Firstly let us consider the left hand side (LHS)
$$\dfrac {-12}{5} + \dfrac {2}{7}$$
Since the denominators are different we need to take the LCM
LCM of the denominators $$5$$ and $$7$$ is $$35$$
So by multiplying the numerator and denominator by $$7$$ we get, $$\dfrac {-12}{5} = \dfrac {-12\times 7}{5\times 7} = \dfrac {-84}{35}$$
And by multiplying the numerator and denominator by $$5$$ we get, $$\dfrac {2}{7} = \dfrac {2\times 5}{7\times 5} = \dfrac {10}{35}$$
$$\therefore LHS= \dfrac {-12}{5} + \dfrac {2}{7} = \dfrac {-84}{35} + \dfrac {10}{35}$$
Similarly,
We have right hand side (RHS):
$$\dfrac {2}{7} + \dfrac {-12}{5}$$
Since the denominators are different we need to take the LCM
LCM of the denominators $$7$$ and $$5$$ is $$35$$
So by multiplying the numerator and denominator by $$5$$ we get, $$\dfrac {2}{7} = \dfrac {2\times 5}{7\times 5} = \dfrac {10}{35}$$
And by multiplying the numerator and denominator by $$7$$ we get, $$\dfrac {-12}{5} = \dfrac {-12\times 7}{5\times 7} = \dfrac {-84}{35}$$
$$\therefore RHS =\dfrac {2}{7} + \dfrac {-12}{5} = \dfrac {10}{35} + \dfrac {-84}{35}$$
Since it has the same denominators we can add directly $$\dfrac {10}{35} + \dfrac {-84}{35} = \dfrac {10 + (-84)}{35} = \dfrac {10 - 84}{35}$$
$$\Rightarrow \dfrac {-74}{35}$$
$$LHS = RHS$$ i.e. $$\dfrac {-75}{35} = \dfrac {-74}{35}\therefore \dfrac {-12}{5} + \dfrac {2}{7} = \dfrac {2}{7} + \dfrac {-12}{5}$$ is verified.