Question

Evaluate :
$\left(i\right)\frac{2}{3}-\frac{4}{5}\left(ii\right)\frac{-4}{9}-\frac{2}{-3}\left(iii\right)-1-\frac{4}{9}\left(iv\right)\frac{-2}{7}-\frac{3}{-14}\left(v\right)\frac{-5}{18}-\frac{-2}{9}\left(vi\right)\frac{5}{21}-\frac{-13}{42}$

Answer 1

$\left(i\right)\frac{2}{3}-\frac{4}{5}$

$\therefore \text{LCM of 3 and 5=15}=\frac{2\times 5}{3\times 5}-\frac{4\times 3}{5\times 3}(\because \text{LCM of and 5=15})=\frac{10-12}{15}=\frac{-2}{15}\left(ii\right)\frac{-4}{9}-\frac{2}{-3}$

$=\frac{-4\times 1}{9\times 1}-\frac{(-2\times 3)}{3\times 3}(\because \text{LCM of 3 and 9=9})=\frac{-4+6}{9}=\frac{2}{9}\left(iii\right)-1-\frac{4}{9}=\frac{-1\times 9}{1\times 9}-\frac{4\times 1}{9\times 1}=\frac{-9-4}{9}=\frac{-13}{9}\left(iv\right)\frac{-2}{7}-\frac{3}{-14}$

$\therefore \text{LCM of 7 and 14=14}=\frac{-2\times 2}{7\times 2}-\frac{(-3\times 1)}{14\times 1}(\because \text{LCM of 7 and 14=14})=\frac{-4+3}{14}=\frac{-1}{14}\left(v\right)\frac{-5}{18}-\frac{-2}{9}$

$\therefore \text{LCM of9 and 18}=2\times 2\times 3\times 3=36=\frac{-5\times 2}{18\times 2}-\frac{(-2\times 4)}{9\times 4}(\because \text{LCM of 18 and 9=36})=\frac{-10+8}{36}=\frac{-1}{18}\left(vi\right)\frac{5}{21}-\frac{-13}{42}$

$\therefore \text{LCM of 21,42=}2\times 3\times 7=42=\frac{5\times 2}{21\times 2}-\frac{(-13\times 1)}{42\times 1}(\because \text{LCM of 21 and 42=42})=\frac{10+13}{42}=\frac{23}{42}$