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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

1 2 + 5 8 2 15 size 12{ { {1} over {2} } + { {5} over {8} } cdot { {2} over {"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.

1 4 + 5 8 2 15 size 12{ { {1} over {4} } + { {5} over {8} } cdot { {2} over {"15"} } } {}

  1. Multiply first.

    1 4 + 5 1 8 4 2 1 15 3 = 1 4 + 1 1 4 3 = 1 4 + 1 12 size 12{ { {1} over {4} } + { { {5} cSup { size 8{1} } } over { {8} cSub { size 8{4} } } } cdot { { {2} cSup { size 8{1} } } over { {"15"} cSub { size 8{3} } } } = { {1} over {4} } + { {1 cdot 1} over {4 cdot 3} } = { {1} over {4} } + { {1} over {"12"} } } {}

  2. Now perform this addition. Find the LCD.

    4 = 2 2 12 = 2 2 3 The LCD = 2 2 3 = 12 .

    1 4 + 1 12 = 1 3 12 + 1 12 = 3 12 + 1 12 = 3 + 1 12 = 4 12 = 1 3

    Thus, 1 4 + 5 8 2 15 = 1 3 size 12{ { {1} over {4} } + { {5} over {8} } cdot { {2} over {"15"} } = { {1} over {3} } } {}

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3 5 + 9 44 5 9 1 4 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {5} over {9} } - { {1} over {4} } right )} {}

  1. Operate within the parentheses first, 5 9 1 4 size 12{ left ( { {5} over {9} } - { {1} over {4} } right )} {} .

    9 = 3 2 4 = 2 2 The LCD = 2 2 3 2 = 4 9 = 36 .


    5 4 36 1 9 36 = 20 36 9 36 = 20 9 36 = 11 36 size 12{ { {5 cdot 4} over {"36"} } - { {1 cdot 9} over {"36"} } = { {"20"} over {"36"} } - { {9} over {"36"} } = { {"20" - 9} over {"36"} } = { {"11"} over {"36"} } } {}


    Now we have


    3 5 + 9 44 11 36 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {"11"} over {"36"} } right )} {}

  2. Perform the multiplication.

    3 5 + 9 1 44 4 11 1 36 4 = 3 5 + 1 1 4 4 = 3 5 + 1 16 size 12{ { {3} over {5} } + { { {9} cSup { size 8{1} } } over { {"44"} cSub { size 8{4} } } } cdot { { {"11"} cSup { size 8{1} } } over { {"36"} cSub { size 8{4} } } } = { {3} over {5} } + { {1 cdot 1} over {4 cdot 4} } = { {3} over {5} } + { {1} over {"16"} } } {}

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

    3 5 + 1 16 = 3 16 80 + 1 5 80 = 48 80 + 5 80 = 48 + 5 80 = 53 80 size 12{ { {3} over {5} } + { {1} over {"16"} } = { {3 cdot "16"} over {"80"} } + { {1 cdot 5} over {"80"} } = { {"48"} over {"80"} } + { {5} over {"80"} } = { {"48"+5} over {"80"} } = { {"53"} over {"80"} } } {}


    Thus, 3 5 + 9 44 5 9 1 4 = 53 80 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {5} over {9} } - { {1} over {4} } right )= { {"53"} over {"80"} } } {}

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8 15 426 2 1 4 15 3 1 5 + 2 1 8 size 12{8 - { {"15"} over {"426"} } left (2 - 1 { {4} over {"15"} } right ) left (3 { {1} over {5} } +2 { {1} over {8} } right )} {}

  1. Work within each set of parentheses individually.

    2 1 4 15 = 2 1 15 + 4 15 = 2 19 15 = 30 15 19 15 = 30 19 15 = 11 15 3 1 5 + 2 1 8 = 3 5 + 1 5 + 2 8 + 1 8 = 16 5 + 17 8 LCD = 40 = 16 8 40 + 17 5 40 = 128 40 + 85 40 = 128 + 85 40 = 213 40


    Now we have

    8 15 426 11 15 213 40 size 12{8 - { {"15"} over {"426"} } left ( { {"11"} over {"15"} } right ) left ( { {"213"} over {"40"} } right )} {}

  2. Now multiply.

    8 15 1 426 2 11 15 1 213 1 40 = 8 1 11 1 2 1 40 = 8 11 80 size 12{8 - { { {"15"} cSup { size 8{1} } } over { {"426"} cSub { size 8{2} } } } cdot { {"11"} over { {"15"} cSub { size 8{1} } } } cdot { { {"213"} cSup { size 8{1} } } over {"40"} } =8 - { {1 cdot "11" cdot 1} over {2 cdot 1 cdot "40"} } =8 - { {"11"} over {"80"} } } {}

