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Addition and Subtraction of Fractions, Comparing Fractions, and Complex Fractions: Addition and Subtraction of Fractions with Unlike Denominators

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to add and subtract fractions with unlike denominators. By the end of the module students should be able to add and subtract fractions with unlike denominators.

Section Overview

  • A Basic Rule
  • Addition and Subtraction of Fractions

A Basic Rule

There is a basic rule that must be followed when adding or subtracting fractions.

A Basic Rule

Fractions can only be added or subtracted conveniently if they have like denomi­nators.

To see why this rule makes sense, let's consider the problem of adding a quarter and a dime.

1 quarter + 1 dime = 35 cents 1 quarter+1 dime=35 cents

Now,

1   quarter = 25 100 1   dime = 10 100 same denominations 35 ¢ = 35 100 1   quarter = 25 100 1   dime = 10 100 same denominations 35 ¢ = 35 100

25 100 + 10 100 = 25 + 10 100 = 35 100 25 100 + 10 100 = 25 + 10 100 = 35 100 size 12{ { {"25"} over {"100"} } + { {"10"} over {"100"} } = { {"25"+"10"} over {"100"} } = { {"35"} over {"100"} } } {}

In order to combine a quarter and a dime to produce 35¢, we convert them to quantities of the same denomination.

Same denomination → same denominator

Addition and Subtraction of Fractions

Least Common Multiple (LCM) and Least Common Denominator (LCD)

In (Reference), we examined the least common multiple (LCM) of a collection of numbers. If these numbers are used as denominators of fractions, we call the least common multiple, the least common denominator (LCD).

Method of Adding or Subtracting Fractions with Unlike Denominators

To add or subtract fractions having unlike denominators, convert each fraction to an equivalent fraction having as a denominator the least common denomina­tor ( LCD) of the original denominators.

Sample Set A

Find the following sums and differences.

Example 1

16+3416+34 size 12{ { {1} over {6} } + { {3} over {4} } } {}. The denominators are not the same. Find the LCD of 6 and 4.

6 = 2 ⋅ 3 4 = 2 2 The LCD = 2 2 ⋅ 3 = 4 ⋅ 3 = 12 6 = 2 ⋅ 3 4 = 2 2 The LCD = 2 2 ⋅ 3 = 4 ⋅ 3 = 12

Write each of the original fractions as a new, equivalent fraction having the common denomina­tor 12.

1 6 + 3 4 = 12 + 12 1 6 + 3 4 = 12 + 12 size 12{ { {1} over {6} } + { {3} over {4} } = { {} over {"12"} } + { {} over {"12"} } } {}

To find a new numerator, we divide the original denominator into the LCD. Since the original denominator is being multiplied by this quotient, we must multiply the original numerator by this quotient.

12 ÷ 6 = 2 12 ÷ 6 = 2 size 12{"12" div 6=2} {} Multiply 1 by 2: 1 times 2 equals 2. The 1 is the original numerator, and the 2, the product of the multiplication is the new numerator.


12 ÷ 4 = 3 12 ÷ 4 = 3 size 12{"12" div 4=3} {} Multiply 3 by 3: 3 times 3 equals 9. 3 is the original numerator. 9 is the new numerator.


1 6 + 3 4 = 1 ⋅ 2 12 + 3 ⋅ 3 12 = 2 12 + 9 12 Now the denominators are the same. = 2 + 9 12 Add the numerators and place the sum over the common denominator. = 11 12 1 6 + 3 4 = 1 ⋅ 2 12 + 3 ⋅ 3 12 = 2 12 + 9 12 Now the denominators are the same. = 2 + 9 12 Add the numerators and place the sum over the common denominator. = 11 12

Example 2

12+2312+23 size 12{ { {1} over {2} } + { {2} over {3} } } {}. The denominators are not the same. Find the LCD of 2 and 3.

LCD = 2 ⋅ 3 = 6 LCD = 2 ⋅ 3 = 6 size 12{ ital "LCD"=2 cdot 3=6} {}

Write each of the original fractions as a new, equivalent fraction having the common denominator 6.

