Review Addition and subtraction of two fractions

### 1.1. Practice theory

a) To add or subtract two fractions with the same denominator, add or subtract the two numerators together and keep the denominator the same.

**Example 1:** \( \frac{3}{7}+\frac{5}{7}=\frac{3+5}{7}=\frac{8}{7}\).

Example 2: \( \frac{10}{15}-\frac{3}{15}=\frac{10-3}{15}=\frac{7}{15}\) .

b) To add or subtract two fractions with different denominators, we reduce the denominator, and then add or subtract two fractions that have the same denominator.

**Example 1:** \( \frac{7}{9}+\frac{3}{10}=\frac{70}{90}+\frac{27}{90}=\frac{97}{90}\) .

Example 2: \( \frac{7}{8}-\frac{7}{9}=\frac{63}{72}-\frac{56}{72}=\frac{7}{72}\) .

### 1.2. Instructions for solving textbook exercises

**Lesson 1 of the textbook page 10:** Calculate:

a) \(\frac{6}{7} + \frac{5}{8}\) b) \(\frac{3}{5} – \frac{3}{8}\) c) \( \frac{1}{4} + \frac{5}{6}\) d) \(\frac{4}{9} – \frac{1}{6}\)

**Solution:**

\(\begin{array}{l}

a)\,\,\frac{6}{7} + \frac{5}{8} = \frac{{48}}{{56}} + \frac{{35}}{{56}} = \frac{{83}}{{56}}\\

b)\,\,\frac{3}{5} – \frac{3}{8} = \frac{{24}}{{40}} – \frac{{15}}{{40}} = \frac{9}{{40}}\\

c)\,\,\frac{1}{4} + \frac{5}{6} = \frac{6}{{24}} + \frac{{20}}{{24}} = \frac {{26}}{{24}}\\

d)\,\,\frac{5}{9} – \frac{1}{6} = \frac{{24}}{{54}} – \frac{9}{{24}} = \frac {{15}}{{54}} = \frac{5}{{18}}

\end{array}\)

**Lesson 2 Textbook page 10: Calculation:**

a) \(3 + \frac{2}{5}\) b) \(4 – \frac{5}{7}\) c) \(1 – \left( {\frac{2}{5} + \frac{1}{3}} \right)\)

Solution:

\(\begin{array}{l}

a)\,\,3 + \frac{2}{5} = \frac{{15}}{5} + \frac{2}{5} = \frac{{17}}{5}\\

b)\,\,4 – \frac{5}{7} = \frac{{28}}{7} – \frac{5}{7} = \frac{{23}}{7}\\

c)\,\,1 – \left( {\frac{2}{5} + \frac{1}{3}} \right) = 1 – \left( {\frac{6}{{15}} + \frac{5}{{15}}} \right) = \frac{{15}}{{15}} – \frac{{11}}{{15}} = \frac{4}{{15 }}

\end{array}\)

**Lesson 3 Textbook page 10:** A box of balls has \(\frac{1}{2}\) number of red balls, \(\frac{1}{3}\) number of blue balls, the rest are yellow balls. Find the yellow ball index fraction

Solution:

The number of golden balls is: \(1 – \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{1}{6}\)

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