Fraction subtraction

### 1.1. Fraction subtraction

**For example :** From \(\frac{5}{6}\) colored paper tape, take \(\frac{3}{6}\) paper tape to cut the text. How many pieces of tape are left?

We have to do the calculation: \(\frac{5}{6} – \frac{3}{6}\).

We have : \(\frac{5}{6} – \frac{3}{6} = \frac{{5 – 3}}{6} = \frac{2}{6}\).

**To subtract two fractions with the same denominator, subtract the numerator of the first fraction by the numerator of the second fraction and keep the denominator the same.**

### 1.2. Fractional Subtraction (continued)

**For example :** A shop has \(\frac{4}{5}\) tons of sugar, the store has sold \(\frac{2}{3}\) tons of sugar. How many pieces of sugar are left in the store?

We have to do the calculation: \(\frac{4}{5} – \frac{2}{3}\).

We need to convert this subtraction to subtracting two fractions with the same denominator:

- Denominator of two fractions:

\(\begin{array}{l}

\frac{4}{5} = \frac{{4 \times 3}}{{5 \times 3}} = \frac{{12}}{{15}}\\

\frac{2}{3} = \frac{{2 \times 5}}{{3 \times 5}} = \frac{{10}}{{15}}

\end{array}\)

- Subtract two fractions : \(\frac{4}{5} – \frac{2}{3} = \frac{{12}}{{15}} – \frac{{10}}{{15}} = \frac{2}{{15}}\).

*To subtract two fractions with different denominators, we reduce the denominators to the two fractions, and then subtract those two fractions.*

### 1.3. Textbook exercise solution page 129

**Lesson 1: **Calculate

a) \(\frac{{15}}{{16}} – \frac{7}{{16}}\) ; b) \(\frac{7}{4} – \frac{3}{4}\) ;

c) \(\frac{9}{5} – \frac{3}{5}\) ; d) \(\frac{{17}}{{49}} – \frac{{12}}{{49}}\).

__Solution guide:__

- To subtract two fractions with the same denominator, subtract the numerator of the first fraction by the numerator of the second fraction and keep the denominator the same. If the resulting fraction can be reduced, we reduce it to the simplest fraction.

a) \(\frac{{15}}{{16}} – \frac{7}{{16}} = \frac{{15 – 7}}{{16}} = \frac{8}{{ 16}} = \frac{1}{2}\)

b) \(\frac{7}{4} – \frac{3}{4} = \frac{{7 – 3}}{4} = \frac{4}{4} = 1\)

c) \(\frac{9}{5} – \frac{3}{5} = \frac{{9 – 3}}{5} = \frac{6}{5}\)

d) \(\frac{{17}}{{49}} – \frac{{12}}{{49}} = \frac{{17 – 12}}{{49}} = \frac{5} {{49}}\)

**Lesson 2: **Shorten and calculate

a) \(\frac{2}{3} – \frac{3}{9}\) ; b) \(\frac{7}{5} – \frac{{15}}{{25}}\) ;

c) \(\frac{3}{2} – \frac{4}{8}\) ; d) \(\frac{{11}}{4} – \frac{6}{8}\).

__Solution guide:__

- Reduce fractions to minimal fractions (if possible), then subtract the two fractions.

a) \(\frac{2}{3} – \frac{3}{9} = \frac{2}{3} – \frac{1}{3} = \frac{{2 – 1}}{ 3} = \frac{1}{3}\)

b) \(\frac{7}{5} – \frac{{15}}{{25}} = \frac{7}{5} – \frac{3}{5} = \frac{{7 – 3}}{5} = \frac{4}{5}\)

c) \(\frac{3}{2} – \frac{4}{8} = \frac{3}{2} – \frac{1}{2} = \frac{{3 – 1}}{ 2} = \frac{2}{2} = 1\)

d) \(\frac{{11}}{4} – \frac{6}{8} = \frac{{11}}{4} – \frac{3}{4} = \frac{{11 – 3}}{4} = \frac{8}{4} = 2\)

**Lesson 3: **At the 6th National Phu Dong Fitness Association in 2004, the number of gold medals of the Dong Thap student union was equal to \(\frac{5}{{19}}\) the total number of gold medals won by the delegation, The rest are silver and bronze medals. The number of silver and bronze medals of Dong Thap delegation is equal to how much of the total number of medals won by the delegation?

__Solution guide:__

Consider the total number of medals won by the team as 1 unit.

To find the fraction of the team’s silver and bronze medal index, subtract 1 from the division’s gold medal fraction.

*Solution*

Consider the total number of medals won by the team to be 1 unit.

Number of silver and bronze medals won by:

\(1 – \frac{5}{{19}} = \frac{{19}}{{19}} – \frac{5}{{19}} = \frac{{14}}{{19} }\) (total medals)

Answer: \(\frac{{14}}{{19}}\) total number of medals.