1 2 + 2 3 = 6 + 6 1 2 + 2 3 = 6 + 6 size 12{ { {1} over {2} } + { {2} over {3} } = { {} over {6} } + { {} over {6} } } {}

To find a new numerator, we divide the original denominator into the LCD. Since the original denominator is being multiplied by this quotient, we must multiply the original numerator by this quotient.

6 ÷ 2 = 36 ÷ 2 = 3 size 12{"6 " div " 2 "=" 3"} {} Multiply the numerator 1 by 3.

6 ÷ 2 = 36 ÷ 2 = 3 size 12{"6 " div " 2 "=" 3"} {} Multiply the numerator 2 by 2.

1 2 + 2 3 = 1 ⋅ 3 6 + 2 ⋅ 3 6 = 3 6 + 4 6 = 3 + 4 6 = 7 6  or  1 1 6 1 2 + 2 3 = 1 ⋅ 3 6 + 2 ⋅ 3 6 = 3 6 + 4 6 = 3 + 4 6 = 7 6  or  1 1 6

Example 3

59−51259−512 size 12{ { {5} over {9} } - { {5} over {"12"} } } {}. The denominators are not the same. Find the LCD of 9 and 12.

9=3⋅3=3212=2⋅6=2⋅2⋅3=22⋅3 LCD = 22⋅32=4⋅9=369=3⋅3=3212=2⋅6=2⋅2⋅3=22⋅3 LCD = 22⋅32=4⋅9=36

5 9 − 5 12 = 36 − 36 5 9 − 5 12 = 36 − 36 size 12{ { {5} over {9} } - { {5} over {"12"} } = { {} over {"36"} } - { {} over {"36"} } } {}

36÷9=436÷9=4 size 12{"36"¸9=4} {} Multiply the numerator 5 by 4.

36÷12=336÷12=3 size 12{"36"¸"12"=3} {} Multiply the numerator 5 by 3.

5 9 − 5 12 = 5 ⋅ 4 36 − 5 ⋅ 3 36 = 20 36 − 15 36 = 20 − 15 36 = 5 36 5 9 − 5 12 = 5 ⋅ 4 36 − 5 ⋅ 3 36 = 20 36 − 15 36 = 20 − 15 36 = 5 36

Example 4

56−18+71656−18+716 size 12{ { {5} over {6} } - { {1} over {8} } + { {7} over {"16"} } } {} The denominators are not the same. Find the LCD of 6, 8, and 16

6 = 2⋅3 8 = 2⋅4=2⋅2⋅2=23 16 = 2⋅8=2⋅2⋅4=2⋅2⋅2⋅2=24  The LCD is  24⋅3=48 6 = 2⋅3 8 = 2⋅4=2⋅2⋅2=23 16 = 2⋅8=2⋅2⋅4=2⋅2⋅2⋅2=24  The LCD is 24⋅3=48

5 6 − 1 8 + 7 16 = 48 − 48 + 48 5 6 − 1 8 + 7 16 = 48 − 48 + 48 size 12{ { {5} over {6} } - { {1} over {8} } + { {7} over {"16"} } = { {} over {"48"} } - { {} over {"48"} } + { {} over {"48"} } } {}

48÷6=848÷6=8 size 12{"48"¸6=8} {} Multiply the numerator 5 by 8

48÷8=648÷8=6 size 12{"48"¸8=6} {} Multiply the numerator 1 by 6

48÷16=348÷16=3 size 12{"48"¸"16"=3} {} Multiply the numerator 7 by 3

5 6 − 1 8 + 7 16 = 5 ⋅ 8 48 − 1 ⋅ 6 48 + 7 ⋅ 3 48 = 40 48 − 6 48 + 21 48 = 40 − 6 + 21 48 = 55 48  or  1 7 48 5 6 − 1 8 + 7 16 = 5 ⋅ 8 48 − 1 ⋅ 6 48 + 7 ⋅ 3 48 = 40 48 − 6 48 + 21 48 = 40 − 6 + 21 48 = 55 48  or  1 7 48

Practice Set A

Find the following sums and differences.

