1.5: Order of Operations
- Page ID
- 22463
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The order in which we evaluate expressions can be ambiguous. Take for example, the expression 4 + 3 · 2. If we do the addition first, then
4+3 · 2=7 · 2
= 14.
On the other hand, if we do the multiplication first, then
4+3 · 2=4+6
= 10.
So, what are we to do? Of course, grouping symbols can remove the ambiguity
Parentheses, brackets, or curly braces can be used to group parts of an expression. Each of the following are equivalent:
(4 + 3) · 2 or [4 + 3] · 2 or {4+3} · 2
In each case, the rule is “evaluate the expression inside the grouping symbols first.” If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first.
Thus, for example,
(4 + 3) · 2=7 · 2
= 14.
Note how the expression contained in the parentheses was evaluated first. Another way to avoid ambiguities in evaluating expressions is to establish an order in which operations should be performed. The following guidelines should always be strictly enforced when evaluating expressions.
When evaluating expressions, proceed in the following order.
- Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first.
- Evaluate all exponents that appear in the expression.
- Perform all multiplications and divisions in the order that they appear in the expression, moving left to right.
- Perform all additions and subtractions in
Evaluate 4 + 3 · 2.
Solution
Because of the established Rules Guiding Order of Operations, this expression is no longer ambiguous. There are no grouping symbols or exponents, so we immediately go to rule three, evaluate all multiplications and divisions in the order that they appear, moving left to right. After that we invoke rule four, performing all additions and subtractions in the order that they appear, moving left to right.
\[ \begin{aligned} 4+3 \dot 2=4+6 \\ = 10 \end{aligned}\nonumber \]
Thus, 4 + 3 · 2 = 10.
Simplify: 8 + 2 · 5.
- Answer
-
18
Evaluate 18 − 2 + 3.
Solution
Follow the Rules Guiding Order of Operations. Addition has no precedence over subtraction, nor does subtraction have precedence over addition. We are to perform additions and subtractions as they occur, moving left to right.
\[ \begin{aligned} 18 − 2 + 3 = 16 + 3 & \textcolor{red}{ \text{ Subtract: 18 − 2 = 16.}} \\ = 19 & \textcolor{red}{ \text{ Add: 16 + 3 = 19. }} \end{aligned}\nonumber \]
Thus, 18 − 2 + 3 = 19.
Simplify: 17 − 8 + 2.
- Answer
-
11
Evaluate 54 ÷ 9 · 2.
Solution
Follow the Rules Guiding Order of Operations. Division has no precedence over multiplication, nor does multiplication have precedence over division. We are to perform divisions and multiplications as they occur, moving left to right.
\[ \begin{aligned} 54 \div 9 \cdot 2=6 \dot 2 & \textcolor{red}{ \text{ Divide: 54 } \div \text{ 9 = 6. }} \\ = 12 & \textcolor{red}{ \text{ Multiply: 6 } \cdot \text{ 2 = 12. }} \end{aligned}\nonumber \]
Thus, 54 ÷ 9 · 2 = 12.
Simplify: 72 ÷ 9 · 2.
- Answer
-
16
Evaluate 2 · 32 − 12.
Solution
Follow the Rules Guiding Order of Operations, exponents first, then multiplication, then subtraction.
\[ \begin{aligned} 2 \cdot 3^2 - 12 = 2 \dot 9 - 12 & \textcolor{red}{ \text{ Evaluate the exponent: 3^2 = 9. }} \\ = 18 - 12 & \textcolor{red}{ \text{ Perform the multiplication: } 2 \cdot 9 = 18. } \\ = 6 & \textcolor{red}{ \text{ Perform the subtraction: } 18 - 12 = 6.} \end{aligned}\nonumber \]
Thus, 2 · 32 − 12 = 6.
Simplify: 14 + 3 · 42
- Answer
-
62
Evaluate 12 + 2(3 + 2 · 5)2.
Solution
Follow the Rules Guiding Order of Operations, evaluate the expression inside the parentheses first, then exponents, then multiplication, then addition.
\[ \begin{aligned} 12 + 2(3 + 5 \cdot 5 )^2 = 12 + 2(3 + 10)^2 ~ & \textcolor{red}{ \text{ Multiply inside parentheses: 2 } \cdot 5 = 10.} \\ = 12 + 2(13)^2 ~ & \textcolor{red}{ \text{ Add inside parentheses: } 3 + 10 = 13.} \\ = 12 + 2(169) ~ & \textcolor{red}{ \text{ Exponents are next: } (13)^2 = 169.} \\ = 12 + 338 ~ & \textcolor{red}{ \text{ Multiplication is next: } 2(169) = 338.} \\ = 350 ~ & \textcolor{red}{ \text{ Time to add: } 12 + 338 = 350.} \end{aligned}\nonumber \]
Thus, 12 + 2(3 + 2 · 5) 2 = 350.
