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Order of Operations for Whole Numbers

Module by: Community College Online Textbook Project. E-mail the author

Summary: In a computation in which more than one operation is involved, grouping symbols indicate which operation to perform first.

Grouping Symbols

Grouping symbols are used to indicate that a particular collection of numbers and meaningful operations are to be grouped together and considered as one number. The grouping symbols commonly used in mathematics are the following:

Parentheses:  (   )

Brackets:     [   ]

Braces:       {   }

Bar:           —

In a computation in which more than one operation is involved, grouping symbols indicate which operation to perform first. If possible, we perform operations inside grouping symbols first.

For example:

(5 · 5) + 20 = 45

whereas:

5 · (5 + 20) = 125

If there are no parentheses, you should always do multiplications and divisions first followed by additions and subtractions. You can always put your own parentheses into equations using this rule to make things easier for yourself, for example:

a × b + c ÷ d = ( a × b ) + ( c ÷ d ) 5 × 5 + 20 ÷ 4 = ( 5 × 5 ) + ( 20 ÷ 4 ) a × b + c ÷ d = ( a × b ) + ( c ÷ d ) 5 × 5 + 20 ÷ 4 = ( 5 × 5 ) + ( 20 ÷ 4 ) alignl { stack { size 12{a times b+c div d= \( a times b \) + \( c div d \) } {} # size 12{5 times 5+"20" div 4= \( 5 times 5 \) + \( "20" div 4 \) } {} } } {} (1)

Grouping Symbols Examples

If possible, determine the value of each of the following.

Example 1

9 + (3 · 8)

Since 3 and 8 are within parentheses, they are to be combined first:

= 9+249+24 size 12{9+"24"} {}

Then add the terms:

= 3333 size 12{"33"} {}

Thus, 9 + (3 · 8) = 33.

Example 2

(10 ÷ 0) · 6

Since (10 ÷ 0) is undefined, this operation is meaningless, and we attach no value to it. We write, “meaningless.”

Grouping Symbols Exercises

If possible, determine the value of each of the following.

Exercise 1

16 – (3 · 2)

Solution

10

Exercise 2

5 + (7 · 9)

Solution

61

Exercise 3

(4 + 8) · 2

Solution

24

Exercise 4

28 ÷ (18 – 11)

Solution

4

Exercise 5

(33 ÷ 3) – 11

Solution

0

Exercise 6

4 + (0 ÷ 0)

Solution

meaningless

Multiple Grouping Symbols

When a set of grouping symbols occurs inside another set of grouping symbols, we perform the operations within the innermost set first.

Multiple Grouping Symbol Examples

Determine the value of each of the following.

Example 1

2 + (8 · 3) – (5 + 6)

Combine 8 and 3 first, then combine 5 and 6.

= 2 + 24 – 11

Now combine left to right.

= 26 –11

= 15

  

Example 2

10 + [ 30 − ( 2 ⋅ 9 ) ] 10 + [ 30 − ( 2 ⋅ 9 ) ] size 12{"10"+ \[ "30" - \( 2 cdot 9 \) \] } {}

Combine 2 and 9 since they occur in the innermost set of parentheses.

= 10+[30−18]10+[30−18] size 12{"10"+ \[ "30" - "18" \] } {}

Now combine 30 and 18.

= 10 + 12

= 22

  

Distributivity

If you see a multiplication outside parentheses like this:

a ( b + c ) 3 ( 4 − 3 ) a ( b + c ) 3 ( 4 − 3 ) alignl { stack { size 12{a \( b+c \) } {} # size 12{3 \( 4 - 3 \) } {} } } {} (2)

then it means you have to multiply each part inside the parentheses by the number outside:

a ( b + c ) = ab + ac 3 ( 4 − 3 ) = 3 × 4 − 3 × 3 = 12 − 9 = 3 a ( b + c ) = ab + ac 3 ( 4 − 3 ) = 3 × 4 − 3 × 3 = 12 − 9 = 3 alignl { stack { size 12{a \( b+c \) = ital "ab"+ ital "ac"} {} # size 12{3 \( 4 - 3 \) =3 times 4 - 3 times 3="12" - 9=3} {} } } {} (3)

Sometimes you can simplify everything inside the parentheses into a single term. In fact, in the above example, it would have been smarter to have done this:

3 ( 4 − 3 ) = 3 × ( 1 ) = 3 3 ( 4 − 3 ) = 3 × ( 1 ) = 3 size 12{3 \( 4 - 3 \) =3 times \( 1 \) =3} {} (4)

This can happen with letters too:

3 ( 4a − 3a ) = 3 × ( a ) = 3a 3 ( 4a − 3a ) = 3 × ( a ) = 3a size 12{3 \( 4a - 3a \) =3 times \( a \) =3a} {} (5)

The fact that a(b+c)=ab+aca(b+c)=ab+ac size 12{a \( b+c \) = ital "ab"+ ital "ac"} {} is know as the distributive property.

