Subtract the first rational number from the second in each of the following:
(i) $ \frac{3}{8}, \frac{5}{8} $
(ii) $ \frac{-7}{9}, \frac{4}{9} $
(iii) $ \frac{-2}{11}, \frac{-9}{11} $
(iv) $ \frac{11}{13}, \frac{-4}{13} $
(v) $ \frac{1}{4}, \frac{-3}{8} $
(vi) $ \frac{-2}{3}, \frac{5}{6} $
(vii) $ \frac{-6}{7}, \frac{-13}{14} $
(viii) $ \frac{-8}{33}, \frac{-7}{22} $

To do:

We have to subtract the first rational number from the second.

Solution:

We know that,

$(-) \times(-)=(+)$

Therefore,

(i) Subtracting $\frac{3}{8}$ from $\frac{5}{8}$, we get,

$\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}$

$=\frac{2}{8}$

$=\frac{1}{4}$

(ii) Subtracting $\frac{-7}{9}$ from $\frac{4}{9}$, we get,

$\frac{4}{9}-(\frac{-7}{9})=\frac{4}{9}+\frac{7}{9}$

$=\frac{4+7}{9}$

$=\frac{11}{9}$

(iii) Subtracting $\frac{-2}{11}$ from $\frac{-9}{11}$, we get,

$\frac{-9}{11}-(\frac{-2}{11})=\frac{-9}{11}+\frac{2}{11}$

$=\frac{-9+2}{11}$

$=\frac{-7}{11}$

(iv) Subtracting $\frac{11}{13}$ from $\frac{-4}{13}$, we get,

$\frac{-4}{13}-\frac{11}{13}=\frac{-4-11}{13}$

$=\frac{-15}{13}$ 

(v) Subtracting $\frac{1}{4}$ from $\frac{-3}{8}$, we get,

$\frac{-3}{8}-(\frac{1}{4})=\frac{-3}{8}-\frac{1}{4}$

$=\frac{-3-2}{8}$

$=\frac{-5}{8}$ 

(vi) Subtracting $\frac{-2}{3}$ from $\frac{5}{6}$, we get,

$\frac{5}{6}-(\frac{-2}{3})=\frac{5}{6}+\frac{2}{3}$

$=\frac{5+2\times2}{6}$

$=\frac{5+4}{6}$

$=\frac{9}{6}$

$=\frac{3}{2}$ 

(vii) Subtracting $\frac{-6}{7}$ from $\frac{-13}{14}$, we get,

$\frac{-13}{14}-(\frac{-6}{7})=\frac{-13}{14}+\frac{6}{7}$

$=\frac{-13+6\times2}{14}$

$=\frac{-13+12}{14}$

$=\frac{-1}{14}$   

(viii) Subtracting $\frac{-8}{33}$ from $\frac{-7}{22}$, we get,

$\frac{-7}{22}-(\frac{-8}{33})=\frac{-7}{22}+\frac{8}{33}$

$=\frac{-7\times3+8\times2}{66}$

$=\frac{-21+16}{66}$

$=\frac{-5}{66}$   

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Updated on: 10-Oct-2022

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