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Add or subtract the following fractions
(i) \[\dfrac{7}{9} + \dfrac{1}{8}\]
( ii ) \[\dfrac{5}{8} - \dfrac{1}{2}\]

Answer
VerifiedVerified
486.9k+ views
Hint: Here we have to find the sum and difference of the given data. Here the data is in the form of fraction. Since in the given data are in fraction the value of denominator is different so we find the LCM for the both the denominators and then we add the numbers. hence, we obtain the required solution for the given question.

Complete step by step answer:
We apply the arithmetic operations on the fractions. Here in this question, we add and subtract the two fractions. the LCM is a least common multiple. The LCM will be common for both the numbers.
 On adding the \[\dfrac{7}{9}\] and \[\dfrac{1}{8}\] we write it as
\[ \Rightarrow \dfrac{7}{9} + \dfrac{1}{8}\]
The denominators are different so we have to take LCM for the numbers 9 and 8. The LCM for the numbers 9 and 8 is 72
Therefore the above inequality is written as
\[ \Rightarrow \dfrac{{\dfrac{7}{9} \times 72 + \dfrac{1}{8} \times 72}}{{72}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{7 \times 8 + 1 \times 9}}{{72}}\]
On further simplifying we have
\[ \Rightarrow \dfrac{{56 + 9}}{{72}}\]
On adding the 48 and 9 we get
\[ \Rightarrow \dfrac{{65}}{{72}}\]
Therefore \[\dfrac{7}{9} + \dfrac{1}{8} = \dfrac{{65}}{{72}}\]
On subtracting the \[\dfrac{1}{2}\] from \[\dfrac{5}{8}\] we write it as
\[ \Rightarrow \dfrac{5}{8} - \dfrac{1}{2}\]
The denominators are different so we have to take LCM for the numbers 8 and 2. The LCM for the numbers 8 and 2 is 8
Therefore the above inequality is written as
\[ \Rightarrow \dfrac{{\dfrac{5}{8} \times 8 - \dfrac{1}{2} \times 8}}{8}\]
On simplifying we have
\[ \Rightarrow \dfrac{{5 - 4}}{8}\]
On further simplifying we have
\[ \Rightarrow \dfrac{1}{8}\]
Therefore \[\dfrac{5}{8} - \dfrac{1}{2} = \dfrac{1}{8}\]

Note: While adding the two fractions we need to check the values of the denominator, if both denominators are having the same value then we can add the numerators. Suppose if the fractions have different denominators, we have to take LCM for the denominators and we simplify for further.