<s>[INST] Express $x=0.\overline{31}_5$ as a fraction in lowest terms 
Given $x=0.\overline{31}_5$, find the value of $x$, expressed as a fraction in lowest terms.

I tried to change it into base $10$, but I don't think it's possible with fractions. So please help I'll appreciate it. Also I'm in 7th grade (easy solutions please) and no copying other people's answer (I've seen it in other problems).
 [/INST] Updated solution:
$0.31_5 = \frac{3}{5} + \frac{1}{25} = \frac{16}{25}$.
$0.\overline{31}_5 = 0.31_5 + 0.0031_5 + 0.000031_5 + \cdots$. Now using the formula for an infinite geometric series:
$$0.\overline{31}_5 = \frac{0.31_5}{1 - 0.01_5} = \frac{16/25}{1-1/25} = \frac{16}{24} = \frac{2}{3}.$$

Old solution:
$x = 0.3_5 + 0.0\overline1_5$.
Now $0.\overline 4_5 = 1$. $0.0\overline1 \cdot 10_5 \cdot 4_5 = 1$, so $0.0\overline1 = \frac{1}{5 \cdot 4} = \frac{1}{20}$
Using the fact that $0.3_5 = \frac{3}{5}$, $x = \frac{3}{5} + \frac{1}{20} = \frac{13}{20}$.
Note:
$0.\overline 4_5 = 1$ is the equivalent of $0.\overline{99} =1$. To prove this you can break it into $0.4_5 + 0.04_5 + 0.004_5 + \cdots$, which is an infinite geometric series and evaluates to $\frac{0.4_5}{1 - 0.1_5} = \frac{4/5}{1 - 1/5} = 1$.
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