $-11\equiv47$. The solution is
$$x\equiv47\cdot 38\equiv (-11)(-20)\equiv 46\mod 58. $$