### 1.4. Textbook exercise solution page 130

**Lesson 1: **Calculate

a) \(\frac{4}{5} – \frac{1}{3}\) ; b) \(\frac{5}{6} – \frac{3}{8}\) ;

c) \(\frac{8}{7} – \frac{2}{3}\) ; d) \(\frac{5}{3} – \frac{3}{5}\).

__Solution guide:__

- To subtract two fractions with different denominators, we reduce the denominators of the two fractions, and then subtract the two fractions.

a) \(\frac{4}{5} – \frac{1}{3}\)

+ Convert the denominator of two fractions:

\(\frac{4}{5} = \frac{{4 \times 3}}{{5 \times 3}} = \frac{{12}}{{15}};\,\,\,\ ,\,\,\,\,\,\,\,\,\frac{1}{3} = \frac{{1 \times 5}}{{3 \times 5}} = \frac{5} {{15}}\)

+ Subtract two fractions: \(\frac{4}{5} – \frac{1}{3} = \frac{{12}}{{15}} – \frac{5}{{15}} = \frac{7}{{15}}\)

b) \(\frac{5}{6} – \frac{3}{8}\)

+ Convert the denominator of two fractions:

\(\frac{5}{6} = \frac{{5 \times 8}}{{6 \times 8}} = \frac{{40}}{{48}};\,\,\,\ ,\,\,\,\,\,\,\,\,\frac{3}{8} = \frac{{3 \times 5}}{{8 \times 5}} = \frac{{15 }}{{48}} = \frac{{18}}{{48}}\)

+ Subtract two fractions : \(\frac{5}{6} – \frac{3}{8} = \frac{{40}}{{48}} – \frac{{18}}{{48} } = \frac{{22}}{{48}} = \frac{{11}}{{24}}\)

c) \(\frac{8}{7} – \frac{2}{3}\)

+ Convert the denominator of two fractions:

\(\frac{8}{7} = \frac{{8 \times 3}}{{7 \times 3}} = \frac{{24}}{{21}};\,\,\,\ ,\,\,\,\,\,\,\,\frac{2}{3} = \frac{{2 \times 7}}{{3 \times 7}} = \frac{{14}} {{21}}\)

+ Subtract two fractions : \(\frac{8}{7} – \frac{2}{3} = \frac{{24}}{{21}} – \frac{{14}}{{21} } = \frac{{10}}{{21}}\)

d) \(\frac{5}{3} – \frac{3}{5}\)

+ Convert the denominator of two fractions:

\(\frac{5}{3} = \frac{{5 \times 5}}{{3 \times 5}} = \frac{{25}}{{15}};\,\,\,\ ,\,\,\,\,\,\,\,\frac{3}{5} = \frac{{3 \times 3}}{{5 \times 3}} = \frac{9}{{ 15}}\)

+ Subtract two fractions : \(\frac{5}{3} – \frac{3}{5} = \frac{{25}}{{15}} – \frac{9}{{15}} = \frac{{16}}{{15}}\)

**Lesson 2: **Calculate

a) \(\frac{{20}}{{16}} – \frac{3}{4}\) ; b) \(\frac{{30}}{{45}} – \frac{2}{5}\) ;

c) \(\frac{{10}}{{12}} – \frac{3}{4}\) ; d) \(\frac{{12}}{9} – \frac{1}{4}\).

__Solution guide:__

Reduce fractions to minimal fractions (if possible), then subtract the two fractions as usual.

Solution:

a) \(\frac{{20}}{{16}} – \frac{3}{4} = \frac{{20:4}}{{16:4}} – \frac{3}{4 } = \frac{5}{4} – \frac{3}{4} = \frac{2}{4} = \frac{1}{2}\)

b) \(\frac{{30}}{{45}} – \frac{2}{5} = \frac{{30:15}}{{45:15}} – \frac{2}{5 } = \frac{2}{3} – \frac{2}{5} = \frac{{10}}{{15}} – \frac{6}{{15}} = \frac{4}{ {15}}\)

c) \(\frac{{10}}{{12}} – \frac{3}{4} = \frac{{10:2}}{{12:2}} – \frac{3}{4 } = \frac{5}{6} – \frac{3}{4} = \frac{{20}}{{24}} – \frac{{18}}{{24}} = \frac{2 }{{24}} = \frac{1}{{12}}\)

d) \(\frac{{12}}{9} – \frac{1}{4} = \frac{{12:3}}{{9:3}} – \frac{1}{4} = \frac{4}{3} – \frac{1}{4} = \frac{{16}}{{12}} – \frac{3}{{12}} = \frac{{13}}{ {twelfth}}\)

**Lesson 3: **In a park there is \(\frac{6}{7}\) the area planted with flowers and trees, where \(\frac{2}{5}\) the area of the park planted with flowers. What is the area of the park to plant trees?

__Solution guide:__

Area planted with trees = total area planted with flowers and trees – area planted with flowers.

*Solution*

The area planted with trees accounting for the number of parts of the park area is:

\(\frac{6}{7} – \frac{2}{5} = \frac{{16}}{{35}}\) (park area)

Answer: \(\frac{{16}}{{35}}\) park area.

.

=====

## Leave a Reply