What should be added to $\dfrac{-5}{6}$ to get $\dfrac{-2}{3}$ ?

Let the required number be x. Then,

$\dfrac{-5}{6} + x = \dfrac{-2}{3} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \left(\dfrac{-5}{6}\right) \\[1em] = \dfrac{-2}{3} + \left(\text{additive inverse of } \dfrac{-5}{6}\right) \\[1em] = \dfrac{-2}{3} + \dfrac{5}{6}$

L.C.M. of denominators 3 and 6 is 6.

Now, expressing each fraction with denominator 6:

$\dfrac{-2 \times 2}{3 \times 2} + \dfrac{5 \times 1}{6 \times 1} \\[1em] = \dfrac{-4}{6} + \dfrac{5}{6} \\[1em] = \dfrac{-4 + 5}{6} \\[1em] = \dfrac{1}{6}$

The required number is $\dfrac{1}{6}$.