# Order of Operations (PEMDAS)

The order of operations determines the order in which you do your calculations. At this point we call the order of operations PEMDAS (initials of Parentheses, Exponents, Multiplication, Division, Addition and Subtraction).

Bootcamps

Rule

### PEMDAS

• First, calculate the parentheses and exponents.

• Next, do all the multiplication and division.

• Finally, do all the addition and subtraction.

In addition, be sure to clean up the parentheses using EMDAS before starting PEMDAS. This is because the calculations will be easier if you perform EMDAS (exponents, multiplication, division, addition and subtraction) inside the parentheses before removing them.

Example 1

Calculate $6+3\cdot 2$.

Think PEMDAS. First, you need to see if there are any parentheses or exponents that need to be calculated. Then you can multiply and divide, and lastly you do the addition and subtraction. It will look like this:

 $6+3\cdot 2=6+6=12$

Example 2

Calculate $\phantom{\rule{-0.17em}{0ex}}\left(6-{2}^{2}\right)\cdot 6+3$.

Think PEMDAS. Before removing the parentheses, you need to clean up inside them. Then you calculate any parentheses or exponents. After that you have to multiply and divide, and lastly you do the addition and subtraction. Then it will look like this: $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(6-{2}^{2}\right)\cdot 6+3& =\left(6-4\right)\cdot 6+3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(2\right)6+3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =12+3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Calculate $12÷\left(4-6\right)+8\cdot 5$.

Think PEMDAS. First, you need to look for parentheses or exponents that need to be calculated. Next you multiply and divide, and in the end you do the addition and subtractions. Then it will look like this:

$\begin{array}{llll}\hfill 12÷\left(4-6\right)+8\cdot 5& =12÷\left(-2\right)+40\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-6+40\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =34\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Here you saw that we did the multiplication in the last term at the same time we cleaned up inside the parentheses. When you get more comfortable with PEMDAS, you can also do this, as the two actions are done in different terms.

Example 4

Let’s calculate the following expression:

 ${2}^{2}-3+5\cdot 8÷\left(1-3\cdot 1\right)$

By following the order of operations in the same way we did it in the other examples, we get

$\begin{array}{llll}\hfill {2}^{2}& -3+5\cdot 8÷\left(1-3\cdot 1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{2}-3+5\cdot 8÷\left(1-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{2}-3+5\cdot 8÷\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3+5\cdot 8÷\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3+5\cdot \left(-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3-20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-19\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {2}^{2}-3+5\cdot 8÷\left(1-3\cdot 1\right)& ={2}^{2}-3+5\cdot 8÷\left(1-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{2}-3+5\cdot 8÷\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3+5\cdot 8÷\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3+5\cdot \left(-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-3-20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-19\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$