If x is a negative integer, then the solution set for the linear inequation $\dfrac{2}{5} + \dfrac{1}{5}(x + 1) \gt 0$ is :

1. {0, -1, -2}

2. {-2, -1}

3. {-1, -2, -3}

4. {-2, -3, -4}

Solving,

$\Rightarrow \dfrac{2}{5} + \dfrac{1}{5}(x + 1) \gt 0 \\[1em] \Rightarrow \dfrac{2}{5} + \dfrac{1}{5}x + \dfrac{1}{5} \gt 0 \\[1em] \Rightarrow \dfrac{1}{5}x + \dfrac{3}{5} \gt 0 \\[1em] \Rightarrow \dfrac{1}{5}x \gt -\dfrac{3}{5} \\[1em] \Rightarrow x \gt -\dfrac{3}{5} \times 5 \\[1em] \Rightarrow x \gt -3.$

Since, x is a negative integer and x > -3.

Solution set = {-2, -1}.

Hence, Option 2 is the correct option.