Question

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i)
(ii)
(iii)
(iv)

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \text{We have:} \\ \frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3} \\ =(\frac{2}{5}+\frac{-4}{5})+(\frac{7}{3}+\frac{-1}{3}) \\ =\left(\frac{2-4}{5}\right)+\left(\frac{7-1}{3}\right) \\ =\frac{-2}{5}+\frac{6}{3} \\ =\frac{-6+30}{15} \\ =\frac{24}{15} \\ =\frac{8}{5} \\ \left(\mathrm{ii}\right) \\ \text{We have:} \\ \frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9} \\ =(\frac{3}{7}+\frac{-11}{7})+(\frac{-4}{9}+\frac{7}{9}) \\ =\left(\frac{3-11}{7}\right)+\left(\frac{-4+7}{9}\right) \\ =\frac{-8}{7}+\frac{3}{9} \\ =\frac{-72+21}{63} \\ =\frac{-51}{63} \\ =\frac{-17}{21} \\ \left(\mathrm{iii}\right) \\ \text{We have:} \\ \frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3} \\ =(\frac{2}{5}+\frac{4}{5})+(+\frac{8}{3}+\frac{-2}{3})+\frac{-11}{15} \\ =\left(\frac{2+4}{5}\right)+\left(\frac{8-2}{3}\right)+\frac{-11}{15} \\ =\frac{6}{5}+\frac{6}{3}+\frac{-11}{15} \\ =\frac{18+30-11}{15} \\ =\frac{37}{15} \\ \left(\mathrm{iv}\right) \\ \text{We have:} \\ \frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21} \\ =(\frac{4}{7}+\frac{-13}{7})+\left(\frac{-8}{9}\right)+\frac{17}{21} \\ =\left(\frac{4-13}{7}\right)+\left(\frac{-8}{9}\right)+\frac{17}{21} \\ =\frac{-9}{7}+\frac{-8}{9}+\frac{17}{21} \\ =\frac{-81-56+51}{63} \\ =\frac{-86}{63} \end{gathered}$$