# 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 $\frac{8}{27}$

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, $\frac{7}{-18}+\frac{8}{27}=\frac{-5}{54}$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

32 Views