Order of operations

The conventions for the order of operations are the same as those for whole numbers.

• Evaluate the expressions inside the brackets first.
• In the absence of brackets, carry out operations in the following order:
1. powers
2. multiplication and division, from left to right
3. addition and subtraction, from left to right.

Example 4

1. \(\dfrac{2}{7}+\dfrac{6}{7}−\dfrac{5}{7}\)
2. \(\dfrac{5}{6}–\dfrac{2}{3}+\dfrac{1}{4}\)

Solution

1. \(\dfrac{2}{7}+\dfrac{6}{7}−\dfrac{5}{7}=\dfrac{3}{7}\)
2. \begin{align}\dfrac{5}{6}–\dfrac{2}{3}+\dfrac{1}{4}&=\dfrac{10}{12}−\dfrac{8}{12}+\dfrac{3}{12}\\\\ &=\dfrac{5}{12}\end{align}

Sometimes it is easier to work with an appropriate pair of related fractions.

Example 5

\(\dfrac{1}{7}+\dfrac{5}{8}+\dfrac{1}{2}−\dfrac{1}{8}\)

Solution

Rearrange the expression thus:

\begin{align}\dfrac{1}{7}+\dfrac{1}{2}+\dfrac{5}{8}–\dfrac{1}{8}&=\dfrac{1}{7}+\dfrac{1}{2}+\dfrac{4}{8}\\\\ &=\dfrac{1}{7}+\dfrac{1}{2}+\dfrac{1}{2}\\\\ &=1\dfrac{1}{7}\end{align}

Example 6

1. \begin{align}\dfrac{1}{2}÷\dfrac{4}{3}+\dfrac{3}{4}&=\dfrac{1}{2}×\dfrac{3}{4}+\dfrac{3}{4}\\\\ &=\dfrac{3}{8}+\dfrac{3}{4}\\\\ &=\dfrac{3}{8}+\dfrac{6}{8}\\\\ &=\dfrac{9}{8}\\\\ &=1\dfrac{1}{8}\end{align}
2. \begin{align}(\dfrac{2}{3}+\dfrac{5}{6})×\dfrac{5}{9}&=(\dfrac{4}{6}+\dfrac{5}{6})×\dfrac{5}{9}\\\\ &=\dfrac{9\hspace{-4mm}\color{darkred}\setminus}{6}×\dfrac{5}{9\hspace{-4mm}\color{darkred}\setminus}\\\\ &=\dfrac{5}{6}\end{align}

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