Using the instant rule from the text (which is actually a bit unwieldy here): $$\frac{1}{10}+\frac{1}{5}=\frac{1\times 5+10\times 1}{10\times 5}=\frac{5+10}{50}=\frac{15}{50}$$ Notice that this can actually be made simpler: it is the same thing as $\frac{3}{10}$, because when we take $\frac{3}{10}$ and split each $\frac{1}{10}$-block into five equal pieces, we obtain $15$ copies of the $\frac{1}{50}$-block.

Alternatively, one could spot that $\frac{1}{10}$-blocks actually can already be used to make $\frac{1}{5}$-blocks: $\frac{1}{5}=\frac{2}{10}$. Therefore we're actually trying to find $\frac{1}{10}+\frac{2}{10}$, which is easy: it's $\frac{3}{10}$.

Using the instant rule from the text: $$\frac{1}{10}+\frac{1}{15}=\frac{1\times 15+10\times 1}{10\times 15}=\frac{25}{150}=\frac{1}{6}$$

Were you expecting a big denominator, or at least a multiple of 5? From this example, we can see that the final answer of a rational addition problem can have a denominator which doesn't even seem related to the others.

Alternatively, one could spot that $\frac{1}{30}$-blocks will make up $\frac{1}{10}$- and $\frac{1}{15}$-blocks. Then we are actually trying to find $\frac{3}{30}+\frac{2}{30}=\frac{5}{30}$; it's a bit easier to see that $\frac{5}{30}=\frac{1}{6}$ than it is to see that $\frac{25}{150}=\frac{1}{6}$.