Divide the sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$ by the difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$.

The sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$

$\dfrac{5}{8} + \dfrac{-11}{12}$

LCM of 8 and 12 is 2 x 2 x 2 x 3 = 24

$= \dfrac{5 \times 3}{8 \times 3} + \dfrac{-11 \times 2}{12 \times 2}\\[1em] = \dfrac{15}{24} + \dfrac{-22}{24}\\[1em] = \dfrac{15 + (-22)}{24}\\[1em] = \dfrac{-7}{24}$

The difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$

$\dfrac{3}{7} - \dfrac{5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$= \dfrac{3 \times 2}{7 \times 2} - \dfrac{5 \times 1}{14 \times 1}\\[1em] = \dfrac{6}{14} - \dfrac{5}{14}\\[1em] = \dfrac{6 - 5}{14}\\[1em] = \dfrac{1}{14}$

Dividing the sum of $\dfrac{5}{8}$ and $\dfrac{-11}{12}$ by the difference of $\dfrac{3}{7}$ and $\dfrac{5}{14}$,

$\dfrac{-7}{24} ÷ \dfrac{1}{14}\\[1em] = \dfrac{-7}{24} \times \dfrac{14}{1}\\[1em] = -\dfrac{7 \times 14}{24 \times 1}\\[1em] = -\dfrac{98}{24}\\[1em] = -\dfrac{49}{12}\\[1em] = -4\dfrac{1}{12}$

$\Big(\dfrac{5}{8} + \dfrac{-11}{12}\Big) ÷ \Big(\dfrac{3}{7} - \dfrac{5}{14}\Big) = -4\dfrac{1}{12}$.