Exercise 1

34+11234+112 size 12{ { {3} over {4} } + { {1} over {"12"} } } {}

Solution

5656 size 12{ { {5} over {6} } } {}

Exercise 2

12−3712−37 size 12{ { {1} over {2} } - { {3} over {7} } } {}

Solution

114114 size 12{ { {1} over {"14"} } } {}

Exercise 3

710-58710-58 size 12{ { {7} over {"10"} } - { {5} over {8} } } {}

Solution

340340 size 12{ { {3} over {"40"} } } {}

Exercise 4

1516+12−341516+12−34 size 12{ { {"15"} over {"16"} } + { {1} over {2} } - { {3} over {4} } } {}

Solution

11161116 size 12{ { {"11"} over {"16"} } } {}

Exercise 5

132−148132−148 size 12{ { {1} over {"32"} } - { {1} over {"48"} } } {}

Solution

196196 size 12{ { {1} over {"96"} } } {}

Exercises

Exercise 6

A most basic rule of arithmetic states that two fractions may be added or subtracted conveniently only if they have
.

Solution

The same denominator

For the following problems, find the sums and differences.

Exercise 7

12+1612+16 size 12{ { {1} over {2} } + { {1} over {6} } } {}

Exercise 8

18+1218+12 size 12{ { {1} over {8} } + { {1} over {2} } } {}

Solution

5858 size 12{ { {5} over {8} } } {}

Exercise 9

34+1334+13 size 12{ { {3} over {4} } + { {1} over {3} } } {}

Exercise 10

58+2358+23 size 12{ { {5} over {8} } + { {2} over {3} } } {}

Solution

31243124 size 12{ { {"31"} over {"24"} } } {}

Exercise 11

112+13112+13 size 12{ { {1} over {"12"} } + { {1} over {3} } } {}

Exercise 12

67−1467−14 size 12{ { {6} over {7} } - { {1} over {4} } } {}

Solution

17281728 size 12{ { {"17"} over {"28"} } } {}

Exercise 13

910−25910−25 size 12{ { {9} over {"10"} } - { {2} over {5} } } {}

Exercise 14

79−1479−14 size 12{ { {7} over {9} } - { {1} over {4} } } {}

Solution

19361936 size 12{ { {"19"} over {"36"} } } {}

Exercise 15

815−310815−310 size 12{ { {8} over {"15"} } - { {3} over {"10"} } } {}

Exercise 16

813−539813−539 size 12{ { {8} over {"13"} } - { {5} over {"39"} } } {}

Solution

19391939 size 12{ { {"19"} over {"39"} } } {}

Exercise 17

1112−251112−25 size 12{ { {"11"} over {"12"} } - { {2} over {5} } } {}

Exercise 18

115+512115+512 size 12{ { {1} over {"15"} } + { {5} over {"12"} } } {}

Solution

29602960 size 12{ { {"29"} over {"60"} } } {}

Exercise 19

1388−141388−14 size 12{ { {"13"} over {"88"} } - { {1} over {4} } } {}

Exercise 20

19−18119−181 size 12{ { {1} over {9} } - { {1} over {"81"} } } {}

Solution

881881 size 12{ { {8} over {"81"} } } {}

Exercise 21

1940+5121940+512 size 12{ { {"19"} over {"40"} } + { {5} over {"12"} } } {}

Exercise 22

2526−7102526−710 size 12{ { {"25"} over {"26"} } - { {7} over {"10"} } } {}

Solution

17651765 size 12{ { {"17"} over {"65"} } } {}

Exercise 23

928−445928−445 size 12{ { {9} over {"28"} } - { {4} over {"45"} } } {}

Exercise 24

2245−16352245−1635 size 12{ { {"22"} over {"45"} } - { {"16"} over {"35"} } } {}

Solution

263263 size 12{ { {2} over {"63"} } } {}

Exercise 25

5663+22335663+2233 size 12{ { {"56"} over {"63"} } + { {"22"} over {"33"} } } {}

Exercise 26

116+34−38116+34−38 size 12{ { {1} over {"16"} } + { {3} over {4} } - { {3} over {8} } } {}

