= Solution
Divide the <divisibility> relation $91\mid26x-39$ by $13$ to obtain $7\mid2x-3$. Since $2\cdot4\equiv1\pmod7$, multiplication by the <modular multiplicative inverse> gives $x\equiv12\equiv5\pmod7$. Therefore all <integer> solutions are $x=5+7q$, $q\in\mathbb Z$.
Modulo $91=13\cdot7$, the choices $q=0,\ldots,12$ give
$$
\boxed{x\equiv5,12,19,26,33,40,47,54,61,68,75,82,89\pmod{91}.}
$$
They are distinct <residue classes>: equality of two such classes would imply $91\mid7(q-q')$, hence $13\mid q-q'$. Every <integer> $q$ is congruent to one of those thirteen choices, so the list is exhaustive.
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