Verify each of the following:

Solution:

(i) $\left(\frac{5}{7} \times \frac{12}{13}\right) \times \frac{7}{18}=\frac{5}{7} \times\left(\frac{12}{13} \times \frac{7}{18}\right)$

$\mathrm{LHS}=\left(\frac{5}{7} \times \frac{12}{13}\right) \times \frac{7}{18}$

$=\frac{5 \times 12}{7 \times 13} \times \frac{7}{18}$

$=\frac{60}{91} \times \frac{7}{18}$

$=\frac{420}{1638}$

$=\frac{10}{39}$

$\operatorname{RHS}=\frac{5}{7} \times\left(\frac{12}{13} \times \frac{7}{18}\right)$

$=\frac{5}{7} \times \frac{12 \times 7}{13 \times 18}$

$=\frac{5}{7} \times \frac{84}{234}$

$=\frac{420}{1638}$

$=\frac{10}{39}$

$\therefore\left(\frac{5}{7} \times \frac{12}{13}\right) \times \frac{7}{18}=\frac{5}{7} \times\left(\frac{12}{13} \times \frac{7}{18}\right)$

(ii) $\frac{-13}{24} \times\left(\frac{-12}{5} \times \frac{35}{36}\right)=\left(\frac{-13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36}$

$\mathrm{LHS}=\frac{-13}{24} \times\left(\frac{-12}{5} \times \frac{35}{36}\right)$

$=\frac{-13}{24} \times \frac{(-12) \times 35}{5 \times 36}$

$=\frac{-13}{24} \times \frac{-420}{180}$

$=\frac{5460}{4320}$

$=\frac{91}{72}$

$\mathrm{RHS}=\left(\frac{-13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36}$

$=\frac{(-13) \times(-12)}{24 \times 5} \times \frac{35}{36}$

$=\frac{156}{120} \times \frac{35}{36}$

$=\frac{156 \times 35}{120 \times 36}$

$=\frac{5460}{4320}$

$=\frac{91}{72}$

$\therefore \frac{-13}{24} \times\left(\frac{-12}{5} \times \frac{35}{36}\right)=\left(\frac{-13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36}$

(iii) $\left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4}=\frac{-9}{5} \times\left(\frac{-10}{3} \times \frac{21}{-4}\right)$

$\mathrm{LHS}=\left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4}$

$=\frac{(-9) \times(-10)}{5 \times 3} \times \frac{21}{-4}$

$=\frac{90}{15} \times \frac{21}{-4}$

$=\frac{90 \times 21}{15 \times(-4)}$

$=-\frac{1890}{60}$

$=-\frac{63}{2}$

$\mathrm{RHS}=\frac{-9}{5} \times\left(\frac{-10}{3} \times \frac{21}{-4}\right)$

$=\frac{-9}{5} \times \frac{(-10) \times 21}{3 \times(-4)}$

$=\frac{-9}{5} \times \frac{210}{12}$

$=\frac{(-9) \times 210}{5 \times 12}$

$=-\frac{1890}{60}$

$=\frac{-63}{2}$

$\therefore\left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4}=\frac{-9}{5} \times\left(\frac{-10}{3} \times \frac{21}{-4}\right)$