5.6: Combinations of Operations with Fractions

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Learning Objectives

gain a further understanding of the order of operations

The Order of Operations

To determine the value of a quantity such as

\(\dfrac{1}{2} + \dfrac{5}{8} \cdot \dfrac{2}{15}\)

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

The Order of Operations:

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

Sample Set A

Determine the value of each of the following quantities.

\(\dfrac{1}{4} + \dfrac{5}{8} \cdot \dfrac{2}{15}\)

Solution

a. Multiply first.

\(\dfrac{1}{4} + \dfrac{\begin{array} {c} {^1} \\ {\cancel{5}} \end{array}}{\begin{array} {c} {\cancel{8}} \\ {^4} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{2}} \end{array}}{\begin{array} {c} {\cancel{15}} \\ {^3} \end{array}} = \dfrac{1}{4} + \dfrac{1 \cdot 1}{4 \cdot 3} = \dfrac{1}{4} + \dfrac{1}{12}\)

b. Now perform this addition. Find the LCD.

\(\left \{ \begin{array} {c} {4 = 2^2} \\ {12 = 2^2 \cdot 3} \end{array} \right \} \text{ The LCD = } 2^2 \cdot 3 = 12\)

\(\begin{array} {rcl} {\dfrac{1}{4} + \dfrac{1}{12}} & = & {\dfrac{1 \cdot 3}{12} + \dfrac{1}{12} = \dfrac{3}{12} + \dfrac{1}{12}} \\ {} & = & {\dfrac{3 + 1}{12} = \dfrac{4}{12} = \dfrac{1}{3}} \end{array}\)

Thus, \(\dfrac{1}{4} + \dfrac{5}{8} \cdot \dfrac{2}{15} = \dfrac{1}{3}\)

Sample Set A

\(\dfrac{3}{5} + \dfrac{9}{44} (\dfrac{5}{9} - \dfrac{1}{4})\)

Solution

a. Operate within the parentheses first, \((\dfrac{5}{9} - \dfrac{1}{4})\)

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

\(\dfrac{5 \cdot 4}{36} - \dfrac{1 \cdot 9}{36} = \dfrac{20}{36} - \dfrac{9}{36} = \dfrac{20 - 9}{36} = \dfrac{11}{36}\)

Now we have

\(\dfrac{3}{5} + \dfrac{9}{44} (\dfrac{11}{36})\)

b. Perform the multiplication.

\(\dfrac{3}{5} + \dfrac{\begin{array} {c} {^1} \\ {\cancel{9}} \end{array}}{\begin{array} {c} {\cancel{44}} \\ {^4} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{11}} \end{array}}{\begin{array} {c} {\cancel{36}} \\ {^4} \end{array}} = \dfrac{3}{5} + \dfrac{1 \cdot 1}{4 \cdot 4} = \dfrac{3}{5} + \dfrac{1}{16}\)

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

\(\dfrac{3}{5} + \dfrac{1}{16} = \dfrac{3 \cdot 16}{80} + \dfrac{1 \cdot 5}{80} = \dfrac{48}{80} + \dfrac{5}{80} = \dfrac{48 + 5}{80} = \dfrac{53}{80}\)

Thus, \(\dfrac{3}{5} + \dfrac{9}{44} (\dfrac{5}{9} - \dfrac{1}{4}) = \dfrac{53}{80}\)

Sample Set A

\(8 - \dfrac{15}{426} (2 - 1 \dfrac{4}{15}) (3 \dfrac{1}{5} + 2 \dfrac{1}{8})\)

Solution

a. Work within each set of parentheses individually.

\(\begin{array} {rcl} {2 - 1 \dfrac{4}{15}} & = & {2 \dfrac{1 \cdot 15 + 4}{15} = 2 - \dfrac{19}{15}} \\ {} & = & {\dfrac{30}{15} - \dfrac{19}{15} = \dfrac{30- 19}{15} = \dfrac{11}{15}} \\ {3 \dfrac{1}{5} + 2 \dfrac{1}{8}} & = & {\dfrac{3 \cdot 5 + 1}{5} + \dfrac{2 \cdot 8 + 1}{8}} \\ {} & = & {\dfrac{16}{5} + \dfrac{17}{8} \text{LCD = 40}} \\ {} & = & {\dfrac{16 \cdot 8}{40} + \dfrac{17 \cdot 5}{40}} \\ {} & = & {\dfrac{128}{40} + \dfrac{85}{40}} \\ {} & = & {\dfrac{128 + 85}{40}} \\ {} & = & {\dfrac{213}{40}} \end{array}\)

Now we have

\(8 - \dfrac{15}{426} (\dfrac{11}{15}) (\dfrac{213}{40})\)

b. Now multiply.

