Find the sum:

$\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3}$

All the denominators are positive but different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 6, 4, 8 and 3 = 24

$\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3} = \dfrac{(-11) \times 4 + (-3) \times 6 + 5 \times 3 + (-7) \times 8}{24} \\[1em] = \dfrac{-44 - 18 + 15 - 56}{24} \\[1em] = \dfrac{15 - 118}{24} \\[1em] = \dfrac{-103}{24} \\[1em] = -4\dfrac{7}{24}$

Hence, $\mathbf{\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3} = -4\dfrac{7}{24}}$.