Solve the inequation given below. Write the solution set and represent it on the number line:

$-\dfrac{2}{3} \lt 1 + \dfrac{x}{3} \le \dfrac{2}{3}, x ∈ R$

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$\Rightarrow 1 + \dfrac{x}{3} \le \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{3 + x}{3} \le \dfrac{2}{3} \\[1em] \Rightarrow x + 3 \le 2 \\[1em] \Rightarrow x \le 2 - 3 \\[1em] \Rightarrow x \le -1 \text{ ……….(2)}$

From (1) and (2) we get,

-5 < x ≤ -1

Since, x ∈ R

Hence, solution set = {x : -5 < x ≤ -1, x ∈ R}.

Solution on the number line is :