Department of Mathematics



Algebra

Worked Examples

Order of Operations

In order to simplify or change algebraic expressions into more convenient forms, we follow the order of operations. To better remember the conventional order, consider the mnemonic device (BEDMAS) described below:

Step 1: B Brackets (then begin steps 1-4 on the expression inside)
Step 2: E Exponents
Step 3: D Division and
M Multiplication
Step 4: A Addition and
S Subtraction

Note:     Although division comes before multiplication in BEDMAS, the convention is to do whichever comes first when reading the problem from left to right. The same is true for addition and subtraction. BEDMAS is just easier to remember than BEMDSA.

Now let’s attempt the following example: $$\frac{2-3(6-2^2)^2-3\cdot 8 \div 2^3}{1+\frac{-3}{-6}\div\frac{-1}{4}}$$ Notice that it can be expressed in other ways: $$\begin{align*}\frac{2-3(6-2^2)^2-3\cdot 8 \div 2^3}{1+\frac{-3}{-6}\div\frac{-1}{4}} &=\frac{(2–3(6−2^2)^2–3 \cdot 8\div 2^3)}{\left(1+ \frac{−3}{−6} \div \frac{-1}{ 4}\right)}\\ &=\left(2-3(6-2^2)^2-3\cdot 8 \div 2^3 \right) \div \left( 1+\frac{-3}{-6}\div\frac{-1}{4}\right) \end{align*}$$ This reminds us of the clarity that brackets can offer and that fractions are simply division problems! Furthermore, because we have equivalent problems when the numerator and denominator are in brackets, BEDMAS requires that we begin by solving them separately. Then, we simply divide the final results. The step-by-step solution is as follows, but note that several operations could have been performed at once to yield a more concise solution:

NUMERATOR:

$$\begin{align*} 2–3(6−2^2)^2–3 \cdot 8\div 2^3 & = 2–3(6−4)^2–3 \cdot 8\div 2^3 \\ & = 2–3(2)^2–3 \cdot 8\div 2^3 \\ & = 2–3(4)–3 \cdot 8\div 2^3 \\ & = 2–3(4)–3 \cdot 8\div 8 \\ & = 2–12–3 \cdot 8 \div 8 \\ & = 2–12– 24 \div 8 \\ & = 2–12– 3 \\ & = –10– 3 \\ & = –13 \end{align*}$$

DENOMINATOR:

$$\begin{align*} 1+\frac{-3}{-6}\div\frac{-1}{4} & = 1+\frac{1}{2}\div\frac{-1}{4} \\ & = 1+\frac{1}{2}\cdot \frac{4}{-1} \\ & = 1+ \frac{-4}{2} \\ & = 1+ (-2) \\ & = −1 \end{align*}$$ Combining the results, we have: $$\frac{numerator}{denominator} = \frac{-13}{-1} = 13$$