Simplify: 3(2 + 3 · 4)2 − 11.
- Answer
-
577
Evaluate 2{2 + 2[2 + 2]}.
Solution
When grouping symbols are nested, evaluate the expression between the pair of innermost grouping symbols first.
\[ \begin{aligned} 2( 2 + 2[2 + 2]) = 2(2 + 2[4]) ~ & \textcolor{red}{ \text{ Innermost grouping first: } 2 + 2 = 4.} \\ = 2(2+8) ~ & \textcolor{red}{ \text{ Multiply next: } 2[4] = 8.} \\ = 2(10) ~ & \textcolor{red}{ \text{ Add inside braces: } 2 + 8 = 10.} \\ = 20 ~ & \textcolor{red}{ \text{ Multiply: } 2(10) = 20} \end{aligned}\nonumber \]
Thus, 2(2 + 2[2 + 2]) = 20.
Simplify: 2{3 + 2[3 + 2]}.
- Answer
-
26
Fraction Bars
Consider the expression
\[ \frac{6^{2}+8^{2}}{(2+3)^{2}}\nonumber \]
Because a fraction bar means division, the above expression is equivalent to
\[\left(6^{2}+8^{2}\right) \div(2+3)^{2}\nonumber \]
The position of the grouping symbols signals how we should proceed. We should simplify the numerator, then the denominator, then divide.
If a fractional expression is present, evaluate the numerator and denominator first, then divide.
Evaluate the expression
\[ \frac{6^{2}+8^{2}}{(2+3)^{2}}.\nonumber \]
Solution
Simplify the numerator and denominator first, then divide.
\[ \begin{aligned} \frac{6^{2}+8^{2}}{(2+3)^{2}}=\frac{6^{2}+8^{2}}{(5)^{2}} ~ & \textcolor{red}{ \text{ Parentheses in denominator first: } 2 + 3 = 5} \\ = \frac{36+64}{25} ~ & \textcolor{red}{ \text{Exponents are next: } 6^2 = 36,~ 8^2 = 64,~ 5^2 = 25.} \\ = \frac{100}{25} ~ & \textcolor{red}{ \text{ Add in numerator: } 36 + 64 = 100} \\ = 4 ~ & \textcolor{red}{ \text{ Divide: } 100 \div 25 = 4.} \end{aligned}\nonumber \]
Thus, \(\frac{6^{2}+8^{2}}{(2+3)^{2}}=4\).
Simplify: \(\frac{12+3 \cdot 2}{6}\)Answer
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3
The Distributive Property
Consider the expression 2 · (3 + 4). If we follow the “Rules Guiding Order of Operations,” we would evaluate the expression inside the parentheses first. 2 · (3 + 4) = 2 · 7 Parentheses first: 3 + 4 = 7. = 14 Multiply: 2 · 7 = 14.
However, we could also choose to “distribute” the 2, first multiplying 2 times each addend in the parentheses.
\[ \begin{aligned} 2 \cdot (3 + 4) = 2 \cdot 3 + 2 \cdot 4 ~ & \textcolor{red}{ \text{ Multiply 2 times both 3 and 4.}} \\ = 6 + 8 ~ & \textcolor{red}{ \text{ Multiply: } 2 \cdot 3 = 6 \text{ and } 2 \cdot 4 = 8.} \\ = 14 ~ & \textcolor{red}{ \text{ Add: } 6 + 8 = 14.} \end{aligned}\nonumber \]
The fact that we get the same answer in the second approach is an illustration of an important property of whole numbers.1
Let a, b, and c be any whole numbers. Then,
a · (b + c) = a · b + a · c.
We say that “multiplication is distributive with respect to addition.”
Multiplication is distributive with respect to addition. If you are not computing the product of a number and a sum of numbers, the distributive property does not apply.
If you are calculating the product of a number and the product of two numbers, the distributive property must not be used. For example, here is a common misapplication of the distributive property.
\[ \begin{aligned} 2 \cdot (3 \cdot 4) = (2 \cdot 3) \cdot (2 \cdot 4) \\ = 6 \cdot 8 \\ = 48 \end{aligned}\nonumber \]
This result is quite distant from the correct answer, which is found by computing the product within the parentheses first.
\[ \begin{aligned} 2 \cdot (3 \cdot 4) = 2 \cdot 12 \\ = 24. \end{aligned}\nonumber \]
In order to apply the distributive property, you must be multiplying times a sum.
Use the distributive property to calculate 4 · (5 + 11).
