How to do it

The equation $a x \equiv b \pmod{m}$ can have no solutions if $(a,m) \nmid b$ since then $m \nmid a x - b$ for any $x \in \mathbb{Z}$. So assume that $(a,m) \mid b$. We first consider the case $(a,m) = 1$. Then we can find $x_0$ and $y_0 \in \mathbb{Z}$ such that $a x_0 + m y_0 = b$ (use the Euclidean algorithm to get $x'$ and $y' \in \mathbb{Z}$ such that $a x' + m y' = 1$). Then put $x_0 = b x'$ so $a x_0 \equiv b \pmod{m}$. Any other solution is congruent to $x_0 \pmod{m}$, as $m \mid a(x_0 - x_1)$ and $(a,m) = 1$. So if $(a,m) = 1$ then a solution exists and is unique modulo $m$.