If $m = -\dfrac{7}{9}$ and $n = \dfrac{5}{6}$, verify that:

-(m + n) = (-m) + (-n)

To prove:

-(m + n) = (-m) + (-n)

LHS:

$-(m + n)\\[1em] -\Big(-\dfrac{7}{9} + \dfrac{5}{6}\Big)$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$= -\Big(-\dfrac{7 \times 2}{9 \times 2} + \dfrac{5 \times 3}{6 \times 3}\Big)\\[1em] = -\Big(-\dfrac{14}{18} + \dfrac{15}{18}\Big)\\[1em] = -\Big(\dfrac{-14 + 15}{18}\Big)\\[1em] = -\Big(\dfrac{1}{18}\Big)$

RHS:

$(-m) + (-n)\\[1em] = -\Big(-\dfrac{7}{9}\Big) + \Big(-\dfrac{5}{6}\Big)\\[1em] = \Big(\dfrac{7}{9}\Big) + \Big(-\dfrac{5}{6}\Big)$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$= \Big(\dfrac{7 \times 2}{9 \times 2}\Big) + \Big(-\dfrac{5 \times 3}{6 \times 3}\Big)\\[1em] = \Big(\dfrac{14}{18}\Big) + \Big(-\dfrac{15}{18}\Big)\\[1em] = \Big(\dfrac{14 + (-15)}{18}\Big)\\[1em] = \Big(\dfrac{-1}{18}\Big)$

Hence, LHS = RHS

$\therefore -(m + n) = (-m) + (-n)$