Performing Calculations Using the Order of Operations
When we multiply a number by itself, we square it or raise it to a power of 2. For example, [latex]{4}^{2}=4\cdot 4=16[/latex]. We can raise any number to any power. In general, the exponential notation [latex]{a}^{n}[/latex] means that the number or variable [latex]a[/latex] is used as a factor [latex]n[/latex] times.
A General Note: Order of Operations
Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS: P(arentheses) E(xponents) M(ultiplication) and D(ivision) A(ddition) and S(ubtraction)How To: Given a mathematical expression, simplify it using the order of operations.
- Simplify any expressions within grouping symbols.
- Simplify any expressions containing exponents or radicals.
- Perform any multiplication and division in order, from left to right.
- Perform any addition and subtraction in order, from left to right.
Example 6: Using the Order of Operations
Use the order of operations to evaluate each of the following expressions.- [latex]{\left(3\cdot 2\right)}^{2}-4\left(6+2\right)[/latex]
- [latex]\frac{{5}^{2}-4}{7}-\sqrt{11 - 2}[/latex]
- [latex]6-|5 - 8|+3\left(4 - 1\right)[/latex]
- [latex]\frac{14 - 3\cdot 2}{2\cdot 5-{3}^{2}}[/latex]
- [latex]7\left(5\cdot 3\right)-2\left[\left(6 - 3\right)-{4}^{2}\right]+1[/latex]
Solution
-
[latex]\begin{array}{cccc}\left(3\cdot 2\right)^{2} \hfill& =\left(6\right)^{2}-4\left(8\right) \hfill& \text{Simplify parentheses} \\ \hfill& =36-4\left(8\right) \hfill& \text{Simplify exponent} \\ \hfill& =36-32 \hfill& \text{Simplify multiplication} \\ \hfill& =4 \hfill& \text{Simplify subtraction}\end{array}[/latex]
-
[latex]\begin{array}{cccc}\frac{5^{2}}{7}-\sqrt{11-2} \hfill& =\frac{5^{2}-4}{7}-\sqrt{9} \hfill& \text{Simplify grouping systems (radical)} \\ \hfill& =\frac{5^{2}-4}{7}-3 \hfill& \text{Simplify radical} \\ \hfill& =\frac{25-4}{7}-3 \hfill& \text{Simplify exponent} \\ \hfill& =\frac{21}{7}-3 \hfill& \text{Simplify subtraction in numerator} \\ \hfill& =3-3 \hfill& \text{Simplify division} \\ \hfill& =0 \hfill& \text{Simplify subtraction}\end{array}[/latex]
Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped. - [latex]\begin{array}{cccc}6-|5-8|+3\left(4-1\right) \hfill& =6-|-3|+3\left(3\right) \hfill& \text{Simplify inside grouping system} \\ \hfill& =6-3+3\left(3\right) \hfill& \text{Simplify absolute value} \\ \hfill& =6-3+9 \hfill& \text{Simplify multiplication} \\ \hfill& =3+9 \hfill& \text{Simplify subtraction} \\ \hfill& =12 \hfill& \text{Simplify addition}\end{array}[/latex]
- [latex]\begin{array}{cccc}\frac{14-3\cdot2}{2\cdot5-3^{2}} \hfill& =\frac{14-3\cdot2}{2\cdot5-9} \hfill& \text{Simplify exponent} \\ \hfill& =\frac{14-6}{10-9} \hfill& \text{Simplify products} \\ \hfill& =\frac{8}{1} \hfill& \text{Simplify quotient} \\ \hfill& =8 \hfill& \text{Simplify quotient}\end{array}[/latex] In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step.
- [latex]\begin{array}{cccc}7\left(5\cdot3\right)-2[\left(6-3\right)-4^{2}]+1 \hfill& =7\left(15\right)-2[\left(3\right)-4^{2}]+1 \hfill& \text{Simplify inside parentheses} \\ \hfill& 7\left(15\right)-2\left(3-16\right)+1 \hfill& \text{Simplify exponent} \\ \hfill& =7\left(15\right)-2\left(-13\right)+1 \hfill& \text{Subtract} \\ \hfill& =105+26+1 \hfill& \text{Multiply} \\ \hfill& =132 \hfill& \text{Add}\end{array}[/latex]
Try It 6
Use the order of operations to evaluate each of the following expressions.a. [latex]\sqrt{{5}^{2}-{4}^{2}}+7{\left(5 - 4\right)}^{2}[/latex] b. [latex]1+\frac{7\cdot 5 - 8\cdot 4}{9 - 6}[/latex] c. [latex]|1.8 - 4.3|+0.4\sqrt{15+10}[/latex] d. [latex]\frac{1}{2}\left[5\cdot {3}^{2}-{7}^{2}\right]+\frac{1}{3}\cdot {9}^{2}[/latex] e. [latex][{\left(3 - 8\right)}^{2}-4]-\left(3 - 8\right)[/latex]
SolutionLicenses & Attributions
CC licensed content, Specific attribution
- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.