Simplify :

(i) $\Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{14}{-15}\Big)$

(ii) $\Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big)$

(iii) $\Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big)$

(iv) $\Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big)$

(i) $\Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{14}{-15}\Big)$

First, express $\dfrac{14}{-15}$ with a positive denominator: $\dfrac{14 \times (-1)}{-15 \times (-1)} = \dfrac{-14}{15}$.

We have:

$\phantom{=} \Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{-14}{15}\Big) \\[1em] = \Big(\dfrac{1}{1} \times \dfrac{1}{4}\Big) + \Big(\dfrac{-1}{1} \times \dfrac{-2}{5}\Big) \\[1em] = \Big(\dfrac{1}{1} \times \dfrac{1}{4}\Big) + \Big(\dfrac{-1}{1} \times \dfrac{-2}{5}\Big) \\[1em] = \dfrac{1}{4} + \dfrac{2}{5}$

L.C.M. of 4 and 5 is 20.

Now, expressing each fraction with denominator 20:

$= \dfrac{1 \times 5}{4 \times 5} + \dfrac{2 \times 4}{5 \times 4} \\[1em] = \dfrac{5}{20} + \dfrac{8}{20} \\[1em] = \dfrac{5 + 8}{20} \\[1em] = \dfrac{13}{20}$

Hence the answer is $\dfrac{13}{20}$

(ii) $\Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big)$

We have:

$\phantom{=} \Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big) \\[1em] = \Big(\dfrac{-2}{1} \times \dfrac{-4}{1}\Big) + \Big(\dfrac{-3}{5} \times \dfrac{3}{4}\Big) \\[1em] = \dfrac{8}{1} + \dfrac{-9}{20}$

L.C.M. of 1 and 20 is 20.

Now, expressing each fraction with denominator 20:

$= \dfrac{8 \times 20}{1 \times 20} + \dfrac{-9}{20} \\[1em] = \dfrac{160}{20} + \dfrac{-9}{20} \\[1em] = \dfrac{160 - 9}{20} \\[1em] = \dfrac{151}{20}$

Hence the answer is $\dfrac{151}{20}$

(iii) $\Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big)$

We have:

$\phantom{=} \Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big) \\[1em] = \Big(\dfrac{3}{5} \times \dfrac{-3}{4}\Big) - \Big(\dfrac{1}{4} \times \dfrac{-1}{2}\Big) \\[1em] = \Big(\dfrac{3 \times (-3)}{5 \times 4}\Big) - \Big(\dfrac{1 \times (-1)}{4 \times 2}\Big) \\[1em] = \Big(\dfrac{-9}{20}\Big) - \Big(\dfrac{-1}{8}\Big)$

L.C.M. of 20 and 8 is 40.

Now, expressing each fraction with denominator 40:

$= \dfrac{-9 \times 2}{20 \times 2} - \dfrac{-1 \times 5}{8 \times 5} \\[1em] = \dfrac{-18}{40} - \Big(\dfrac{-5}{40}\Big) \\[1em] = \dfrac{-18}{40} + \dfrac{5}{40} \\[1em] = \dfrac{-18 + 5}{40} \\[1em] = \dfrac{-13}{40}$

Hence the answer is $\dfrac{-13}{40}$.

(iv) $\Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big)$

We have:

$\phantom{=} \Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big) \\[1em] = \Big(\dfrac{-2}{1} \times \dfrac{-2}{1}\Big) - \Big(\dfrac{-1}{3} \times \dfrac{1}{2}\Big) \\[1em] = \Big(\dfrac{-2 \times (-2)}{1 \times 1}\Big) - \Big(\dfrac{-1 \times 1}{3 \times 2}\Big) \\[1em] = \dfrac{4}{1} - \Big(\dfrac{-1}{6}\Big) \\[1em] = \dfrac{4}{1} + \dfrac{1}{6}$

L.C.M. of 1 and 6 is 6.

Now, expressing each fraction with denominator 6:

$= \dfrac{4 \times 6}{1 \times 6} + \dfrac{1}{6} \\[1em] \dfrac{24}{6} + \dfrac{1}{6} \\[1em] = \dfrac{24 + 1}{6} \\[1em] = \dfrac{25}{6}$

Hence the answer is $\dfrac{25}{6}$.