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 $\frac{25}{150}=\frac{1}{6}$.

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

You may notice that we've basically done the same calculations as in the second exercise. In fact, addition doesn't care which way round the numbers go: $\frac{1}{10}+\frac{1}{15}=\frac{1}{15}+\frac{1}{10}$, even if we don't already know that that number is $\frac{1}{6}$.

This is intuitive from the fact that addition is the idea of "place the apples next to each other and count up the total": just putting the apples down in a different order doesn't change the total amount of apple.

Notice that both the denominators are the same (namely $5$), so we can just combine the $\frac{1}{5}$-sized pieces straight away. We have $0$ pieces and $2$ pieces, so the total is $2$ pieces.

That is, the answer is $\frac{2}{5}$.

If you spot that this is "no $\frac{1}{7}$-pieces" next to "two $\frac{1}{5}$-pieces", then you might just immediately write down that the answer is $\frac{2}{5}$ because there aren't any $\frac{1}{7}$-pieces to change the answer; and you'd be correct.

To use the instant rule from the text: $$\frac{0}{7}+\frac{2}{5}=\frac{0\times 5+2\times 7}{5\times 7}=\frac{0+14}{35}=\frac{14}{35}$$

But that is the same as $\frac{2}{5}$ (simply expressed with each $\frac{1}{5}$-piece subdivided further into sevenths).

Using the instant rule from the text: $$\frac{1}{15}+\frac{-1}{10}=\frac{1\times 10+15\times (-1)}{15\times 10}=\frac{10-15}{150}=\frac{-5}{150}=\frac{-1}{30}$$

What has happened here? What have we "really done" with our chunks of apple? Have a think; we'll see a lot more of this when we get to subtraction.

Something you can do to make this question easier is to notice that both the numbers have the same chunk-size (namely $\frac{1}{8}$), so we might try adding some number of $\frac{1}{8}$-chunks. Then we're trying to get from $7$ chunks to $13$ chunks, so we need to add $6$ chunks.

That is, the final number is $\frac{6}{8}$ (which is also $\frac{3}{4}$).

Now, the numbers are no longer of the same chunk-size, so we should make it so that they are of the same size.

The chunk-size to use is $\frac{1}{8\times 7}=\frac{1}{56}$. The reason for this is the same as the reasoning we saw when working out how to add $\frac{1}{8}$ and $\frac{1}{7}$.

Then the two numbers are $\frac{7\times 7}{7\times 8}=\frac{49}{56}$ and $\frac{8\times 13}{8\times 7}=\frac{104}{56}$, so the answer is that we need to get from $49$ to $104$; to do that, we need to add $55$ chunks of size $\frac{1}{56}$, so the answer is $\frac{55}{56}$.