If there are two sets of parentheses multiplied by each other, then you can do it one step at a time:

( a + b ) ( c + d ) = a ( c + d ) + b ( c + d ) = ac + ad + bc + bd ( a + 3 ) ( 4 + d ) = a ( 4 + d ) + 3 ( 4 + d ) = 4a + ad + 12 + 3d ( a + b ) ( c + d ) = a ( c + d ) + b ( c + d ) = ac + ad + bc + bd ( a + 3 ) ( 4 + d ) = a ( 4 + d ) + 3 ( 4 + d ) = 4a + ad + 12 + 3d alignl { stack { size 12{ \( a+b \) \( c+d \) =a \( c+d \) +b \( c+d \) } {} # size 12{ {}= ital "ac"+ ital "ad"+ ital "bc"+ ital "bd"} {} # size 12{ \( a+3 \) \( 4+d \) =a \( 4+d \) +3 \( 4+d \) } {} # size 12{ {}=4a+ ital "ad"+"12"+3d} {} } } {} (6)

Multiple Grouping Symbol Exercises

Determine the value of each of the following:

Exercise 7

( 17 + 8 ) + ( 9 + 20 ) ( 17 + 8 ) + ( 9 + 20 ) size 12{ \( "17"+8 \) + \( 9+"20" \) } {}

Solution

54

Exercise 8

( 55 − 6 ) + ( 13 ⋅ 2 ) ( 55 − 6 ) + ( 13 ⋅ 2 ) size 12{ \( "55" - 6 \) + \( "13" cdot 2 \) } {}

Solution

23

Exercise 9

23 + ( 12 ÷ 4 ) + ( 11 ⋅ 2 ) 23 + ( 12 ÷ 4 ) + ( 11 ⋅ 2 ) size 12{"23"+ \( "12" div 4 \) + \( "11" cdot 2 \) } {}

Solution

48

Exercise 10

86 + [ 14 + ( 10 − 8 ) ] 86 + [ 14 + ( 10 − 8 ) ] size 12{"86"+ \[ "14"+ \( "10" - 8 \) \] } {}

Solution

102

Exercise 11

31 + ( 9 + [ 1 + ( 35 − 2 ) ] ) 31 + ( 9 + [ 1 + ( 35 − 2 ) ] ) size 12{"31"+ \( 9+ \[ 1+ \( "35" - 2 \) \] \) } {}

Solution

74

Exercise 12

{6 – [24 ÷ (4 · 2)]}3

Solution

9

Order of Operations

Sometimes there are no grouping symbols indicating which operations to perform first. For example, suppose we wish to find the value of 3 + 5⋅ 23 + 5⋅ 2 size 12{"3 "+" 5" cdot " 2"} {}. We could do either of two things:

Add 3 and 5, then multiply this sum by 2.

  3+5⋅2=8⋅2=163+5⋅2=8⋅2=16alignl { stack { size 12{`3+5 cdot 2} {} # size 12{`=8 cdot 2} {} # size 12{`="16"} {} } } {}

Multiply 5 and 2, then add 3 to this product.

  3+5⋅2=3+10=133+5⋅2=3+10=13alignl { stack { size 12{`3+5 cdot 2} {} # size 12{`=3+"10"} {} # size 12{`="13"} {} } } {}

We now have two values for the same expression.

We need a set of rules to guide anyone to one unique value for this kind of expression. Some of these rules are based on convention, while other are forced on up by mathematical logic.

The universally agreed-upon accepted order of operations for evaluating a mathematical expression is as follows:

1. Parentheses (grouping symbols) from the inside out.

By parentheses we mean anything that acts as a grouping symbol, including anything inside symbols such as [  ], {  }, |  |, and size 12{` sqrt {`} } {}. Any expression in the numerator or denominator of a fraction or in an exponent is also considered grouped, and should be simplified before carrying out further operations.

If there are nested parentheses (parentheses inside parentheses), you work from the innermost parentheses outward.

