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

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

$\Rightarrow -2\dfrac{2}{3} \le x + \dfrac{1}{3}\lt 3\dfrac{1}{3} \\[1em] \Rightarrow -\dfrac{8}{3} \le x+\dfrac{1}{3} \lt \dfrac{10}{3}$

Solving L.H.S. of the inequation,

$\Rightarrow -\dfrac{8}{3} \le x+\dfrac{1}{3} \\[1em] \Rightarrow x+\dfrac{1}{3}\ge-\dfrac{8}{3} \\[1em] \Rightarrow x \ge -\dfrac{8}{3}-\dfrac{1}{3} \\[1em] \Rightarrow x \ge -\dfrac{9}{3} \\[1em] \Rightarrow x \ge -3 \text{ …..(1)}$

Solving R.H.S. of the inequation,

$\Rightarrow x + \dfrac{1}{3} \lt \dfrac{10}{3} \\[1em] \Rightarrow x \lt \dfrac{10}{3} - \dfrac{1}{3} \\[1em] \Rightarrow x \lt \dfrac{9}{3} \\[1em] \Rightarrow x \lt 3 \text{ ….(2)}$

From (1) and (2) we get,

-3 ≤ x < 3

Since, x ∈ R

Hence, solution set = {x : -3 ≤ x < 3, x ∈ R}.

Solution on the number line is :