Adding Fractions With Unlike Denominators

$\tfrac{5}{7}$

The bottom numbers match, so add the tops: $2 + 3 = 5$, giving $\tfrac{5}{7}$.

$4$ is already a multiple of $2$.

$\tfrac{3}{4}$

$\tfrac{1}{2} = \tfrac{2}{4}$, so $\tfrac{2}{4} + \tfrac{1}{4} = \tfrac{3}{4}$.

$\tfrac{8}{15}$

$3$ and $5$ share no factors, so the LCD is $15$.

$\tfrac{5}{15} + \tfrac{3}{15} = \tfrac{8}{15}$.

$\tfrac{7}{8}$

$\tfrac{1}{4} = \tfrac{2}{8}$, so $\tfrac{5}{8} + \tfrac{2}{8} = \tfrac{7}{8}$.

$\tfrac{5}{6}$

$\tfrac{2}{3} = \tfrac{4}{6}$, so $\tfrac{4}{6} + \tfrac{1}{6} = \tfrac{5}{6}$.

$5$ and $6$ share no factors, so it is already in simplest form.

Both are over a half, so expect an answer past $1$.

$1\tfrac{7}{12}$

The LCD of $4$ and $6$ is $12$: $\tfrac{9}{12} + \tfrac{10}{12} = \tfrac{19}{12}$.

$19 \div 12 = 1$ remainder $7$, so the answer is $1\tfrac{7}{12}$.

$3\tfrac{3}{4}$

Whole numbers: $1 + 2 = 3$.

Fractions: $\tfrac{1}{4} + \tfrac{2}{4} = \tfrac{3}{4}$.

Together: $3\tfrac{3}{4}$.

The fraction part will spill past one whole.

$4\tfrac{1}{3}$

Whole numbers: $2 + 1 = 3$.

Fractions over an LCD of $6$: $\tfrac{5}{6} + \tfrac{3}{6} = \tfrac{8}{6} = 1\tfrac{2}{6} = 1\tfrac{1}{3}$.

Carry the whole across: $3 + 1\tfrac{1}{3} = 4\tfrac{1}{3}$.

$\tfrac{7}{10}$ of a mile

$10$ is a multiple of $5$, so the LCD is $10$: $\tfrac{2}{5} = \tfrac{4}{10}$.

$\tfrac{3}{10} + \tfrac{4}{10} = \tfrac{7}{10}$ of a mile.

Compare $\tfrac{2}{9}$ with $\tfrac{1}{4}$.

It is smaller than $\tfrac{1}{4}$, and a total cannot be smaller than one of its parts. The answer is $\tfrac{9}{20}$.

Adding a positive amount to $\tfrac{1}{4}$ must give something bigger. But $\tfrac{2}{9}$ is less than $\tfrac{2}{8}$, and $\tfrac{2}{8} = \tfrac{1}{4}$ — so his total is smaller than one of the parts he started with.

Theo added the bottom numbers. Doing it correctly, the LCD of $4$ and $5$ is $20$:

$\tfrac{5}{20} + \tfrac{4}{20} = \tfrac{9}{20}$, which is a little under a half — and that is a sensible size for the total.