2. Exponents and other special functions, such as log, sin, cos etc.

3. Multiplications and divisions, from left to right.

4. Additions and subtractions, from left to right.

For example, given: 3 + 15 ÷ 3 + 5 × 22+3

The exponent is an implied grouping, so the 2+3 must be evaluated first:

 = 3 + 15 ÷ 3 + 5 × 25

Now the exponent is carried out:

 = 3 +15 ÷ 3 + 5 × 32

Then the multiplication and division, left to right using 15 ÷ 3 = 5 and 5 × 32 = 160:

 = 3 + 5 + 160

Finally, the addition, left to right:

 = 168

Examples, Order of Operation

Determine the value of each of the following.

Example 3

21+3⋅1221+3⋅12 size 12{"21"+3 cdot "12"} {}.

Multiply first:

= 21+3621+36 size 12{"21"+"36"} {}

Add.

= 57

  

Example 4

(15−8)+5(6+4)(15−8)+5(6+4) size 12{ \( "15" - 8 \) +5 \( 6+4 \) } {}.

Simplify inside parentheses first.

= 7+5⋅107+5⋅10 size 12{7+5 cdot "10"} {}

Multiply.

= 7+507+50 size 12{7+"50"} {}

Add.

= 57

  

Example 5

63−(4+6⋅3)+76−463−(4+6⋅3)+76−4 size 12{"63" - \( 4+6 cdot 3 \) +"76" - 4} {}.

Simplify first within the parentheses by multiplying, then adding:

= 63−(4+18)+76−463−(4+18)+76−4 size 12{"63" - \( 4+"18" \) +"76" - 4} {}

= 63−22+76−463−22+76−4 size 12{"63" - "22"+"76" - 4} {}

Now perform the additions and subtractions, moving left to right:

= 41+76−441+76−4 size 12{"41"+"76" - 4} {}

= 117−4117−4 size 12{"117" - 4} {}

= 113.

  

Example 6

7 ⋅ 6 − 4 2 + 1 5 7 ⋅ 6 − 4 2 + 1 5 size 12{7 cdot 6 - 4 rSup { size 8{2} } +1 rSup { size 8{5} } } {}

Evaluate the exponential forms, moving from left to right:

= 7⋅6−16+17⋅6−16+1 size 12{7 cdot 6 - "16"+1} {}

Multiply 7 · 6:

= 42−16+142−16+1 size 12{"42" - "16"+1} {}

Subtract 16 from 42:

= 26 + 1

Add 26 and 1:

= 27.

  

Example 7

6 ⋅ ( 3 2 + 2 2 ) + 4 2 6 ⋅ ( 3 2 + 2 2 ) + 4 2 size 12{6 cdot \( 3 rSup { size 8{2} } +2 rSup { size 8{2} } \) +4 rSup { size 8{2} } } {}

Evaluate the exponential forms in the parentheses:

= 6⋅(9+4)+426⋅(9+4)+42 size 12{6 cdot \( 9+4 \) +4 rSup { size 8{2} } } {}

Add 9 and 4 in the parentheses:

= 6⋅(13)+426⋅(13)+42 size 12{6 cdot \( "13" \) +4 rSup { size 8{2} } } {}

Evaluate the exponential form 4242 size 12{4 rSup { size 8{2} } } {}:

= 6⋅(13)+166⋅(13)+16 size 12{6 cdot \( "13" \) +"16"} {}

Multiply 6 and 13:

= 78+1678+16 size 12{"78"+"16"} {}

Add 78 and 16:

= 94

  

Example 8

62+2242+6⋅22+13+82102−19⋅562+2242+6⋅22+13+82102−19⋅5 size 12{ { {6 rSup { size 8{2} } +2 rSup { size 8{2} } } over {4 rSup { size 8{2} } +6 cdot 2 rSup { size 8{2} } } } + { {1 rSup { size 8{3} } +8 rSup { size 8{2} } } over {"10" rSup { size 8{2} } - "19" cdot 5} } } {}.

