# 5.6 Combinations of operations with fractions

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses combinations of operations with fractions. By the end of the module students should gain a further understanding of the order of operations.

## Section overview

• The Order of Operations

## The order of operations

To determine the value of a quantity such as

$\frac{1}{2}+\frac{5}{8}\cdot \frac{2}{\text{15}}$

where we have a combination of operations (more than one operation occurs), we must use the accepted order of operations.

## The order of operations:

1. In the order (2), (3), (4) described below, perform all operations inside group­ing symbols: ( ), [ ], ( ), . Work from the innermost set to the outermost set.
2. Perform exponential and root operations.
3. Perform all multiplications and divisions moving left to right.
4. Perform all additions and subtractions moving left to right.

## Sample set a

Determine the value of each of the following quantities.

$\frac{1}{4}+\frac{5}{8}\cdot \frac{2}{\text{15}}$

1. Multiply first.

$\frac{1}{4}+\frac{\stackrel{1}{\overline{)5}}}{\underset{4}{\overline{)8}}}\cdot \frac{\stackrel{1}{\overline{)2}}}{\underset{3}{\overline{)\text{15}}}}=\frac{1}{4}+\frac{1\cdot 1}{4\cdot 3}=\frac{1}{4}+\frac{1}{\text{12}}$

2. Now perform this addition. Find the LCD.

$\left(\begin{array}{}4={2}^{2}\\ \text{12}={2}^{2}\cdot 3\end{array}}\text{The LCD}={2}^{2}\cdot 3=\text{12}\text{.}$

$\begin{array}{ccc}\hfill \frac{1}{4}+\frac{1}{\text{12}}& =& \frac{1\cdot 3}{\text{12}}+\frac{1}{\text{12}}=\frac{3}{\text{12}}+\frac{1}{\text{12}}\hfill \\ & =& \frac{3+1}{\text{12}}=\frac{4}{\text{12}}=\frac{1}{3}\hfill \end{array}$

Thus, $\frac{1}{4}+\frac{5}{8}\cdot \frac{2}{\text{15}}=\frac{1}{3}$

$\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{5}{9}-\frac{1}{4}\right)$

1. Operate within the parentheses first, $\left(\frac{5}{9}-\frac{1}{4}\right)$ .

$\left(\begin{array}{}9={3}^{2}\\ 4={2}^{2}\end{array}}\text{The LCD}={2}^{2}\cdot {3}^{2}=4\cdot 9=\text{36}\text{.}$

$\frac{5\cdot 4}{\text{36}}-\frac{1\cdot 9}{\text{36}}=\frac{\text{20}}{\text{36}}-\frac{9}{\text{36}}=\frac{\text{20}-9}{\text{36}}=\frac{\text{11}}{\text{36}}$

Now we have

$\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{\text{11}}{\text{36}}\right)$

2. Perform the multiplication.

$\frac{3}{5}+\frac{\stackrel{1}{\overline{)9}}}{\underset{4}{\overline{)\text{44}}}}\cdot \frac{\stackrel{1}{\overline{)\text{11}}}}{\underset{4}{\overline{)\text{36}}}}=\frac{3}{5}+\frac{1\cdot 1}{4\cdot 4}=\frac{3}{5}+\frac{1}{\text{16}}$

3. Now perform the addition. The LCD=80.

$\frac{3}{5}+\frac{1}{\text{16}}=\frac{3\cdot \text{16}}{\text{80}}+\frac{1\cdot 5}{\text{80}}=\frac{\text{48}}{\text{80}}+\frac{5}{\text{80}}=\frac{\text{48}+5}{\text{80}}=\frac{\text{53}}{\text{80}}$

Thus, $\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{5}{9}-\frac{1}{4}\right)=\frac{\text{53}}{\text{80}}$

$8-\frac{\text{15}}{\text{426}}\left(2-1\frac{4}{\text{15}}\right)\left(3\frac{1}{5}+2\frac{1}{8}\right)$

