Table of Contents
Last modified on August 3rd, 2023
Just like solving equations with whole numbers, solving equations with fractions follows the same PEMDAS rule.
Let us recap the PEMDAS rule before we solve some expressions involving fractions.
We can simplify expressions having both like and unlike fractions involving PEMDAS.
Let us consider a simple example with like fractions (having the same denominators)
Simplify: ${\begin{aligned}\left( \dfrac{2}{3}\times \dfrac{4}{3}\right) +\dfrac{6}{3}\\ .\end{aligned}}$
Solving this expression using PEMDAS rule, we will first perform the operation within the parenthesis involving multiplication and then we will add the result with 6/3
= ${\dfrac{8}{3}+\dfrac{6}{3}}$ (∵ Multiplication of ${\begin{aligned}\dfrac{2}{3}\times \dfrac{4}{3}=\dfrac{8}{3}\\ .\end{aligned}}$)
= ${\dfrac{14}{3}}$ (∵ Addition of ${\dfrac{8}{3}+\dfrac{6}{3}=\dfrac{14}{3}}$)
Let us solve some more expressions using PEMDAS having like fraction
Simplify: ${\dfrac{1}{3}\times \left( \dfrac{2}{3}\right) ^{2}-\dfrac{1}{3}}$
Using PEMDAS,
= ${\dfrac{1}{3}\times \dfrac{4}{9}-\dfrac{1}{3}}$ (Evaluating the exponents)
= ${\dfrac{4}{27}-\dfrac{1}{3}}$ (Multiplying)
= ${\dfrac{4}{27}-\dfrac{9}{27}}$ (Making the fractions equivalent)
= ${-\dfrac{5}{27}}$ (Subtracting)
Answer: ${-\dfrac{5}{27}}$
Solve: ${\dfrac{1}{14}\div \dfrac{2}{14}-\dfrac{1}{2}}$
Using PEMDAS,
= ${\dfrac{1}{14}\times \dfrac{14}{2}-\dfrac{1}{2}}$ (Dividing: Inverting the 2nd fraction)
= ${\dfrac{1}{2}-\dfrac{1}{2}}$ (Multiplying)
= 0 (Subtracting)
When, we consider an expression involving improper fractions, we follow the same rule but an extra step is involved, making the fractions equivalent
Simplify: ${-\dfrac{1}{3}+\dfrac{1}{3}\left( -\dfrac{1}{5}\right)}$
Here, we will use the rule of PEMDAS
${-\dfrac{1}{3}-\dfrac{1}{15}}$ (∵ Multiplication of ${\dfrac{1}{3}\left( -\dfrac{1}{5}\right) =-\dfrac{1}{15}}$)
= ${-1\times \dfrac{5}{3}\times 5-\dfrac{1}{15}}$ (Making the fractions equivalent, Lowest Common Denominator (LCD) = 15)
= ${-\dfrac{5}{15}-\dfrac{1}{15}}$ (Subtracting the like fractions)
= ${-\dfrac{6}{15}}$
Let us solve another expression using PEMDAS having unlike fraction
Simplify: ${\dfrac{1}{3}\div \dfrac{4}{5}-\dfrac{1}{7}}$
Using PEMDAS,
= ${\dfrac{1}{3}\times \dfrac{5}{4}-\dfrac{1}{6}}$ (Dividing: Inverting the 2nd fraction)
= ${\dfrac{5}{12}-\dfrac{1}{6}}$ (Multiplying)
= ${\dfrac{5}{12}-1\times \dfrac{2}{6}\times 2}$ (Making the fractions equivalent)
= ${\dfrac{5}{12}-\dfrac{2}{12}}$ (Subtracting
= ${\dfrac{3}{12}}$
Last modified on August 3rd, 2023