Solution

716716 size 12{ { {7} over {"16"} } } {}

Exercise 27

512−1120+1920512−1120+1920 size 12{ { {5} over {"12"} } - { {1} over {"120"} } + { {"19"} over {"20"} } } {}

Exercise 28

83−14+73683−14+736 size 12{ { {8} over {3} } - { {1} over {4} } + { {7} over {"36"} } } {}

Solution

47184718 size 12{ { {"47"} over {"18"} } } {}

Exercise 29

119−17+1663119−17+1663 size 12{ { {"11"} over {9} } - { {1} over {7} } + { {"16"} over {"63"} } } {}

Exercise 30

125−23+1710125−23+1710 size 12{ { {"12"} over {5} } - { {2} over {3} } + { {"17"} over {"10"} } } {}

Solution

1033010330 size 12{ { {"103"} over {"30"} } } {}

Exercise 31

49+1321−91449+1321−914 size 12{ { {4} over {9} } + { {"13"} over {"21"} } - { {9} over {"14"} } } {}

Exercise 32

34−322+52434−322+524 size 12{ { {3} over {4} } - { {3} over {"22"} } + { {5} over {"24"} } } {}

Solution

217264217264 size 12{ { {"217"} over {"264"} } } {}

Exercise 33

2548−788+5242548−788+524 size 12{ { {"25"} over {"48"} } - { {7} over {"88"} } + { {5} over {"24"} } } {}

Exercise 34

2740+4748−1191262740+4748−119126 size 12{ { {"27"} over {"40"} } + { {"47"} over {"48"} } - { {"119"} over {"126"} } } {}

Solution

511720511720 size 12{ { {"511"} over {"720"} } } {}

Exercise 35

4144−599−111754144−599−11175 size 12{ { {"41"} over {"44"} } - { {5} over {"99"} } - { {"11"} over {"175"} } } {}

Exercise 36

512+118+124512+118+124 size 12{ { {5} over {"12"} } + { {1} over {"18"} } + { {1} over {"24"} } } {}

Solution

37723772 size 12{ { {"37"} over {"72"} } } {}

Exercise 37

59+16+71559+16+715 size 12{ { {5} over {9} } + { {1} over {6} } + { {7} over {"15"} } } {}

Exercise 38

2125+16+7152125+16+715 size 12{ { {"21"} over {"25"} } + { {1} over {6} } + { {7} over {"15"} } } {}

Solution

221150221150 size 12{ { {"221"} over {"150"} } } {}

Exercise 39

518−136+79518−136+79 size 12{ { {5} over {"18"} } - { {1} over {"36"} } + { {7} over {9} } } {}

Exercise 40

1114−136−1321114−136−132 size 12{ { {"11"} over {"14"} } - { {1} over {"36"} } - { {1} over {"32"} } } {}

Solution

1,4652,0161,4652,016 size 12{ { {1,"465"} over {2,"016"} } } {}

Exercise 41

2133+1222+15552133+1222+1555 size 12{ { {"21"} over {"33"} } + { {"12"} over {"22"} } + { {"15"} over {"55"} } } {}

Exercise 42

551+234+1168551+234+1168 size 12{ { {5} over {"51"} } + { {2} over {"34"} } + { {"11"} over {"68"} } } {}

Solution

6520465204 size 12{ { {"65"} over {"204"} } } {}

Exercise 43

87−1614+192187−1614+1921 size 12{ { {8} over {7} } - { {"16"} over {"14"} } + { {"19"} over {"21"} } } {}

Exercise 44

715+310−3460715+310−3460 size 12{ { {7} over {"15"} } + { {3} over {"10"} } - { {"34"} over {"60"} } } {}

Solution

1515 size 12{ { {1} over {5} } } {}

Exercise 45

1415−310−625+7201415−310−625+720 size 12{ { {"14"} over {"15"} } - { {3} over {"10"} } - { {6} over {"25"} } + { {7} over {"20"} } } {}

