General solution of a linear congruence $\,ax\equiv b\pmod m$

$t\,$ is a root of $(7),$ which, scaled by $d,\,$ shows $t$ is root of $(5)$, so $\,at\equiv b\equiv ay\Rightarrow at\equiv ay,\,$ by transitivity of congruence (being an equivalence relation).

Alas, the proof is poorly presented. Below is a more conceptual presentation that generalizes widely. Similar to solutions of linear differential and difference equations (recurrences), for any linear equation $\,ax \equiv b,\,$ it's easy to show (see here or here) that the $\rm\color{darkorange}{general}$ solution is the sum of any $\rm\color{#0a0}{\text{particular solution }\,x_0}\,$ plus the solutions of the associated homogeneous equation $\,ax \equiv 0,\,$ here $\, m\mid ax\!\iff m/d\mid (a/d)x\!\iff m/d\mid x,\,$ by Euclid & $\,(m/d,a/d)\! =\! (m,a)/d = 1.\,$ Viewed $\!\bmod m\,$ such $\,x\,$ are the $\,d\,$ multiples of $\,m/d,\,$ i.e. $\, m/d\,\color{#c00}{\{0,1,2,\ldots, d\!-\!1\}}^{\phantom{|^|}}\!$ [to be rigorous $\!\bmod m\!:\ x\equiv k(m/d)\equiv k(m/d)\bmod m\,\overset{\small\rm\color{#90f}{DL}}\equiv\, m/d\,\color{#c00}{(k\bmod d)}\,$ by $\,{\small\rm\color{#90f}{DL}}\!=\!$ $\!\bmod\!$ Distributive Law].
So the $\rm\color{darkorange}{general}$ solution is $\!\bmod m\!:\ x\equiv \color{#0a0}{x_0} + \color{#c00}k\,m/d^{\phantom{|^{|^.}}}\!\!\!,\ \color{#c00}{0\!\le\! k\!<\! d}\ \ $ [$x_0 = t\,$ in the OP]

Said structure of the solution space will be clarified when one studies linear algebra and modules (= linear algebra where the coefficient algebra is a ring vs. field)

See here for more on switching back-and-forth between the equivalent linear congruence and its associated Bezout linear equation, and links to handy algorithms for solving such equations.