Explanation:
first off, we'll move the non-repeating part in the decimal to the left-side, by doing a division by a power of 10.
then we'll equate the value to some variable, and move the repeating part over to the left as well.
anyhow, the idea being, we can just use that variable, say "x" for the repeating bit, let's proceed,
\begin{gathered}\bf 0.580\overline{80}\implies \boxed{\cfrac{5.80\overline{80}}{10}}\qquad \textit{now, let's say }x= 5.80\overline{80}\\\\ -------------------------------\end{gathered}
0.580
80
⟹
10
5.80
80
now, let’s say x=5.80
80
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
\begin{gathered}\bf thus\qquad \begin{array}{llll} 100\cdot x&=&580.80\overline{80}\\ &&575+5.80\overline{80}\\ &&575+x \end{array}\qquad \implies 100x=575+x \\\\\\ 99x=575\implies x=\cfrac{575}{99}\qquad therefore\qquad \boxed{\cfrac{5.80\overline{80}}{10}}\implies \cfrac{\quad \frac{575}{99}\quad }{10} \\\\\\ \cfrac{\quad \frac{575}{99}\quad }{\frac{10}{1}}\implies \cfrac{575}{99}\cdot \cfrac{1}{10}\implies \cfrac{575}{990}\implies \stackrel{simplified}{\cfrac{115}{198}}\end{gathered}
thus
100⋅x
=
580.80
80
575+5.80
80
575+x
⟹100x=575+x
99x=575⟹x=
99
575
therefore
10
5.80
80
⟹
10
99
575
1
10
99
575
⟹
99
575
⋅
10
1
⟹
990
575
⟹
198
115
simplified