Question

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) $\frac{-8}{33},\frac{-7}{22}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}=\frac{2}{8}=\frac{1}{4} \\ \left(\mathrm{ii}\right) \\ \frac{4}{9}-\frac{-7}{9}=\frac{4-(-7)}{9}=\frac{4+7}{9}=\frac{11}{9} \\ \left(\mathrm{iii}\right) \\ \frac{-9}{11}-\frac{-2}{11}=\frac{(-9)-(-2)}{11}=\frac{-9+2}{11}=\frac{-7}{11} \\ \left(\mathrm{iv}\right) \\ \frac{-4}{13}-\frac{11}{13}=\frac{-4-11}{13}=\frac{-15}{13} \\ \left(\mathrm{v}\right) \\ \frac{-3}{8}-\frac{1}{4}=\frac{-3-2}{8}=\frac{-5}{8} \\ \left(\mathrm{vi}\right) \\ \frac{5}{6}-\frac{-2}{3}=\frac{5}{6}-\frac{-4}{6}=\frac{5-(-4)}{6}=\frac{5+4}{6}=\frac{9}{6}=\frac{3}{2} \\ \left(\mathrm{vii}\right) \\ \frac{-13}{14}-\frac{-6}{7}=\frac{-13-(-12)}{14}=\frac{-13+12}{14}=\frac{-1}{14} \\ \left(\mathrm{viii}\right) \\ \frac{-7}{22}-\frac{-8}{33}=\frac{-21}{66}-\frac{-16}{66}=\frac{(-21)-(-16)}{66}=\frac{-21+16}{66}=\frac{-5}{66} \end{gathered}$$