1. Work within each set of parentheses individually.

$\begin{array}{ccc}2-1\frac{4}{\text{15}}& =& 2\frac{1\cdot \text{15}+4}{\text{15}}=2-\frac{\text{19}}{\text{15}}\hfill \\ & =& \frac{\text{30}}{\text{15}}-\frac{\text{19}}{\text{15}}=\frac{\text{30}-\text{19}}{\text{15}}=\frac{\text{11}}{\text{15}}\hfill \\ 3\frac{1}{5}+2\frac{1}{8}& =& \frac{3\cdot 5+1}{5}+\frac{2\cdot 8+1}{8}\hfill \\ & =& \frac{\text{16}}{5}+\frac{\text{17}}{8}\text{LCD}=\text{40}\hfill \\ & =& \frac{\text{16}\cdot 8}{\text{40}}+\frac{\text{17}\cdot 5}{\text{40}}\hfill \\ & =& \frac{\text{128}}{\text{40}}+\frac{\text{85}}{\text{40}}\hfill \\ & =& \frac{\text{128}+\text{85}}{\text{40}}\hfill \\ & =& \frac{\text{213}}{\text{40}}\hfill \end{array}$

Now we have

$8-\frac{\text{15}}{\text{426}}\left(\frac{\text{11}}{\text{15}}\right)\left(\frac{\text{213}}{\text{40}}\right)$

2. Now multiply.

$8-\frac{\stackrel{1}{\overline{)\text{15}}}}{\underset{2}{\overline{)\text{426}}}}\cdot \frac{\text{11}}{\underset{1}{\overline{)\text{15}}}}\cdot \frac{\stackrel{1}{\overline{)\text{213}}}}{\text{40}}=8-\frac{1\cdot \text{11}\cdot 1}{2\cdot 1\cdot \text{40}}=8-\frac{\text{11}}{\text{80}}$

3. Now subtract.

$8-\frac{\text{11}}{\text{80}}=\frac{\text{80}\cdot 8}{\text{80}}-\frac{\text{11}}{\text{80}}=\frac{\text{640}}{\text{80}}-\frac{\text{11}}{\text{80}}=\frac{\text{640}-\text{11}}{\text{80}}=\frac{\text{629}}{\text{80}}\phantom{\rule{4px}{0ex}}\text{or}\phantom{\rule{4px}{0ex}}7\frac{\text{69}}{\text{80}}$

Thus, $8-\frac{15}{426}\left(2-1\frac{4}{15}\right)\left(3\frac{1}{5}+2\frac{1}{8}\right)=7\frac{69}{80}$

${\left(\frac{3}{4}\right)}^{2}\cdot \frac{8}{9}-\frac{5}{\text{12}}$

1. Square $\frac{3}{4}$ .

${\left(\frac{3}{4}\right)}^{2}=\frac{3}{4}\cdot \frac{3}{4}=\frac{3\cdot 3}{4\cdot 4}=\frac{9}{\text{16}}$

Now we have

$\frac{9}{\text{16}}\cdot \frac{8}{9}-\frac{5}{\text{12}}$

2. Perform the multiplication.

$\frac{\stackrel{1}{\overline{)9}}}{\underset{2}{\overline{)\text{16}}}}\cdot \frac{\stackrel{1}{\overline{)8}}}{\underset{1}{\overline{)9}}}-\frac{5}{\text{12}}=\frac{1\cdot 1}{2\cdot 1}-\frac{5}{\text{12}}=\frac{1}{2}-\frac{5}{\text{12}}$

3. Now perform the subtraction.

$\frac{1}{2}-\frac{5}{\text{12}}=\frac{6}{\text{12}}-\frac{5}{\text{12}}=\frac{6-5}{\text{12}}=\frac{1}{\text{12}}$

Thus, ${\left(\frac{4}{3}\right)}^{2}\cdot \frac{8}{9}-\frac{5}{\text{12}}=\frac{1}{\text{12}}$

$2\frac{7}{8}+\sqrt{\frac{\text{25}}{\text{36}}}÷\left(2\frac{1}{2}-1\frac{1}{3}\right)$

