Find the range of values of x which satisfies

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

Graph these values on number line.

Given,

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

Solving L.H.S. of the equation,

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

Solving R.H.S. of the equation,

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

From (i) and (ii) we get,

-3 ≤ x < 3.

Solution set = {x : x ∈ R and -3 ≤ x < 3}.

Solution on the number line is :