\(8 - \dfrac{\begin{array} {c} {^1} \\ {\cancel{15}} \end{array}}{\begin{array} {c} {\cancel{426}} \\ {^2} \end{array}} \cdot \dfrac{11}{\begin{array} {c} {\cancel{15}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{213}} \end{array}}{40} = 8 - \dfrac{1 \cdot 11 \cdot 1}{2 \cdot 1 \cdot 40} = 8 - \dfrac{11}{80}\)

c. Now subtract.

\(8 - \dfrac{11}{80} = \dfrac{80 \cdot 8}{80} - \dfrac{11}{80} = \dfrac{640}{80} - \dfrac{11}{80} = \dfrac{640 - 11}{80} = \dfrac{629}{80} \text{ or } 7 \dfrac{69}{80}\)

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

Sample Set A

\((\dfrac{3}{4})^2 \cdot \dfrac{8}{9} - \dfrac{5}{12}\)

Solution

a. Square \(\dfrac{3}{4}\).

\((\dfrac{3}{4})^2 = \dfrac{3}{4} \cdot \dfrac{3}{4} = \dfrac{3 \cdot 3}{4 \cdot 4} = \dfrac{9}{16}\)

Now we have

\(\dfrac{9}{16} \cdot \dfrac{8}{9} - \dfrac{5}{12}\)

b. Perform the multiplication.

\(\dfrac{\begin{array} {c} {^1} \\ {\cancel{9}} \end{array}}{\begin{array} {c} {\cancel{16}} \\ {^2} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{8}} \end{array}}{\begin{array} {c} {\cancel{9}} \\ {^1} \end{array}} - \dfrac{5}{12} = \dfrac{1 \cdot 1}{2 \cdot 1} - \dfrac{5}{12} = \dfrac{1}{2} - \dfrac{5}{12}\)

c. Now perform the subtraction.

\(\dfrac{1}{2} - \dfrac{5}{12} = \dfrac{6}{12} - \dfrac{5}{12} = \dfrac{6 - 5}{12} = \dfrac{1}{12}\)

Thus, \((\dfrac{4}{3})^2 \cdot \dfrac{8}{9} - \dfrac{5}{12} = \dfrac{1}{12}\)

Sample Set A

\(2 \dfrac{7}{8} + \sqrt{\dfrac{25}{36}} \div (2 \dfrac{1}{2} - 1 \dfrac{1}{3})\)

Solution

a. Begin by operating inside the parentheses.

\(\begin{array} {rcl} {2 \dfrac{1}{2} - 1 \dfrac{1}{3}} & = & {\dfrac{2 \cdot 2 + 1}{2} - \dfrac{1 \cdot 3 + 1}{3} = \dfrac{5}{2} - \dfrac{4}{3}} \\ {} & = & {\dfrac{15}{6} - \dfrac{8}{6} = \dfrac{15 - 8}{6} = \dfrac{7}{6}} \end{array}\)

b. Now simplify the square root.

\(\sqrt{\dfrac{25}{36}} = \dfrac{5}{6} (\text{since} (\dfrac{5}{6})^2 = \dfrac{25}{36})\)

Now we have

\(2 \dfrac{7}{8} + \dfrac{5}{6} \div \dfrac{7}{6}\)

c. Perform the division.

\(2 \dfrac{7}{8} + \dfrac{5}{\begin{array} {c} {\cancel{6}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{6}} \end{array}}{7} = 2 \dfrac{7}{8} + \dfrac{5 \cdot 1}{1 \cdot 7} = 2 \dfrac{7}{8} + \dfrac{5}{7}\)

d. Now perform the addition.

