Question

How do you simplify $\dfrac{3}{4}+\left( \dfrac{1}{4}-\dfrac{1}{6} \right)$?

Answer 1

Hint: In this question, we need to simplify the expression given in fractions. For this, we will first solve the bracket term. We will make the fraction like by making their denominator the same. Then we will perform operation on the numerator and keep the denominator the same. After that we will solve the found fraction by fraction outside the term $\left( \dfrac{3}{4} \right)$ the same way, making the denominator the same and adding numerator. To make the denominator the same we will take the LCM of the existing denominator and then make the denominator same as LCM by suitable multiplication.

Complete step by step answer:
Here we have the expression given as $\dfrac{3}{4}+\left( \dfrac{1}{4}-\dfrac{1}{6} \right)$.
Let us simplify the term inside the bracket first. We know the fraction as $\dfrac{1}{4}-\dfrac{1}{6}$.
As we can see both terms have different denominators (unlike fraction). So we cannot subtract them directly. We need to make their denominator the same. Let us take the least common multiple of 4 and 6 we get, $2\times 2\times 3=12$.
\[\begin{align}
  & 2\left| \!{\underline {\,
  4,6 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2,3 \,}} \right. \\
 & 3\left| \!{\underline {\,
  1,3 \,}} \right. \\
 & 0\left| \!{\underline {\,
  1,1 \,}} \right. \\
\end{align}\]
So we need to make the denominator of both fractions as 12. We know that, if we multiply 4 by 3, we get 12. So multiplying by 3 in the numerator and denominator of $\dfrac{1}{4}$ we get $\dfrac{1\times 3}{4\times 3}=\dfrac{3}{12}$.
Also $6\times 2=12$ so multiplying 2 in the numerator and denominator of $\dfrac{1}{6}$ we get $\dfrac{1\times 2}{6\times 2}=\dfrac{2}{12}$.
Therefore, our bracket term reduces to $\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{3}{12}-\dfrac{2}{12}$.
For subtracting like fractions, we subtract their numerator and keep denominator same so we have $\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1}{12}$.
Our original expression becomes $\dfrac{3}{4}+\dfrac{1}{12}$.
Again the denominators are different, so let us change their denominator to the LCM of 4 and 12.
Least common multiple of 4 and 12 = $2\times 2\times 3=12$.
\[\begin{align}
  & 2\left| \!{\underline {\,
  4,12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2,6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  1,3 \,}} \right. \\
 & 0\left| \!{\underline {\,
  1,1 \,}} \right. \\
\end{align}\]
We know $4\times 3=12$ so multiplying the numerator and denominator by 3 in $\dfrac{3}{4}$ we get $\dfrac{3\times 3}{4\times 3}=\dfrac{9}{12}$.
Fraction $\dfrac{1}{12}$ already has denominator 12 so our fraction becomes $\dfrac{9}{12}+\dfrac{1}{12}$.
Adding the numerator and keeping the denominator same we get $\dfrac{10}{12}$.
Dividing the numerator and the denominator by 2 we get $\dfrac{10\div 2}{12\div 2}=\dfrac{5}{6}$.
Therefore $\dfrac{3}{4}+\left( \dfrac{1}{4}-\dfrac{1}{6} \right)=\dfrac{5}{6}$.

Hence $\dfrac{5}{6}$ is the required answer.

Note:
Students should note that numerators are added/subtracted when fractions are added/subtracted. Make sure to multiply the numerator by the same number as that of the denominator to make an equivalence relation.