  3. Now subtract.

    8 11 80 = 80 8 80 11 80 = 640 80 11 80 = 640 11 80 = 629 80 or 7 69 80 size 12{8 - { {"11"} over {"80"} } = { {"80" cdot 8} over {"80"} } - { {"11"} over {"80"} } = { {"640"} over {"80"} } - { {"11"} over {"80"} } = { {"640" - "11"} over {"80"} } = { {"629"} over {"80"} } " or "7 { {"69"} over {"80"} } } {}


    Thus, 8 - 15 426 2 - 1 4 15 3 1 5 + 2 1 8 = 7 69 80

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3 4 2 8 9 5 12 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } cdot { {8} over {9} } - { {5} over {"12"} } } {}

  1. Square 3 4 size 12{ { {3} over {4} } } {} .

    3 4 2 = 3 4 3 4 = 3 3 4 4 = 9 16 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } = { {3} over {4} } cdot { {3} over {4} } = { {3 cdot 3} over {4 cdot 4} } = { {9} over {"16"} } } {}

    Now we have

    9 16 8 9 5 12 size 12{ { {9} over {"16"} } cdot { {8} over {9} } - { {5} over {"12"} } } {}

  2. Perform the multiplication.

    9 1 16 2 8 1 9 1 5 12 = 1 1 2 1 5 12 = 1 2 5 12 size 12{ { { {9} cSup { size 8{1} } } over { {"16"} cSub { size 8{2} } } } cdot { { {8} cSup { size 8{1} } } over { {9} cSub { size 8{1} } } } - { {5} over {"12"} } = { {1 cdot 1} over {2 cdot 1} } - { {5} over {"12"} } = { {1} over {2} } - { {5} over {"12"} } } {}

  3. Now perform the subtraction.

    1 2 5 12 = 6 12 5 12 = 6 5 12 = 1 12 size 12{ { {1} over {2} } - { {5} over {"12"} } = { {6} over {"12"} } - { {5} over {"12"} } = { {6 - 5} over {"12"} } = { {1} over {"12"} } } {}

    Thus, 4 3 2 8 9 5 12 = 1 12 size 12{ left ( { {4} over {3} } right ) rSup { size 8{2} } cdot { {8} over {9} } - { {5} over {"12"} } = { {1} over {"12"} } } {}

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2 7 8 + 25 36 ÷ 2 1 2 1 1 3 size 12{2 { {7} over {8} } + sqrt { { {"25"} over {"36"} } } div left (2 { {1} over {2} } - 1 { {1} over {3} } right )} {}

  1. Begin by operating inside the parentheses.

    2 1 2 1 1 3 = 2 2 + 1 2 1 3 + 1 3 = 5 2 4 3 = 15 6 8 6 = 15 8 6 = 7 6

  2. Now simplify the square root.

    25 36 = 5 6 since 5 6 2 = 25 36 size 12{ sqrt { { {"25"} over {"36"} } } = { {5} over {6} } left ("since " left ( { {5} over {6} } right ) rSup { size 8{2} } = { {"25"} over {"36"} } right )} {}

    Now we have

    2 7 8 + 5 6 ÷ 7 6 size 12{2 { {7} over {8} } + { {5} over {6} } div { {7} over {6} } } {}

  3. Perform the division.

    2 7 8 + 5 6 1 6 1 7 = 2 7 8 + 5 1 1 7 = 2 7 8 + 5 7 size 12{2 { {7} over {8} } + { {5} over { {6} cSub { size 8{1} } } } cdot { { {6} cSup { size 8{1} } } over {7} } =2 { {7} over {8} } + { {5 cdot 1} over {1 cdot 7} } =2 { {7} over {8} } + { {5} over {7} } } {}

  4. Now perform the addition.

    2 7 8 + 5 7 = 2 8 + 7 8 + 5 7 = 23 8 + 5 7 LCD = 56 . = 23 7 56 + 5 8 56 = 161 56 + 40 56 = 161 + 40 56 = 201 56  or  3 33 56

    Thus, 2 7 8 + 25 36 ÷ 2 1 2 1 1 3 = 3 33 56 size 12{2 { {7} over {8} } + sqrt { { {"25"} over {"36"} } } div left (2 { {1} over {2} } - 1 { {1} over {3} } right )=3 { {"33"} over {"56"} } } {}

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Practice set a

Find the value of each of the following quantities.

