Subtract the first rational number from the second in each of the following:

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

Solution:

(i) $\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}=\frac{2}{8}=\frac{1}{4}$

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

(iii) $\frac{-9}{11}-\frac{-2}{11}=\frac{(-9)-(-2)}{11}=\frac{-9+2}{11}=\frac{-7}{11}$

(iv) $\frac{-4}{13}-\frac{11}{13}=\frac{-4-11}{13}=\frac{-15}{13}$

(v) $\frac{-3}{8}-\frac{1}{4}=\frac{-3-2}{8}=\frac{-5}{8}$

(vi) $\frac{5}{6}-\frac{-2}{3}=\frac{5}{6}-\frac{-4}{6}=\frac{5-(-4)}{6}=\frac{5+4}{6}=\frac{9}{6}=\frac{3}{2}$

(vii) $\frac{-13}{14}-\frac{-6}{7}=\frac{-13-(-12)}{14}=\frac{-13+12}{14}=\frac{-1}{14}$

(viii) $\frac{-7}{22}-\frac{-8}{33}=\frac{-21}{66}-\frac{-16}{66}=\frac{(-21)-(-16)}{66}=\frac{-21+16}{66}=\frac{-5}{66}$