# Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:(i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$(ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$(iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$(iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$

To do:

We have to express each of the given sums as a rational number.

Solution:

(i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}=(\frac{2}{5}+\frac{-4}{5})+(\frac{7}{3}+\frac{-1}{3})$

$=(\frac{2-4}{5})+(\frac{7-1}{3})$

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

$=\frac{-2}{5}+\frac{2}{1}$

$=\frac{2}{1}-\frac{2}{5}$

$=\frac{10-2}{5}$

$=\frac{8}{5}$

(ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}=(\frac{3}{7}+\frac{-11}{7})+(\frac{-4}{9}+\frac{7}{9})$

$=\frac{3-11}{7}+\frac{-4+7}{9}$

$=\frac{-8}{7}+\frac{3}{9}$

$=\frac{-8}{7}+\frac{1}{3}$

$=\frac{-24+7}{21}$

$=\frac{-17}{21}$

(iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}=(\frac{2}{5}+\frac{4}{5})+(\frac{8}{3}+\frac{-2}{3})+\frac{-11}{15}$

$=\frac{2+4}{5}+\frac{8-2}{3}+\frac{-11}{15}$

$=\frac{6}{5}+\frac{6}{3}+\frac{-11}{15}$

$=\frac{18+30-11}{15}$

$=\frac{48-11}{15}$

$=\frac{37}{15}$

(iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}=(\frac{4}{7}+\frac{-13}{7})+(\frac{-8}{9}+\frac{17}{21})+0$

$=\frac{4-13}{7}+\frac{-56+51}{63}$

$=\frac{-9}{7}+\frac{-5}{63}$

$=\frac{-81-5}{63}$

$=\frac{-86}{63}$

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Updated on: 10-Oct-2022

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