1. Begin by operating inside the parentheses.

$\begin{array}{ccc}2\frac{1}{2}-1\frac{1}{3}& =& \frac{2\cdot 2+1}{2}-\frac{1\cdot 3+1}{3}=\frac{5}{2}-\frac{4}{3}\hfill \\ & =& \frac{\text{15}}{6}-\frac{8}{6}=\frac{\text{15}-8}{6}=\frac{7}{6}\hfill \end{array}$

2. Now simplify the square root.

$\sqrt{\frac{\text{25}}{\text{36}}}=\frac{5}{6}\left(\text{since}{\left(\frac{5}{6}\right)}^{2}=\frac{\text{25}}{\text{36}}\right)$

Now we have

$2\frac{7}{8}+\frac{5}{6}÷\frac{7}{6}$

3. Perform the division.

$2\frac{7}{8}+\frac{5}{\underset{1}{\overline{)6}}}\cdot \frac{\stackrel{1}{\overline{)6}}}{7}=2\frac{7}{8}+\frac{5\cdot 1}{1\cdot 7}=2\frac{7}{8}+\frac{5}{7}$

4. Now perform the addition.

Thus, $2\frac{7}{8}+\sqrt{\frac{\text{25}}{\text{36}}}÷\left(2\frac{1}{2}-1\frac{1}{3}\right)=3\frac{\text{33}}{\text{56}}$

## Practice set a

Find the value of each of the following quantities.

$\frac{5}{\text{16}}\cdot \frac{1}{\text{10}}-\frac{1}{\text{32}}$

0

$\frac{6}{7}\cdot \frac{\text{21}}{\text{40}}÷\frac{9}{\text{10}}+5\frac{1}{3}$

$\frac{\text{35}}{6}$ or $5\frac{5}{\text{6}}$

$8\frac{7}{\text{10}}-2\left(4\frac{1}{2}-3\frac{2}{3}\right)$

$\frac{\text{211}}{\text{30}}$ or $7\frac{1}{\text{30}}$

$\frac{\text{17}}{\text{18}}-\frac{\text{58}}{\text{30}}\left(\frac{1}{4}-\frac{3}{\text{32}}\right)\left(1-\frac{\text{13}}{\text{29}}\right)$

$\frac{7}{9}$

$\left(\frac{1}{\text{10}}+1\frac{1}{2}\right)÷\left(1\frac{4}{5}-1\frac{6}{\text{25}}\right)$

$2\frac{6}{7}$

$\frac{\frac{2}{3}-\frac{3}{8}\cdot \frac{4}{9}}{\frac{7}{\text{16}}\cdot 1\frac{1}{3}+1\frac{1}{4}}$

$\frac{3}{\text{11}}$

${\left(\frac{3}{8}\right)}^{2}+\frac{3}{4}\cdot \frac{1}{8}$

$\frac{\text{15}}{\text{64}}$

$\frac{2}{3}\cdot 2\frac{1}{4}-\sqrt{\frac{4}{\text{25}}}$

$\frac{\text{11}}{\text{10}}$

## Exercises

Find each value.

$\frac{4}{3}-\frac{1}{6}\cdot \frac{1}{2}$

$\frac{5}{4}$

$\frac{7}{9}-\frac{4}{5}\cdot \frac{5}{\text{36}}$

$2\frac{2}{7}+\frac{5}{8}÷\frac{5}{\text{16}}$

$4\frac{2}{7}$

$\frac{3}{\text{16}}÷\frac{9}{\text{14}}\cdot \frac{\text{12}}{\text{21}}+\frac{5}{6}$

$\frac{4}{\text{25}}÷\frac{8}{\text{15}}-\frac{7}{\text{20}}÷2\frac{1}{\text{10}}$

$\frac{2}{\text{15}}$

$\frac{2}{5}\cdot \left(\frac{1}{\text{19}}+\frac{3}{\text{38}}\right)$

$\frac{3}{7}\cdot \left(\frac{3}{\text{10}}-\frac{1}{\text{15}}\right)$

$\frac{1}{\text{10}}$

$\frac{\text{10}}{\text{11}}\cdot \left(\frac{8}{9}-\frac{2}{5}\right)+\frac{3}{\text{25}}\cdot \left(\frac{5}{3}+\frac{1}{4}\right)$

