1.4: Order of Operations

Last updated

Jul 2, 2024
Save as PDF
- 1.3: Multiplying and Dividing Whole Numbers
- 1.5: Positive and Negative Numbers

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

↵

The order in which we evaluate expressions matters. 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? If this expression had grouping symbols, we would start there.

Grouping Symbols

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.

Rules Guiding Order of Operations

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 the order that they appear in the expression, moving left to right.

Example 1

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.

Exercise

Simplify: 8 + 2 · 5.

Answer: 18

Example 2

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.

Exercise

Simplify: 17 − 8 + 2.

Answer: 11

Example 3

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.

Exercise

Simplify: 72 ÷ 9 · 2.

Answer: 16

Example 4

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.

Exercise

Simplify: 14 + 3 · 4²

Answer: 62

Example 5

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.

Exercise

Simplify: 3(2 + 3 · 4)² − 11.

Answer: 577

Example 6

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.

Exercise

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.

Fractional Expressions

If a fractional expression is present, evaluate the numerator and denominator first, then divide.

Example 7

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, $\frac{6^{2}+8^{2}}{(2+3)^{2}}=4$ .

Exercise

Simplify: $\frac{12+3 \cdot 2}{6}$

Answer: 3

Exercises

In Exercises 1-12, simplify the given expression.

7. 6 · 5+4 · 3

12. 24 − 2 · 5

18. 27 ÷ 3 · 3

25. 22 − 10 + 7

27. 20 · 10 + 15 ÷ 5 − 7 · 6

31. 7 · [8 − 5] − 10

39. 9+6 · (12 + 3)

48. 9 − 8[6 − (2 + 3)]

53. 3{8[6 + 5] − 8[7 + 3]}

57. (5 − 2)²

65. 12 · 5² + 8 · 9+4

67. 9 − 3 · 2 + 12 · 10²

73. 19 + 3[12 − (2³ + 1)]

84. $\frac{19-(4 \cdot 3-2)}{6 \cdot 3-9}$

Answers

7. 42

12. 14

18. 27

25. 19

27. 161

31. 11

39. 99

48. 1

53. 24

57. 9

65. 376

67. 1203

73. 28

84. 1

¹Later, we’ll see that this property applies to all numbers, not just whole numbers

Search

Text Color

Text Size

Margin Size

Font Type

Fraction Bars

Exercises

Answers

Support Center

How can we help?