= 36+416+6⋅4+1+64100−19⋅536+416+6⋅4+1+64100−19⋅5 size 12{ { {"36"+4} over {"16"+6 cdot 4} } + { {1+"64"} over {"100" - "19" cdot 5} } } {}

= 36+416+24+1+64100−9536+416+24+1+64100−95 size 12{ { {"36"+4} over {"16"+"24"} } + { {1+"64"} over {"100" - "95"} } } {}

= 4040+6554040+655 size 12{ { {"40"} over {"40"} } + { {"65"} over {5} } } {}

= 1+13

= 14

Recall that the bar is a grouping symbol. The fraction 62+2242+6⋅2262+2242+6⋅22 size 12{ { {6 rSup { size 8{2} } +2 rSup { size 8{2} } } over {4 rSup { size 8{2} } +6 cdot 2 rSup { size 8{2} } } } } {} is equivalent to 62+22÷42+6⋅2262+22÷42+6⋅22 size 12{ left (6 rSup { size 8{2} } +2 rSup { size 8{2} } right ) div left (4 rSup { size 8{2} } +6 cdot 2 rSup { size 8{2} } right )} {}

Exercises, Order of Operations

Determine the value of the following:

Exercise 13

8 + (32 – 7)

Solution

66

Exercise 14

(34 + 18 – 2 · 3) + 11

Solution

57

Exercise 15

8(10) + 4(2 + 3) – (20 + 3 · 15 + 40 – 5)

Solution

0

Exercise 16

5 · 8 + 42 – 22

Solution

52

Exercise 17

4(62 – 33) ÷ (42 – 4)

Solution

9

Exercise 18

(8 + 9 · 3) ÷ 7 + 5 · (8 ÷ 4 + 7 + 3 · 5)

Solution

125

Exercise 19

3 3 + 2 3 6 2 − 29 + 5 8 2 + 2 4 7 2 − 3 2 ÷ 8 ⋅ 3 + 1 8 2 3 − 3 3 3 + 2 3 6 2 − 29 + 5 8 2 + 2 4 7 2 − 3 2 ÷ 8 ⋅ 3 + 1 8 2 3 − 3 size 12{ { {3 rSup { size 8{3} } +2 rSup { size 8{3} } } over {6 rSup { size 8{2} } - "29"} } +5 left ( { {8 rSup { size 8{2} } +2 rSup { size 8{4} } } over {7 rSup { size 8{2} } - 3 rSup { size 8{2} } } } right ) div { {8 cdot 3+1 rSup { size 8{8} } } over {2 rSup { size 8{3} } - 3} } } {}

Solution

7

Module Review Exercises

For the following problems, find each value.

Exercise 20

2 + 3 ⋅ ( 8 ) 2 + 3 ⋅ ( 8 ) size 12{2+3 cdot \( 8 \) } {}

Solution

48

Exercise 21

1 − 5 ( 8 − 8 ) 1 − 5 ( 8 − 8 ) size 12{1 - 5 \( 8 - 8 \) } {}

Solution

meaningless

Exercise 22

37 − 1 ⋅ 6 2 37 − 1 ⋅ 6 2 size 12{"37" - 1 cdot 6 rSup { size 8{2} } } {}

Solution

1

Exercise 23

98 ÷ 2 ÷ 7 2 98 ÷ 2 ÷ 7 2 size 12{"98" div 2 div 7 rSup { size 8{2} } } {}

Solution

1

Exercise 24

( 4 2 − 2 ⋅ 4 ) − 2 3 ( 4 2 − 2 ⋅ 4 ) − 2 3 size 12{ \( 4 rSup { size 8{2} } - 2 cdot 4 \) - 2 rSup { size 8{3} } } {}

Solution

0

Exercise 25

61 − 22 + 4 [ 3 ⋅ ( 10 ) + 11 ] 61 − 22 + 4 [ 3 ⋅ ( 10 ) + 11 ] size 12{"61" - "22"+4 \[ 3 cdot \( "10" \) +"11" \] } {}

Solution

203

Exercise 26

121 − 4 ⋅ [ ( 4 ) ⋅ ( 5 ) − 12 ] + 16 2 121 − 4 ⋅ [ ( 4 ) ⋅ ( 5 ) − 12 ] + 16 2 size 12{"121" - 4 cdot \[ \( 4 \) cdot \( 5 \) - "12" \] + { {"16"} over {2} } } {}

Solution

97

Exercise 27

2 2 ⋅ 3 + 2 3 ( 6 − 2 ) − ( 3 + 17 ) + 11 ( 6 ) 2 2 ⋅ 3 + 2 3 ( 6 − 2 ) − ( 3 + 17 ) + 11 ( 6 ) size 12{2 rSup { size 8{2} } cdot 3+2 rSup { size 8{3} } \( 6 - 2 \) - \( 3+"17" \) +"11" \( 6 \) } {}

Solution

90

Exercise 28

8 ( 6 + 20 ) 8 + 3 ( 6 + 16 ) 22 8 ( 6 + 20 ) 8 + 3 ( 6 + 16 ) 22 size 12{ { {8 \( 6+"20" \) } over {8} } + { {3 \( 6+"16" \) } over {"22"} } } {}