\(\begin{array} {rcl} {2 \dfrac{7}{8} + \dfrac{5}{7}} & = & {\dfrac{2 \cdot 8 + 7}{8} + \dfrac{5}{7} = \dfrac{23}{8} + \dfrac{5}{7} \text{ LCD = }56.} \\ {} & = & {\dfrac{23 \cdot 7}{56} + \dfrac{5 \cdot 8}{56} = \dfrac{161}{56} + \dfrac{40}{56}} \\ {} & = & {\dfrac{161 + 40}{56} = \dfrac{201}{56} \text{ or } 3 \dfrac{33}{56}} \end{array}\)

Thus, \(2 \dfrac{7}{8} + \sqrt{\dfrac{25}{36}} \div (2 \dfrac{1}{2} - 1 \dfrac{1}{3}) = 3 \dfrac{33}{56}\)

Practice Set A

Find the value of each of the following quantities.

\(\dfrac{5}{16} \cdot \dfrac{1}{10} - \dfrac{1}{32}\)

Answer: 0

Practice Set A

\(\dfrac{6}{7} \cdot \dfrac{21}{40} \div \dfrac{9}{10} + 5 \dfrac{1}{3}\)

Answer: \(\dfrac{35}{6}\) or \(5 \dfrac{5}{6}\)

Practice Set A

\(8\dfrac{7}{10} - 2(4 \dfrac{1}{2} - 3 \dfrac{2}{3})\)

Answer: \(\dfrac{211}{30}\) or \(7 \dfrac{1}{30}\)

Practice Set A

\(\dfrac{17}{18} - \dfrac{58}{30} (\dfrac{1}{4} - \dfrac{3}{32}) (1 - \dfrac{13}{29})\)

Answer: \(\dfrac{7}{9}\)

Practice Set A

\((\dfrac{1}{10} + 1 \dfrac{1}{2}) \div (1 \dfrac{4}{5} - 1 \dfrac{6}{25})\)

Answer: \(2 \dfrac{6}{7}\)

Practice Set A

\(\dfrac{\dfrac{2}{3} - \dfrac{3}{8} \cdot \dfrac{4}{9}}{\dfrac{7}{16} \cdot 1 \dfrac{1}{3} + 1 \dfrac{1}{4}}\)

Answer: \(\dfrac{3}{11}\)

Practice Set A

\((\dfrac{3}{8})^2 + \dfrac{3}{4} \cdot \dfrac{1}{8}\)

Answer: \(\dfrac{15}{64}\)

Practice Set A

\(\dfrac{2}{3} \cdot 2 \dfrac{1}{4} - \sqrt{\dfrac{4}{25}}\)

Answer: \(\dfrac{11}{10}\)

Exercises

Find each value.

Exercise \(\PageIndex{1}\)

\(\dfrac{4}{3} - \dfrac{1}{6} \cdot \dfrac{1}{2}\)

Answer: \(\dfrac{5}{4}\)

Exercise \(\PageIndex{2}\)

\(\dfrac{7}{9} - \dfrac{4}{5} \cdot \dfrac{5}{36}\)

Exercise \(\PageIndex{3}\)

\(2 \dfrac{2}{7} + \dfrac{5}{8} \div \dfrac{5}{16}\)

Answer: \(4 \dfrac{2}{7}\)

Exercise \(\PageIndex{4}\)

\(\dfrac{3}{16} \div \dfrac{9}{14} \cdot \dfrac{12}{21} + \dfrac{5}{6}\)

Exercise \(\PageIndex{5}\)

\(\dfrac{4}{25} \div \dfrac{8}{15} - \dfrac{7}{20} \div 2 \dfrac{1}{10}\)

Answer: \(\dfrac{2}{15}\)

Exercise \(\PageIndex{6}\)

\(\dfrac{2}{5} \cdot (\dfrac{1}{19} + \dfrac{3}{38})\)

Exercise \(\PageIndex{7}\)

\(\dfrac{3}{7} \cdot (\dfrac{3}{10} - \dfrac{1}{15})\)

Answer: \(\dfrac{1}{10}\)

Exercise \(\PageIndex{8}\)

\(\dfrac{10}{11} \cdot (\dfrac{8}{9} - \dfrac{2}{5}) + \dfrac{3}{25} \cdot (\dfrac{5}{3} + \dfrac{1}{4})\)

Exercise \(\PageIndex{9}\)

\(\dfrac{2}{7} \cdot (\dfrac{6}{7} - \dfrac{3}{28}) + 5 \dfrac{1}{3} \cdot (1 \dfrac{1}{4} - \dfrac{1}{8})\)

Answer: \(6 \dfrac{3}{14}\)

