Here is the slacker's way:
S = 0.929292..... eqn (1)
100S = 92.9292.... eqn (2)
Eqn (2) - eqn (1): 99S = 92 => S = 92/99.
--------------------------------------------------------------------------------------
S = 52.31272727.... eqn (1)
100S = 5231.272727.... eqn (2)
10000S = 523127.2727..... eqn (3)
Eqn (3) - eqn (2): etc.
1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:
The exponent of the base 10 corresponds to n.
3. Now subtract the first equation from the second cloumnwise and you'll get:
Solve for x and simplify if possible:
4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:
Subtract the last two equations:
I'll leave the rest for you.
EDIT: ... too late - again.
Simpler maybe, but I don't think it's fair to call it better. I think they're equally good.
More info on the Wikipedia page on geometric series.