Add the following rational numbers:
(i)M/b> $ \frac{3}{4} $ and $ \frac{-5}{8} $
(ii) $ \frac{5}{-9} $ and $ \frac{7}{3} $
(iii) $ -3 $ and $ \frac{3}{5} $
(iv) $ \frac{-7}{27} $ and $ \frac{11}{18} $
(v) $ \frac{31}{-4} $ and $ \frac{-5}{8} $
(vi) $ \frac{5}{36} $ and $ \frac{-7}{12} $
(vii) $ \frac{-5}{16} $ and $ \frac{7}{24} $
(viii) $ \frac{7}{-18} $ and $ \frac{8}{27} $

Academic Mathematics NCERT Class 8

To do:
We have to add the given rational numbers.
Solution:
(i) $\frac{3}{4}+\frac{-5}{8}=\frac{3\times2-5\times1}{8}$ (LCM of 4 and 8 is 8)
$=\frac{6-5}{8}$
$=\frac{1}{8}$
Therefore, $\frac{3}{4}+\frac{-5}{8}=\frac{1}{8}$.
(ii) $\frac{5}{-9}+\frac{7}{3}=\frac{-5}{9}+\frac{7}{3}$
$=\frac{-5\times1+7\times3}{9}$ (LCM of 9 and 3 is 9)
$=\frac{-5+21}{9}$
$=\frac{16}{9}$
Therefore, $\frac{5}{-9}+\frac{7}{3}=\frac{16}{9}$.
(iii) $-3+\frac{3}{5}=\frac{-3\times5}{1\times5}+\frac{3}{5}$
$=\frac{-15+3}{5}$
$=\frac{-(15-3)}{5}$
$=\frac{-12}{5}$
Therefore, $-3+\frac{3}{5}=\frac{-12}{5}$.
(iv) $\frac{-7}{27}+\frac{11}{18}=\frac{-7\times2}{27\times2}+\frac{11\times3}{18\times3}$ (LCM of 27 and 18 is 54)
$=\frac{-14}{54}+\frac{33}{54}$
$=\frac{-14+33}{54}$
$=\frac{19}{54}$
Therefore, $\frac{-7}{27}+\frac{11}{18}=\frac{19}{54}$.
(v) $\frac{31}{-4}+\frac{-5}{8}=\frac{-31\times2}{4\times2}+\frac{-5\times1}{8\times1}$ (LCM of 4 and 8 is 8)
$=\frac{-62}{8}+\frac{-5}{8}$
$=\frac{-62-5}{8}$
$=\frac{-67}{8}$
Therefore, $\frac{31}{-4}+\frac{-5}{8}=\frac{-67}{8}$.
(vi) $\frac{5}{36}+\frac{-7}{12}=\frac{5\times1}{36\times1}+\frac{-7\times3}{12\times3}$ (LCM of 36 and 12 is 36)
$=\frac{5}{36}+\frac{-21}{36}$
$=\frac{5-21}{36}$
$=\frac{-16}{36}$
$=\frac{-4}{9}$
Therefore, $\frac{5}{36}+\frac{-7}{12}=\frac{-4}{9}$.
(vii) $\frac{-5}{16}+\frac{7}{24}=\frac{-5\times3}{16\times3}+\frac{7\times2}{24\times2}$ (LCM of 16 and 24 is 48)
$=\frac{-15}{48}+\frac{14}{48}$
$=\frac{-15+14}{48}$
$=\frac{-(15-14)}{48}$
$=\frac{-1}{48}$
Therefore, $\frac{-5}{16}+\frac{7}{24}=\frac{-1}{48}$.
(viii) $\frac{7}{-18}+\frac{8}{27}=\frac{-7\times3}{18\times3}+\frac{8\times2}{27\times2}$ (LCM of 18 and 27 is 54)
$=\frac{-21}{54}+\frac{16}{54}$
$=\frac{-21+16}{54}$
$=\frac{-(21-16)}{54}$
$=\frac{-5}{54}$
Therefore, $\frac{7}{-18}+\frac{8}{27}=\frac{-5}{54}$.

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Updated on: 10-Oct-2022

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