For each pair of rational number, verify commutative property of addition of rational numbers.

3 and $\dfrac{-2}{7}$

To prove:

$3 + \dfrac{-2}{7} = \dfrac{-2}{7} + 3$

Taking LHS: $3 + \dfrac{-2}{7} \\[1em] = \dfrac{3}{1} + \dfrac{-2}{7} \\[1em]$

LCM of 1 and 7 is 7

$= \dfrac{3 \times 7}{1 \times 7} + \dfrac{-2 \times 1}{7 \times 1} \\[1em] = \dfrac{21}{7} + \dfrac{-2}{7} \\[1em] = \dfrac{21 + (-2)}{7} \\[1em] = \dfrac{19}{7} \\[1em]$

Taking RHS: $\dfrac{-2}{7} + 3 \\[1em] = \dfrac{-2}{7} + \dfrac{3}{1} \\[1em]$

LCM of 7 and 1 is 7

$= \dfrac{-2 \times 1}{7 \times 1} + \dfrac{3 \times 7}{1 \times 7} \\[1em] = \dfrac{-2}{7} + \dfrac{21}{7} \\[1em] = \dfrac{(-2) + 21}{7} \\[1em] = \dfrac{19}{7} \\[1em]$

∴ LHS = RHS

Hence, $3 + \dfrac{-2}{7} = \dfrac{-2}{7} + 3$

So, the commutative property for the addition of the rational number is verified.