Order of operations

To correctly solve an equation or simplify an expression, the order in which you perform each operation is important. You must first solve any expressions within Parentheses, then solve Exponents, then Multiplication/Division, and finally Addition/Subtraction.

You can remember this order by the acronym PEMDAS.

1. P arentheses
2. E xponents
3. M ultiplication
4. D ivision
5. A ddition
6. S ubtraction

The mnemonic ‘Please Excuse My Dear Aunt Sally’ will help you remember the acronym.

Here is one example of simplifying an expression with many operations. Each line shows the next step in the process:

$$\begin{align*} & 3\cdot2^3/(2+2^2)+2 \\ & =3\cdot2^3/(2+4)+2\\ & =3\cdot2^3/6+2\\ & =3\cdot8/6+2\\ & =24/6+2\\ & =4+2\\ & =6 \end{align*}$$

Below, notice how we get a different (and incorrect) solution when we improperly perform an addition before the division.

$$\begin{align*} & 3\cdot2^3/(2+2^2)+2 \\ & =3\cdot2^3/(2+4)+2 \\ & =3\cdot2^3/6+2 \\ & =3\cdot8/6+2 \\ & =24/6+2 \\ & \neq 24/8 \qquad\qquad \text{Incorrect order} \\ & =3 \qquad \qquad \quad \; \text{Wrong answer} \end{align*}$$

The previous example shows why order is important. Here is another example of correctly simplifying an expression with several operations:

$$\begin{align*} & (7\cdot10-2\cdot3^{2+1})/2^3-1 \\ & =(7\cdot10-2\cdot3^3)/2^3-1 \\ & =(7\cdot10-2\cdot27)/2^3-1 \\ & =(70-54)/2^3-1 \\ & =16/2^3-1 \\ & =16/8-1 \\ & =2-1 \\ & =1 \end{align*}$$

Notice how we began by adding the 2 and 1 in the exponent position of the 3? Any operations in an exponent should be computed first, they are in unwritten but implied parenthesis. For example:

$$\begin{align*} & 2^{3\cdot2+1} \\ & =2^{6+1} \\ & =2^7 \\ & =128 \end{align*}$$

Division can occur before multiplication, and subtraction can come before addition (PEDMSA would work too), but parenthesis must come first overall, then exponents, then division or multiplication, then addition or subtraction.

$$\begin{align*} & 3\cdot6/2+11-7 \\ & =18/2+11-7 \\ & =9+11-7 \\ & =20-7 \\ & =13 \end{align*}$$

Notice we also get the same result when we perform division before multiplication and subtraction before addition.

$$\begin{align*} & 3\cdot6/2+11-7 \\ & =3\cdot3+11-7 \\ & =9+11-7 \\ & =9+4 \\ & =13 \end{align*}$$

Practice questions

Solve

1. \( 4\cdot3^2/(2\cdot3)^{1+1}+2^2 \)
2. \( (2\cdot5/(4+1)+4)\cdot3 \)
3. \( 2\cdot4+1-3-5\cdot6 \)
4. \( 4\cdot(3+2)^{3}/25 \)
5. \( (4\cdot5-6\cdot3)/2^{3} \)

Solutions

---

1. Solution $$ \begin{align*} & 4\cdot3^{2}/(2\cdot3)^{1+1}+2^{2} \\ & =4\cdot9/6^{2}+4 \\ & =4\cdot9/36+4 \\ & =36/36+4 \\ & =1+4 \\ & =5 \end{align*} $$
2. Solution: $$ \begin{align*} & (2\cdot5/(4+1)+4)\cdot3 \\ & =(10/5+4)\cdot3 \\ & =(2+4)\cdot3 \\ & =6\cdot3 \\ & =18 \end{align*} $$
3. Solution: $$ \begin{align*} & 2\cdot4+1-3-5\cdot6 \\ & =8+1-3-30 \\ & =9-33 \\ & =-24 \end{align*} $$
4. Solution: $$ \begin{align*} & 4\cdot(3+2)^{3}/25 \\ & =4\cdot5^{3}/25 \\ & =4\cdot125/25 \\ & =4\cdot5 \\ & =20 \end{align*} $$
5. Solution: $$ \begin{align*} & (4\cdot5-6\cdot3)/2^{3} \\ & =(20-18)/8 \\ & =2/8 \\ & =1/4 \end{align*} $$

---