Solution
This is the product of a number and a sum, so the distributive property may be applied.
\[ \begin{aligned} 4 \cdot (5 + 11) = 4 \cdot 5 + 4 \cdot 11 ~ & \textcolor{red}{ \text{ Distribute the 4 times addend in the sum.}} \\ = 20 + 44 ~ & \textcolor{red}{ \text{ Multiply: } 4 \cdot 5 = 20 \text{ and } 4 \cdot 11 = 44.} \\ = 64 ~ & \textcolor{red}{ \text{ Add: } 20 + 44 = 64.} \end{aligned}\nonumber \]
Readers should check that the same answer is found by computing the sum within the parentheses first.
Distribute: 5 · (11 + 8).
- Answer
-
95
The distributive property is the underpinning of the multiplication algorithm learned in our childhood years.
Multiply: 6 · 43.
Solution
We’ll express 43 as sum, then use the distributive property.
\[ \begin{aligned} 6 \cdot 43 = 6 \cdot (40 + 3) ~ & \textcolor{red}{ \text{ Express 43 as a sum: } 43 = 40 + 3} \\ = 6 \cdot 40 + 6 \cdot 3 ~ & \textcolor{red}{ \text{ Distribute the 6.}} \\ = 240 + 18 ~ & \textcolor{red}{ \text{ Multiply: } 6 \cdot 40 = 240 \text{ and } 6 \cdot 3 = 18.} \\ = 258 ~ & \textcolor{red}{ \text{ Add: } 240 + 18 = 258.} \end{aligned}\nonumber \]
Readers should be able to see this application of the distributive property in the more familiar algorithmic form:
\( \begin{array}{r}{43} \\ { \times 6} \\ \hline 18 \\ {\frac{240}{258}}\end{array}\)
Or in the even more condensed form with “carrying:”
\( \begin{array}{r}{^{1} 43} \\ {\frac{ \times 6}{258}}\end{array}\)
Use the distributive property to evaluate 8 · 92.
- Answer
-
736
Multiplication is also distributive with respect to subtraction.
Let a, b, and c be any whole numbers. Then,
a · (b − c) = a · b − a · c.
We say the multiplication is “distributive with respect to subtraction.”
Use the distributive property to simplify: 3 · (12 − 8).
Solution
This is the product of a number and a difference, so the distributive property may be applied.
\[ \begin{aligned} 3 \cdot (12 - 8) = 3 \cdot 12 - 3 \cdot 8 ~ & \textcolor{red}{ \text{ Distribute the 3 times each term in the difference.}} \\ = 36 - 24 ~ & \textcolor{red}{ \text{Multiply: } 3 \cdot 12 = 36 \text{ and } 3 \cdot 8 = 24.} \\ = 12 ~ & \textcolor{red}{ \text{Subtract: } 36 - 24 = 12.} \end{aligned}\nonumber \]
Alternate solution
Note what happens if we use the usual “order of operations” to evaluate the expression.
\[ \begin{aligned} 3 \cdot (12 - 8) = 3 \cdot 4 ~ & \textcolor{red}{ \text{ Parentheses first: } 12 - 8 = 4.} \\ = 12 ~ & \textcolor{red}{ \text{ Multiply: } 3 \cdot 4 = 12.} \end{aligned}\nonumber \]
Same answer.
Distribute: 8 · (9 − 2).
- Answer
-
56
Exercises
In Exercises 1-12, simplify the given expression.
1. 5+2 · 2
2. 5+2 · 8
3. 23 − 7 · 2
4. 37 − 3 · 7
5. 4 · 3+2 · 5
6. 2 · 5+9 · 7
7. 6 · 5+4 · 3
8. 5 · 2+9 · 8
9. 9+2 · 3
10. 3+6 · 6
11. 32 − 8 · 2
12. 24 − 2 · 5
In Exercises 13-28, simplify the given expression.
13. 45 ÷ 3 · 5
14. 20 ÷ 1 · 4
15. 2 · 9 ÷ 3 · 18
16. 19 · 20 ÷ 4 · 16
17. 30 ÷ 2 · 3
18. 27 ÷ 3 · 3
19. 8 − 6+1
20. 15 − 5 + 10
21. 14 · 16 ÷ 16 · 19
22. 20 · 17 ÷ 17 · 14
23. 15 · 17 + 10 ÷ 10 − 12 · 4
24. 14 · 18 + 9 ÷ 3 − 7 · 13
25. 22 − 10 + 7
26. 29 − 11 + 1
27. 20 · 10 + 15 ÷ 5 − 7 · 6
28. 18 · 19 + 18 ÷ 18 − 6 · 7
In Exercises 29-40, simplify the given expression.