Solution

29

Exercise 29

( 1 + 16 ) − 3 7 + 5 ( 12 ) ( 1 + 16 ) − 3 7 + 5 ( 12 ) size 12{ { { \( 1+"16" \) - 3} over {7} } +5 \( "12" \) } {}

Solution

62

Exercise 30

1 6 + 0 8 + 5 2 ( 2 + 8 ) 3 1 6 + 0 8 + 5 2 ( 2 + 8 ) 3 size 12{1 rSup { size 8{6} } +0 rSup { size 8{8} } +5 rSup { size 8{2} } \( 2+8 \) rSup { size 8{3} } } {}

Solution

25,001

Exercise 31

5 ( 8 2 − 9 ⋅ 6 ) 2 5 − 7 + 7 2 − 4 2 2 4 − 5 5 ( 8 2 − 9 ⋅ 6 ) 2 5 − 7 + 7 2 − 4 2 2 4 − 5 size 12{ { {5 \( 8 rSup { size 8{2} } - 9 cdot 6 \) } over {2 rSup { size 8{5} } - 7} } + { {7 rSup { size 8{2} } - 4 rSup { size 8{2} } } over {2 rSup { size 8{4} } - 5} } } {}

Solution

5

Exercise 32

6 { 2 ⋅ 8 + 3 } − ( 5 ) ⋅ ( 2 ) + 8 4 + ( 1 + 8 ) ⋅ ( 1 + 11 ) 6 { 2 ⋅ 8 + 3 } − ( 5 ) ⋅ ( 2 ) + 8 4 + ( 1 + 8 ) ⋅ ( 1 + 11 ) size 12{6 lbrace 2 cdot 8+3 rbrace - \( 5 \) cdot \( 2 \) + { {8} over {4} } + \( 1+8 \) cdot \( 1+"11" \) } {}

Solution

214

Exercise 33

26 − 2 ⋅ 6 + 20 13 26 − 2 ⋅ 6 + 20 13 size 12{"26"` - `2` cdot ` left lbrace { {6+"20"} over {"13"} } right rbrace } {}

Solution

22

Exercise 34

( 10 + 5 ) ⋅ ( 10 + 5 ) − 4 ⋅ ( 60 − 4 ) ( 10 + 5 ) ⋅ ( 10 + 5 ) − 4 ⋅ ( 60 − 4 ) size 12{ \( "10"+5 \) ` cdot ` \( "10"+5 \) ` - `4 cdot \( "60" - 4 \) } {}

Solution

1

Exercise 35

6 2 − 1 2 3 − 3 + 4 3 + 2 ⋅ 3 2 ⋅ 5 6 2 − 1 2 3 − 3 + 4 3 + 2 ⋅ 3 2 ⋅ 5 size 12{ { {6 rSup { size 8{2} } - 1} over {2 rSup { size 8{3} } - 3} } `+` { {4 rSup { size 8{3} } +2` cdot `3} over {2` cdot `5} } } {}

Solution

14

Exercise 36

51 17 + 7 − 2 ⋅ 5 ⋅ 12 3 51 17 + 7 − 2 ⋅ 5 ⋅ 12 3 size 12{ { {"51"} over {"17"} } `+`7` - `2` cdot `5` cdot ` left ( { {"12"} over {3} } right )} {}

Solution

-30

Exercise 37

( 21 − 3 ) ⋅ ( 6 − 1 ) ⋅ 6 + 4 ( 6 + 3 ) ( 21 − 3 ) ⋅ ( 6 − 1 ) ⋅ 6 + 4 ( 6 + 3 ) size 12{ \( "21" - 3 \) ` cdot ` \( 6 - 1 \) ` cdot ` left (6 right )+4 \( 6+3 \) } {}

Solution

576

Exercise 38

( 2 + 1 ) 3 + 2 3 + 1 10 6 2 − 15 2 − [ 2 ⋅ 5 ] 2 5 ⋅ 5 2 ( 2 + 1 ) 3 + 2 3 + 1 10 6 2 − 15 2 − [ 2 ⋅ 5 ] 2 5 ⋅ 5 2 size 12{ { { \( 2+1 \) rSup { size 8{3} } +2 rSup { size 8{3} } +1 rSup { size 8{"10"} } } over {6 rSup { size 8{2} } } } ` - ` { {"15" rSup { size 8{2} } - \[ 2` cdot `5 \] rSup { size 8{2} } } over {5` cdot `5 rSup { size 8{2} } } } } {}

Solution

0

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