Exercise \(\PageIndex{10}\)

\(\dfrac{(\dfrac{6}{11} - \dfrac{1}{3}) \cdot (\dfrac{1}{21} + 2 \dfrac{13}{42})}{1 \dfrac{1}{5} + \dfrac{7}{40}}\)

Exercise \(\PageIndex{11}\)

\((\dfrac{1}{2})^2 + \dfrac{1}{8}\)

Answer: \(\dfrac{3}{8}\)

Exercise \(\PageIndex{12}\)

\((\dfrac{3}{5})^2 - \dfrac{3}{10}\)

Exercise \(\PageIndex{13}\)

\(\sqrt{\dfrac{36}{81}} + \dfrac{1}{3} \cdot \dfrac{2}{9}\)

Answer: \(\dfrac{20}{27}\)

Exercise \(\PageIndex{14}\)

\(\sqrt{\dfrac{49}{64}} - \sqrt{\dfrac{9}{4}}\)

Exercise \(\PageIndex{15}\)

\(\dfrac{2}{3} \cdot \sqrt{\dfrac{9}{4}} - \dfrac{15}{4} \cdot \sqrt{\dfrac{16}{225}}\)

Answer: 0

Exercise \(\PageIndex{16}\)

\((\dfrac{3}{4})^2 + \sqrt{dfrac{25}{16}}\)

Exercise \(\PageIndex{17}\)

\((\dfrac{1}{3})^2 \cdot \sqrt{\dfrac{81}{25}} + \dfrac{1}{40} \div \dfrac{1}{8}\)

Answer: \(\dfrac{2}{5}\)

Exercise \(\PageIndex{18}\)

\((\sqrt{\dfrac{4}{49}})^2 + \dfrac{3}{7} \div 1 \dfrac{3}{4}\)

Exercise \(\PageIndex{19}\)

\((\sqrt{\dfrac{100}{121}})^2 + \dfrac{21}{(11)^2}\)

Answer: 1

Exercise \(\PageIndex{20}\)

\(\sqrt{\dfrac{3}{8} + \dfrac{1}{64}} - \dfrac{1}{2} \div 1 \dfrac{1}{3}\)

Exercise \(\PageIndex{21}\)

\(\sqrt{\dfrac{1}{4}} \cdot (\dfrac{5}{6})^2 + \dfrac{9}{14} \cdot 2 \dfrac{1}{3} - \sqrt{\dfrac{1}{81}}\)

Answer: \(\dfrac{215}{72}\)

Exercise \(\PageIndex{22}\)

\(\sqrt{\dfrac{1}{9}} \cdot \sqrt{\dfrac{6 \dfrac{3}{8} + 2 \dfrac{5}{8}}{16}} + 7 \dfrac{7}{10}\)

Exercise \(\PageIndex{23}\)

\(\dfrac{3 \dfrac{3}{4} + \dfrac{4}{5} \cdot (\dfrac{1}{2})^3}{\dfrac{67}{240} + (\dfrac{1}{3})^4 \cdot (\dfrac{9}{10})}\)

Answer: \(\dfrac{252}{19}\)

Exercise \(\PageIndex{24}\)

\(\sqrt{\sqrt{\dfrac{16}{81}}} + \dfrac{1}{4} \cdot 6\)

Exercise \(\PageIndex{25}\)

\(\sqrt{\sqrt{\dfrac{81}{256}}} - \dfrac{3}{32} \cdot 1 \dfrac{1}{8}\)

Answer: \(\dfrac{165}{256}\)

Exercises for Review

Exercise \(\PageIndex{26}\)

True or false: Our number system, the Hindu-Arabic number system, is a positional number system with base ten.

Exercise \(\PageIndex{27}\)

The fact that 1 times any whole number = that particular whole number illustrates which property of multiplication?

Answer: multiplicative identity

Exercise \(\PageIndex{28}\)

Convert \(8 \dfrac{6}{7}\) to an improper fraction.

Exercise \(\PageIndex{29}\)

Find the sum. \(\dfrac{3}{8} + \dfrac{4}{5} + \dfrac{5}{6}\).

Answer: \(\dfrac{241}{120}\) or \(2 \dfrac{1}{120}\)

Exercise \(\PageIndex{30}\)

Simplify \(\dfrac{6 + \dfrac{1}{8}}{6 - \dfrac{1}{8}}\).

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