Exercise 46

116−512+1730+2518116−512+1730+2518 size 12{ { {"11"} over {6} } - { {5} over {"12"} } + { {"17"} over {"30"} } + { {"25"} over {"18"} } } {}

Solution

607180607180 size 12{ { {"607"} over {"180"} } } {}

Exercise 47

19+2221−518−14519+2221−518−145 size 12{ { {1} over {9} } + { {"22"} over {"21"} } - { {5} over {"18"} } - { {1} over {"45"} } } {}

Exercise 48

726+2865−51104+0726+2865−51104+0 size 12{ { {7} over {"26"} } + { {"28"} over {"65"} } - { {"51"} over {"104"} } +0} {}

Solution

109520109520 size 12{ { {"109"} over {"520"} } } {}

Exercise 49

A morning trip from San Francisco to Los Angeles took 13121312 size 12{ { {"13"} over {"12"} } } {} hours. The return trip took 57605760 size 12{ { {"57"} over {"60"} } } {} hours. How much longer did the morning trip take?

Exercise 50

At the beginning of the week, Starlight Publishing Company's stock was selling for 11581158 size 12{ { {"115"} over {8} } } {} dollars per share. At the end of the week, analysts had noted that the stock had gone up 114114 size 12{ { {"11"} over {4} } } {} dollars per share. What was the price of the stock, per share, at the end of the week?

Solution

$ 137 8 or $ 17 1 8 $ 137 8 or $ 17 1 8 size 12{$ { {"137"} over {8} } " or "$"17" { {1} over {8} } } {}

Exercise 51

A recipe for fruit punch calls for 233233 size 12{ { {"23"} over {3} } } {} cups of pineapple juice, 1414 size 12{ { {1} over {4} } } {} cup of lemon juice, 152152 size 12{ { {"15"} over {2} } } {} cups of orange juice, 2 cups of sugar, 6 cups of water, and 8 cups of carbonated non-cola soft drink. How many cups of ingredients will be in the final mixture?

Exercise 52

The side of a particular type of box measures 834834 size 12{8 { {3} over {4} } } {} inches in length. Is it possible to place three such boxes next to each other on a shelf that is 26152615 size 12{"26" { {1} over {5} } } {} inches in length? Why or why not?

Solution

No; 3 boxes add up to 26 1′′ 4 26 1′′ 4 , which is larger than 25 1′′ 5 25 1′′ 5 .

Exercise 53

Four resistors, 3838 size 12{ { {3} over {8} } } {} ohm, 1414 size 12{ { {1} over {4} } } {} ohm, 3535 size 12{ { {3} over {5} } } {} ohm, and 7878 size 12{ { {7} over {8} } } {} ohm, are connected in series in an electrical circuit. What is the total resistance in the circuit due to these resistors? ("In series" implies addition.)

Exercise 54

A copper pipe has an inside diameter of 23162316 size 12{2 { {3} over {"16"} } } {} inches and an outside diameter of 25342534 size 12{2 { {5} over {"34"} } } {} inches. How thick is the pipe?

Solution

No pipe at all; inside diameter is greater than outside diameter

Exercise 55

The probability of an event was originally thought to be 15321532 size 12{ { {"15"} over {"32"} } } {}. Additional information decreased the probability by 314314 size 12{ { {3} over {"14"} } } {}. What is the updated probability?

Exercises for Review

Exercise 56

((Reference)) Find the difference between 867 and 418.

Solution

449

Exercise 57

((Reference)) Is 81,147 divisible by 3?

Exercise 58

((Reference)) Find the LCM of 11, 15, and 20.

Solution

660

Exercise 59

((Reference)) Find 3434 size 12{ { {3} over {4} } } {} of 429429 size 12{4 { {2} over {9} } } {}.

Exercise 60

((Reference)) Find the value of 815−315+215815−315+215 size 12{ { {8} over {"15"} } - { {3} over {"15"} } + { {2} over {"15"} } } {}.

Solution

715715 size 12{ { {7} over {"15"} } } {}

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