29. 9+8 ÷ {4+4}
30. 10 + 20 ÷ {2+2}
31. 7 · [8 − 5] − 10
32. 11 · [12 − 4] − 10
33. (18 + 10) ÷ (2 + 2)
34. (14 + 7) ÷ (2 + 5)
35. 9 · (10 + 7) − 3 · (4 + 10)
36. 9 · (7 + 7) − 8 · (3 + 8)
37. 2 · {8 + 12} ÷ 4
38. 4 · {8+7} ÷ 3
39. 9+6 · (12 + 3)
40. 3+5 · (10 + 12)
In Exercises 41-56, simplify the given expression.
41. 2+9 · [7 + 3 · (9 + 5)]
42. 6+3 · [4 + 4 · (5 + 8)]
43. 7+3 · [8 + 8 · (5 + 9)]
44. 4+9 · [7 + 6 · (3 + 3)]
45. 6 − 5[11 − (2 + 8)]
46. 15 − 1[19 − (7 + 3)]
47. 11 − 1[19 − (2 + 15)]
48. 9 − 8[6 − (2 + 3)]
49. 4{7[9 + 3] − 2[3 + 2]}
50. 4{8[3 + 9] − 4[6 + 2]}
51. 9 · [3 + 4 · (5 + 2)]
52. 3 · [4 + 9 · (8 + 5)]
53. 3{8[6 + 5] − 8[7 + 3]}
54. 2{4[6 + 9] − 2[3 + 4]}
55. 3 · [2 + 4 · (9 + 6)]
56. 8 · [3 + 9 · (5 + 2)]
In Exercises 57-68, simplfiy the given expression.
57. (5 − 2)2
58. (5 − 3)4
59. (4 + 2)2
60. (3 + 5)2
61. 23 + 33
62. 54 + 24
63. 23 − 13
64. 32 − 12
65. 12 · 52 + 8 · 9+4
66. 6 · 32 + 7 · 5 + 12
67. 9 − 3 · 2 + 12 · 102
68. 11 − 2 · 3 + 12 · 42
In Exercises 69-80, simplify the given expression.
69. 42 − (13 + 2)
70. 33 − (7 + 6)
71. 33 − (7 + 12)
72. 43 − (6 + 5)
73. 19 + 3[12 − (23 + 1)]
74. 13 + 12[14 − (22 + 1)]
75. 17 + 7[13 − (22 + 6)]
76. 10 + 1[16 − (22 + 9)]
77. 43 − (12 + 1)
78. 53 − (17 + 15)
79. 5 + 7[11 − (22 + 1)]
80. 10 + 11[20 − (22 + 1)]
In Exercises 81-92, simplify the given expression.
81. \( \frac{13+35}{3(4)}\)
82. \( \frac{35+28}{7(3)}\)
83. \( \frac{64-(8 \cdot 6-3)}{4 \cdot 7-9}\)
84. \( \frac{19-(4 \cdot 3-2)}{6 \cdot 3-9}\)
85. \(\frac{2+13}{4-1}\)
86. \( \frac{7+1}{8-4}\)
87. \( \frac{17+14}{9-8}\)
88. \( \frac{16+2}{13-11}\)
89. \( \frac{37+27}{8(2)}\)
90. \( \frac{16+38}{6(3)}\)
91. \( \frac{40-(3 \cdot 7-9)}{8 \cdot 2-2}\)
92. \( \frac{60-(8 \cdot 6-3)}{5 \cdot 4-5}\)
In Exercises 93-100, use the distributive property to evaluate the given expression.
93. 5 · (8 + 4)
94. 8 · (4 + 2)
95. 7 · (8 − 3)
96. 8 · (9 − 7)
97. 6 · (7 − 2)
98. 4 · (8 − 6)
99. 4 · (3 + 2)
100. 4 · (9 + 6)
In Exercises 101-104, use the distributive property to evaluate the given expression using the technique shown in Example 9.
101. 9 · 62
102. 3 · 76
103. 3 · 58
104. 7 · 57
Answers
1. 9
3. 9
5. 22
7. 42
9. 15
11. 16
13. 75
15. 108
17. 45
19. 3
21. 266
23. 208
25. 19
27. 161
29. 10
31. 11
33. 7
35. 111
37. 10
39. 99
41. 443
43. 367
45. 1
47. 9
49. 296
51. 279
53. 24
55. 186
57. 9
59. 36
61. 35
63. 7
65. 376
67. 1203
69. 1
71. 8
73. 28
75. 38
77. 51
79. 47
81. 4
83. 1
85. 5
87. 31
89. 4
91. 2
93. 60
95. 35
97. 30
99. 20
101. 558
103. 174
1Later, we’ll see that this property applies to all numbers, not just whole numbers