5 16 1 10 1 32 size 12{ { {5} over {"16"} } cdot { {1} over {"10"} } - { {1} over {"32"} } } {}

0

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6 7 21 40 ÷ 9 10 + 5 1 3 size 12{ { {6} over {7} } cdot { {"21"} over {"40"} } div { {9} over {"10"} } +5 { {1} over {3} } } {}

35 6 size 12{ { {"35"} over {6} } } {} or 5 5 6 size 12{5 { {5} over {"6"} } } {}

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8 7 10 2 4 1 2 3 2 3 size 12{8 { {7} over {"10"} } - 2 left (4 { {1} over {2} } - 3 { {2} over {3} } right )} {}

211 30 size 12{ { {"211"} over {"30"} } } {} or 7 1 30 size 12{7 { {1} over {"30"} } } {}

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17 18 58 30 1 4 3 32 1 13 29 size 12{ { {"17"} over {"18"} } - { {"58"} over {"30"} } left ( { {1} over {4} } - { {3} over {"32"} } right ) left (1 - { {"13"} over {"29"} } right )} {}

7 9 size 12{ { {7} over {9} } } {}

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1 10 + 1 1 2 ÷ 1 4 5 1 6 25 size 12{ left ( { {1} over {"10"} } +1 { {1} over {2} } right ) div left (1 { {4} over {5} } - 1 { {6} over {"25"} } right )} {}

2 6 7 size 12{2 { {6} over {7} } } {}

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2 3 3 8 4 9 7 16 1 1 3 + 1 1 4 size 12{ { { { {2} over {3} } - { {3} over {8} } cdot { {4} over {9} } } over { { {7} over {"16"} } cdot 1 { {1} over {3} } +1 { {1} over {4} } } } } {}

3 11 size 12{ { {3} over {"11"} } } {}

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3 8 2 + 3 4 1 8 size 12{ left ( { {3} over {8} } right ) rSup { size 8{2} } + { {3} over {4} } cdot { {1} over {8} } } {}

15 64 size 12{ { {"15"} over {"64"} } } {}

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2 3 2 1 4 4 25 size 12{ { {2} over {3} } cdot 2 { {1} over {4} } - sqrt { { {4} over {"25"} } } } {}

11 10 size 12{ { {"11"} over {"10"} } } {}

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Exercises

Find each value.

4 3 1 6 1 2 size 12{ { {4} over {3} } - { {1} over {6} } cdot { {1} over {2} } } {}

5 4 size 12{ { {5} over {4} } } {}

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7 9 4 5 5 36 size 12{ { {7} over {9} } - { {4} over {5} } cdot { {5} over {"36"} } } {}

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2 2 7 + 5 8 ÷ 5 16 size 12{2 { {2} over {7} } + { {5} over {8} } div { {5} over {"16"} } } {}

4 2 7 size 12{4 { {2} over {7} } } {}

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3 16 ÷ 9 14 12 21 + 5 6 size 12{ { {3} over {"16"} } div { {9} over {"14"} } cdot { {"12"} over {"21"} } + { {5} over {6} } } {}

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4 25 ÷ 8 15 7 20 ÷ 2 1 10 size 12{ { {4} over {"25"} } div { {8} over {"15"} } - { {7} over {"20"} } div 2 { {1} over {"10"} } } {}

2 15 size 12{ { {2} over {"15"} } } {}

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2 5 1 19 + 3 38 size 12{ { {2} over {5} } cdot left ( { {1} over {"19"} } + { {3} over {"38"} } right )} {}

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3 7 3 10 1 15 size 12{ { {3} over {7} } cdot left ( { {3} over {"10"} } - { {1} over {"15"} } right )} {}

1 10 size 12{ { {1} over {"10"} } } {}

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10 11 8 9 2 5 + 3 25 5 3 + 1 4 size 12{ { {"10"} over {"11"} } cdot left ( { {8} over {9} } - { {2} over {5} } right )+ { {3} over {"25"} } cdot left ( { {5} over {3} } + { {1} over {4} } right )} {}

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2 7 6 7 3 28 + 5 1 3 1 1 4 1 8 size 12{ { {2} over {7} } cdot left ( { {6} over {7} } - { {3} over {"28"} } right )+5 { {1} over {3} } cdot left (1 { {1} over {4} } - { {1} over {8} } right )} {}

6 3 14 size 12{6 { {3} over {"14"} } } {}

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6 11 1 3 1 21 + 2 13 42 1 1 5 + 7 40 size 12{ { { left ( { {6} over {"11"} } - { {1} over {3} } right ) cdot left ( { {1} over {"21"} } +2 { {"13"} over {"42"} } right )} over {1 { {1} over {5} } + { {7} over {"40"} } } } } {}

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1 2 2 + 1 8 size 12{ left ( { {1} over {2} } right ) rSup { size 8{2} } + { {1} over {8} } } {}

3 8 size 12{ { {3} over {8} } } {}

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3 5 2 3 10 size 12{ left ( { {3} over {5} } right ) rSup { size 8{2} } - { {3} over {"10"} } } {}

