Solving linear congruences by hand: modular fractions and inverses
Generally, if $\,b\,$ is coprime to the modulus $m$ then (by Bezout) it is invertible $\!\bmod m,\,$ so scaling $\,bx\equiv a\,$ by $\,b^{-1}\,$ we obtain the unique solution $\,x\equiv b^{-1}a =: a/b.\,$ We can quickly compute $\,b^{-1}\pmod{\!m}\,$ by the extended Euclidean algorithm, but there are often more convenient ways for smaller numbers (e.g. see here and here for a handful of methods applied). We describe a few of these methods below, where we view $\, x\equiv b^{-1}a \equiv a/b\,$ as a modular fraction.
The first, Gauss's algorithm, is based on Gauss's proof of Euclid's lemma via the descent $\,p\mid ab\,\Rightarrow\, p\mid a(p\bmod b).\,$ Generally it only works for prime moduli, but we can also execute the general extended Euclidean algorithm in fraction form too (using multi-valued "fractions").
It works by repeatedly scaling $\rm\:\color{#C00}{\frac{A}B}\overset{\times\ N} \to \frac{AN}{BN}\: $ by the least $\rm\,N\,$ with $\rm\, BN \ge 13,\, $ then reducing mod $13$
$$\rm\displaystyle \ mod\ 13\!:\,\ \color{#C00}{\frac{7}9} \,\overset{\times\ 2}\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\overset{\times \ 3}\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\overset{\times\ 7}\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}\qquad\!\! $$
Denominators of the $\color{#c00}{\rm reduced}$ fractions decrease $\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$ so reach $\color{#C00}{1}\,$ (not $\,0\,$ else the denominator would be a proper factor of the prime modulus; it may fail for composite modulus)
Or, simpler, allowing negative residues $\displaystyle\ \ \frac{7}9\,\equiv\, \frac{7}{\!-4\!\ \,}\,\equiv\,\frac{21}{\!\!-12\ \ \ \!\!}\,\equiv\, \frac{8}1$
This optimization using least magnitude residues $0,\pm 1, \pm 2.\ldots$ often simplifies modular arithmetic. Here we can also optimize by (sometimes) cancelling obvious common factors, or by pulling out obvious factors of denominators, etc. For example
$$\frac{7}9\,\equiv\, \frac{\!-6\,}{\!-4\,}\,\equiv\frac{\!-3\,}{\!-2\,}\,\equiv\frac{10}{\!-2\,}\,\equiv\,-5$$
$$\frac{7}9\,\equiv\,\frac{\!-1\cdot 6}{\ \ 3\cdot 3}\,\equiv\,\frac{\!\,12\cdot 6\!}{\ \ \,3\cdot 3}\,\equiv\, 4\cdot 2$$
Or as you did: $ $ check if the quotient $\rm\,a/b\equiv (a\pm\!13\,i)/(b\pm\!13\,j)\,$ is exact for small $\rm\,i,j,\,$ e.g.
$$ \frac{1}7\,\equiv \frac{\!-12}{-6}\,\equiv\, 2;\ \ \ \frac{5}7\,\equiv\,\frac{18}{\!-6\!\,}\,\equiv -3$$
When working with smaller numbers there is a higher probability of such optimizations being applicable (the law of small numbers), so it's well-worth looking for such in manual calculations.
More generally we can make the quotient exact by using Inverse Reciprocity.
$\bmod 13\!:\ \dfrac{a}{b}\equiv \dfrac{a-13\left[\color{#0a0}{\dfrac{a}{13}}\bmod b\right]}b\,\ $ e.g. $\,\ \dfrac{8}9\equiv \dfrac{8-13\overbrace{\left[\dfrac{8}{\color{#c00}{13}}\bmod 9\right]}^{\large\color{#c00}{ 13\ \,\equiv\,\ 4\ }}}9\equiv\dfrac{8-13[2]}9\equiv-2$
Note that the value $\,\color{#0a0}{x\equiv a/13}\,$ is what is needed to make the numerator divisible by $b,\,$ i.e.
$\qquad\quad\bmod b\!:\,\ a-13\,[\color{#0a0}x]\equiv 0\iff 13x\equiv a\iff \color{#0a0}{x\equiv a/13}$
This can be viewed as an optimization of the Extended Euclidean Algorithm in the case when it terminates in two steps.
Note $ $ Gauss' algorithm is my name for a special case of the Euclidean algorithm that's implicit in Gauss' Disquisitiones Arithmeticae, Art. 13, 1801. I don't know if Gauss explicitly used this algorithm elsewhere (apparently he chose to avoid use or mention of the Euclidean algorithm in Disq. Arith.). Gauss does briefly mention modular fractions in Art. 31 in Disq. Arith.
The reformulation above in terms of fractions does not occur in Gauss' work as far as I know. I devised it in my youth before I had perused Disq. Arith. It is likely very old but I don't recall seeing it in any literature. I'd be very grateful for any historical references.
See here for further discussion, including a detailed comparison with the descent employed by Gauss, and a formal proof of correctness of the algorithm.
Beware $ $ Modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus. See here for further discussion.
When the prime is a reasonably small one I'd rather find directly the inverse: $$9^{-1}=\frac{1}{9}=3\pmod {13}\Longrightarrow 9x=7\Longrightarrow x=7\cdot 9^{-1}=7\cdot 3= 21=8\pmod {13}$$ But...I try Gauss's method when the prime is big and/or evaluating inverses is messy.