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.