Opening the book…
- 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.
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, \(\dfrac{1}{4} + \dfrac{5}{8} \cdot \dfrac{2}{15} = \dfrac{1}{3}\)
\(\dfrac{3}{5} + \dfrac{9}{44} (\dfrac{5}{9} - \dfrac{1}{4})\)
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, \(\dfrac{3}{5} + \dfrac{9}{44} (\dfrac{5}{9} - \dfrac{1}{4}) = \dfrac{53}{80}\)
\(8 - \dfrac{15}{426} (2 - 1 \dfrac{4}{15}) (3 \dfrac{1}{5} + 2 \dfrac{1}{8})\)
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, \(8 - \dfrac{15}{426} (2 - 1 \dfrac{4}{15}) (3 \dfrac{1}{5} + 2 \dfrac{1}{8}) = 7 \dfrac{69}{80}\)
\((\dfrac{3}{4})^2 \cdot \dfrac{8}{9} - \dfrac{5}{12}\)
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, \((\dfrac{4}{3})^2 \cdot \dfrac{8}{9} - \dfrac{5}{12} = \dfrac{1}{12}\)
\(2 \dfrac{7}{8} + \sqrt{\dfrac{25}{36}} \div (2 \dfrac{1}{2} - 1 \dfrac{1}{3})\)
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 \(\dfrac{6 + \dfrac{1}{8}}{6 - \dfrac{1}{8}}\).
This page titled 5.6: Combinations of Operations with Fractions was last modified on Sat, 08 Jun 2024 18:05:14 GMT and is shared under a CC BY license and was authored, remixed, and/or curated by Denny Burzynski and Wade Ellis, Jr. (OpenStax CNX) via source content that was edited to the style and standards of the LibreTexts platform.
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