$\frac{2}{7}\cdot \left(\frac{6}{7}-\frac{3}{\text{28}}\right)+5\frac{1}{3}\cdot \left(1\frac{1}{4}-\frac{1}{8}\right)$

$6\frac{3}{\text{14}}$

$\frac{\left(\frac{6}{\text{11}}-\frac{1}{3}\right)\cdot \left(\frac{1}{\text{21}}+2\frac{\text{13}}{\text{42}}\right)}{1\frac{1}{5}+\frac{7}{\text{40}}}$

${\left(\frac{1}{2}\right)}^{2}+\frac{1}{8}$

$\frac{3}{8}$

${\left(\frac{3}{5}\right)}^{2}-\frac{3}{\text{10}}$

$\sqrt{\frac{\text{36}}{\text{81}}}+\frac{1}{3}\cdot \frac{2}{9}$

$\frac{\text{20}}{\text{27}}$

$\sqrt{\frac{\text{49}}{\text{64}}}-\sqrt{\frac{9}{4}}$

$\frac{2}{3}\cdot \sqrt{\frac{9}{4}}-\frac{\text{15}}{4}\cdot \sqrt{\frac{\text{16}}{\text{225}}}$

0

${\left(\frac{3}{4}\right)}^{2}+\sqrt{\frac{\text{25}}{\text{16}}}$

${\left(\frac{1}{3}\right)}^{2}\cdot \sqrt{\frac{\text{81}}{\text{25}}}+\frac{1}{\text{40}}÷\frac{1}{8}$

$\frac{2}{5}$

${\left(\sqrt{\frac{4}{\text{49}}}\right)}^{2}+\frac{3}{7}÷1\frac{3}{4}$

${\left(\sqrt{\frac{\text{100}}{\text{121}}}\right)}^{2}+\frac{\text{21}}{{\left(\text{11}\right)}^{2}}$

1

$\sqrt{\frac{3}{8}+\frac{1}{\text{64}}}-\frac{1}{2}÷1\frac{1}{3}$

$\sqrt{\frac{1}{4}}\cdot {\left(\frac{5}{6}\right)}^{2}+\frac{9}{\text{14}}\cdot 2\frac{1}{3}-\sqrt{\frac{1}{\text{81}}}$

$\frac{\text{125}}{\text{72}}$

$\sqrt{\frac{1}{9}}\cdot \sqrt{\frac{6\frac{3}{8}+2\frac{5}{8}}{\text{16}}}+7\frac{7}{\text{10}}$

$\frac{3\frac{3}{4}+\frac{4}{5}\cdot {\left(\frac{1}{2}\right)}^{3}}{\frac{\text{67}}{\text{240}}+{\left(\frac{1}{3}\right)}^{4}\cdot \left(\frac{9}{\text{10}}\right)}$

$\frac{\text{252}}{\text{19}}$

$\sqrt{\sqrt{\frac{\text{16}}{\text{81}}}}+\frac{1}{4}\cdot 6$

$\sqrt{\sqrt{\frac{\text{81}}{\text{256}}}}-\frac{3}{\text{32}}\cdot 1\frac{1}{8}$

$\frac{\text{165}}{\text{256}}$

## Exercises for review

( [link] ) True or false: Our number system, the Hindu-Arabic number system, is a positional number system with base ten.

( [link] ) The fact that 1 times any whole number = that particular whole number illustrates which property of multiplication?

multiplicative identity

( [link] ) Convert $8\frac{6}{7}$ to an improper fraction.

( [link] ) Find the sum. $\frac{3}{8}+\frac{4}{5}+\frac{5}{6}$ .

$\frac{\text{241}}{\text{120}}$ or $2\frac{1}{\text{120}}$

( [link] ) Simplify $\frac{6+\frac{1}{8}}{6-\frac{1}{8}}$ .

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