Turning a fraction with repeating decimals into a mixed number: why doesn't this work? Problem:
Turn $\frac{0.\overline{48}}{0.\overline{15}} $ into a mixed number.
My solution:
$0.\overline{15}$ goes into $0.\overline{48}$ 3 times, with a remainder of $0.\overline{48} - 3 x 0.\overline{15} = 0.\overline{48}-0.\overline{45}= 0.\overline{03}$
$100x -x = 3.\overline{03} - 0.\overline{03} = 99 x = 3$, hence $x = \frac{99x}{99} = \frac{3}{99}= \frac{1}{33}$. Since the remainder is $\frac{1}{33}$, the mixed number I'm looking for is $3\frac{1}{33}$ but the book gives $3\frac{1}{5}$ as a result, where am I wrong? Does it have anything to do with the way I multiplied $0.\overline{15}$ by 3?

When you want to compute, say,
$$
\frac{25}{7}
$$
you say "$7$ goes into $25$ three times, with a remainder of $4$."
But does that mean that
$$
\frac{25}{7} = 3 + 4?
$$
Not at all! It means that
$$
\frac{25}{7} = 3 + \frac{4}{7}.
$$
By analogy, in your case, you have
$$
\frac{0.\overline{48}}{0.\overline{15}}
$$
is $3$, with a remainder of $0\overline{.03}$, which means that
\begin{align}
\frac{0.\overline{48}}{0.\overline{15}} = 3 + \frac{0.\overline{03}}{0.\overline{15}}
\end{align}
You still have to simplify that last fraction but that's relatively easy: You can write
\begin{align}
\frac{0.\overline{03}}{0.\overline{15}} 
&= \frac{1}{10} \frac{0.\overline{30}}{0.\overline{15}} \\
&= \frac{1}{10} 2 \\
&= \frac{2}{10}\\
&= \frac{1}{5},
\end{align}
although your text may have some other way of reducing that to get the same answer --- I just happened to notice that the "3" and the 15" could be made to cancel nicely if it was "30" and "15" instead.
