"Borrow" 1 from the 8 and subtract vertically:

\(\displaystyle 8 - 3 \frac{3}{7} = 7 \frac{7}{7} - 3 \frac{3}{7}\):

   \(\displaystyle 7 \frac{7}{7}\)

\(\displaystyle \underline{- 3 \frac{3}{7}}\)

   \(\displaystyle 4 \frac{4}{7}\)

\(\displaystyle 4 \frac{4}{7} < 5 \frac{3}{7}\)

"Borrow" 1 from the 6 and subtract vertically:

\(\displaystyle 6 \frac{2}{7} - 2 \frac{6}{7} = 5 \frac{9}{7} - 2 \frac{6}{7}\)

   \(\displaystyle 5 \frac{9}{7}\)

\(\displaystyle \underline{ - 2 \frac{6}{7}}\)

   \(\displaystyle 3 \frac{3}{7}\)

\(\displaystyle 3 \frac{3}{7} < 4 \frac{4}{7}\)

(a) \(\displaystyle a + 3.7 =8.2\)

\(\displaystyle a + 3.7- 3.7 = 8.2 - 3.7\)

\(\displaystyle a = 4.5\)

(b) \(\displaystyle b + 4 \frac{2}{3} = 9\)

\(\displaystyle b + 4 \frac{2}{3} - 4 \frac{2}{3} = 9- 4 \frac{2}{3}\)

\(\displaystyle b = 9- 4 \frac{2}{3} = 8\frac{3}{3}- 4\frac{2}{3} = 4\frac{1}{3}\)

\(\displaystyle 4\frac{1}{3} = 4.333... < 4.5\), so \(\displaystyle a>b\).

Subtract both sides of the two equations:

\(\displaystyle x = \frac{1}{2} + \frac{1}{3}\)

\(\displaystyle y = \frac{1}{2} + \frac{2}{3}\)

\(\displaystyle x - y = \frac{1}{2} + \frac{1}{3}- \left ( \frac{1}{2} + \frac{2}{3} \right )\)

\(\displaystyle x - y = \frac{1}{2} + \frac{1}{3}-\frac{1}{2} - \frac{2}{3}\)

\(\displaystyle x - y = \frac{1}{2} -\frac{1}{2}+ \frac{1}{3} - \frac{2}{3}\)

\(\displaystyle x - y = - \frac{1}{3}\)

Since  \(\displaystyle \frac{1}{3} < \frac{1}{2}\), then \(\displaystyle x-y = -\frac{1}{3} > - \frac{1}{2}\)

When subtracting fractions with different denominators, first change the fractions so that they have the same denominator. Do this by finding the least common multiple of both 4 and 8. Some multiples of 4 and 8 are:

4: 4, 8, 12, 16...

8: 8, 16, 24, 32...

Since 8 is the first common multiple between 4 and 8, change the fractions accordingly so that their denominators equal 8. Since \(\displaystyle \tfrac{7}{8}\) already has a denominator of 8, it does not need to change. Change \(\displaystyle \tfrac{3}{4}\), however, accordingly. 

\(\displaystyle \frac{3}{4}\times\frac{2}{2}\rightarrow\frac{6}{8}\)

The problem now looks like this:

\(\displaystyle \frac{7}{8}-\frac{6}{8}=\)

Subtract the numerators. The result is your answer.

\(\displaystyle \frac{7}{8}-\frac{6}{8}=\frac{7-6}{8}=\frac{1}{8}\)

When dealing with fractions and mixed numbers, first convert the mixed numbers to improper fractions.

\(\displaystyle 2\tfrac{1}{2} \rightarrow\tfrac{5}{2}\)

\(\displaystyle 1\tfrac{3}{5}\rightarrow\tfrac{8}{5}\)

The next thing you must do is change the denominators so that they are equal. Do this by finding the least common multiple of 2 and 5. Some multiples of 2 and 5 are:

2: 2, 4, 6, 8, 10...

5: 5, 10, 15, 20...

10 is the first common multiple of both 2 and 5, so change the fractions accordingly so that their denominators both equal 10.

\(\displaystyle \tfrac{5}{2}\times\tfrac{5}{5}\rightarrow\tfrac{25}{10}\)

\(\displaystyle \tfrac{8}{5}\times\tfrac{2}{2}\rightarrow\tfrac{16}{10}\)

The problem now looks like this:

\(\displaystyle \tfrac{25}{10}-\tfrac{16}{10}=\)

Subtract the numerators of the fraction. The result is your answer. 

\(\displaystyle \tfrac{25}{10}-\tfrac{16}{10}=\tfrac{25-16}{10}=\tfrac{9}{10}\)

The phrase, "how much more" tells as that we want to find the difference in how much they've grown. 

\(\displaystyle \frac{2}{6}-\frac{1}{6}=\frac{1}{6}\)

The phrase, "how much more" tells as that we want to find the difference in how much they've grown. 

\(\displaystyle \frac{5}{9}-\frac{3}{9}=\frac{2}{9}\)

The phrase, "how much more" tells as that we want to find the difference in how much they've grown. 

\(\displaystyle \frac{4}{6}-\frac{1}{6}=\frac{3}{6}\)

The phrase, "how much more" tells as that we want to find the difference in how much they've grown. 

\(\displaystyle \frac{5}{6}-\frac{3}{6}=\frac{2}{6}\)