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36 81 + 1 3 2 9 size 12{ sqrt { { {"36"} over {"81"} } } + { {1} over {3} } cdot { {2} over {9} } } {}

20 27 size 12{ { {"20"} over {"27"} } } {}

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49 64 9 4 size 12{ sqrt { { {"49"} over {"64"} } } - sqrt { { {9} over {4} } } } {}

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2 3 9 4 15 4 16 225 size 12{ { {2} over {3} } cdot sqrt { { {9} over {4} } } - { {"15"} over {4} } cdot sqrt { { {"16"} over {"225"} } } } {}

0

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3 4 2 + 25 16 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } + sqrt { { {"25"} over {"16"} } } } {}

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1 3 2 81 25 + 1 40 ÷ 1 8 size 12{ left ( { {1} over {3} } right ) rSup { size 8{2} } cdot sqrt { { {"81"} over {"25"} } } + { {1} over {"40"} } div { {1} over {8} } } {}

2 5 size 12{ { {2} over {5} } } {}

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4 49 2 + 3 7 ÷ 1 3 4 size 12{ left ( sqrt { { {4} over {"49"} } } right ) rSup { size 8{2} } + { {3} over {7} } div 1 { {3} over {4} } } {}

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100 121 2 + 21 11 2 size 12{ left ( sqrt { { {"100"} over {"121"} } } right ) rSup { size 8{2} } + { {"21"} over { left ("11" right ) rSup { size 8{2} } } } } {}

1

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3 8 + 1 64 1 2 ÷ 1 1 3 size 12{ sqrt { { {3} over {8} } + { {1} over {"64"} } } - { {1} over {2} } div 1 { {1} over {3} } } {}

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1 4 5 6 2 + 9 14 2 1 3 1 81 size 12{ sqrt { { {1} over {4} } } cdot left ( { {5} over {6} } right ) rSup { size 8{2} } + { {9} over {"14"} } cdot 2 { {1} over {3} } - sqrt { { {1} over {"81"} } } } {}

125 72 size 12{ { {"125"} over {"72"} } } {}

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1 9 6 3 8 + 2 5 8 16 + 7 7 10 size 12{ sqrt { { {1} over {9} } } cdot sqrt { { {6 { {3} over {8} } +2 { {5} over {8} } } over {"16"} } } +7 { {7} over {"10"} } } {}

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3 3 4 + 4 5 1 2 3 67 240 + 1 3 4 9 10 size 12{ { {3 { {3} over {4} } + { {4} over {5} } cdot left ( { {1} over {2} } right ) rSup { size 8{3} } } over { { {"67"} over {"240"} } + left ( { {1} over {3} } right ) rSup { size 8{4} } cdot left ( { {9} over {"10"} } right )} } } {}

252 19 size 12{ { {"252"} over {"19"} } } {}

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16 81 + 1 4 6 size 12{ sqrt { sqrt { { {"16"} over {"81"} } } } + { {1} over {4} } cdot 6} {}

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81 256 3 32 1 1 8 size 12{ sqrt { sqrt { { {"81"} over {"256"} } } } - { {3} over {"32"} } cdot 1 { {1} over {8} } } {}

165 256 size 12{ { {"165"} over {"256"} } } {}

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Exercises for review

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

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

multiplicative identity

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( [link] ) Convert 8 6 7 size 12{8 { {6} over {7} } } {} to an improper fraction.

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( [link] ) Find the sum. 3 8 + 4 5 + 5 6 size 12{ { {3} over {8} } + { {4} over {5} } + { {5} over {6} } } {} .

241 120 size 12{ { {"241"} over {"120"} } } {} or 2 1 120 size 12{2 { {1} over {"120"} } } {}

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( [link] ) Simplify 6 + 1 8 6 1 8 size 12{ { {6+ { {1} over {8} } } over {6 - { {1} over {8} } } } } {} .

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Questions & Answers

explain and give four Example hyperbolic function
Lukman Reply
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
how do I set up the problem?
Harshika Reply
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Shirley Reply
please can go further on polynomials quadratic
Abdullahi
hi mam
Mark
I need quadratic equation link to Alpa Beta
Abdullahi Reply
find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
Seidu Reply
please help me prove quadratic formula
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
Shirley Reply
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
In the number 779,844,205 how many ten millions are there?
TELLY Reply
From 1973 to 1979, in the United States, there was an increase of 166.6% of Ph.D. social scien­tists to 52,000. How many were there in 1973?
Khizar Reply
7hours 36 min - 4hours 50 min
Tanis Reply

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Source:  OpenStax, Fundamentals of mathematics. OpenStax CNX. Aug 18, 2010 Download for free at http://cnx.org/content/col10615/1.4
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