Yes, it's well-known and occurs here many times, e.g. here where it is special case $\,b\! =\! 1\,$ below
$\quad\ \, \exists\, x\in\Bbb Z\!:\ ax\equiv b\pmod{\! m}\!\iff\! \exists\, x,y\in\Bbb Z\!:\ ax\!+\!my = b\!\overset{\rm\ Bezout}\iff{\color{#c00}{\overbrace{\gcd(a,m)}^{\large d}}}\mid b$
Proof $\ \ (\Rightarrow)\ \ ax\equiv b\pmod{\!m}\Rightarrow ax\!+\!my = b\,$ for $\,y\in\Bbb Z\,$ so $\,\color{#c00}{d\mid a,m}\Rightarrow\,d\mid \color{#c00}ax\!+\!\color{#c00}my = b.\ $
$(\Leftarrow)\ \ $ By Bezout: $\,\exists\,\bar x,\bar y\in\Bbb Z\!:\,$ $\,a\bar x\!+\!m\bar y = d\,$ $\Rightarrow\, a(c\bar x)\!+\!\color{#0a0}m(c\bar y) = b,\,$ via scaling by $\,c = b/d,\ $ hence reducing the prior equation $\!\bmod \color{#0a0}m\,$ yields $\,a(\color{#90f}{c\bar x})\equiv b\pmod{\!\color{#0a0}m},\,$ so let $\,x=\color{#90f}{c\bar x}$.
Remark $ $ Switching back-and-forth between a linear congruence and its associated linear Diophantine equation is something so basic and ubiquitous that it is rarely explicitly mentioned - just as for use of other basic laws (associative, commutative, distributive etc.)
By the above arrows, computing modular fractions (= inverses when $\,b=1)\,$ is equivalent to solving the associated linear Diophantine equation.
There are various ways to solve such congruences and equations, e.g. see here & here for a handful of methods applicable when the solution is unique.
See here for the general method when the solution is not unique, which includes a handy fractional view of above, and its use in the extended Euclidean algorithm. As is often the case, use of fractions may yield significant simplification and conceptual insight.
See here for the equivalence of the above and CRT = Chinese Remainder Theorem.
Best Answer
For this example it is simpler to note that
$$\rm mod\ 71\!:\ \ \frac{22}{144}\: \equiv\: \frac{11}{72}\:\equiv\: \frac{11} 1 $$
When the modulus is prime one can employ Gauss's algorithm, for example
$$\rm mod\ 29\!:\ \ \frac{1}8\: \equiv \frac{4}{32}\: \equiv\: \frac{4}{3}\:\equiv\: \frac{40}{30}\: \equiv\: \frac{11}{1}$$
I.e. scale $\rm A/B\ \to AN/BN\ $ by the least $\rm\:N\:$ so that $\rm\ BN > 29\:.\ $ Then reduce the numerator and denominator $\rm\ mod\ 29,\:$ and iterate. You will eventually obtain a denominator of $1$ since each step reduces the denominator. Isn't that sweet? That's they key idea that led Gauss to the first proof of the Fundamental Theorem of Arithmetic, i.e. unique factorization of integers.
See here and here for many methods to compute modular inverses and fractions.