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

$-3 + x \le \dfrac{8x}{3} + 2 \le \dfrac{14}{3} + 2x$ ,x ∈ I

Given,

$-3 + x \le \dfrac{8x}{3} + 2 \le \dfrac{14}{3} + 2x$

Solving L.H.S. of the inequation,

$\Rightarrow -3 + x \le \dfrac{8x}{3} + 2 \\[1em] \Rightarrow x - \dfrac{8x}{3} \le 2 + 3 \\[1em] \Rightarrow x - \dfrac{8x}{3} \le 5 \\[1em] \Rightarrow \dfrac{3x-8x}{3} \le 5 \\[1em] \Rightarrow 3x - 8x \le 5 \times 3 \\[1em] \Rightarrow -5x \le 15 \\[1em]$

Dividing by -5 on both sides we get,

⇒ x ≥ -3 (As on dividing by negative number the sign reverses) ……..(1)

Solving R.H.S. of the inequation,

$\Rightarrow \dfrac{8x}{3} + 2 \le \dfrac{14}{3} + 2x \\[1em] \Rightarrow \dfrac{8x + 6}{3} \le \dfrac{14 + 6x}{3} \\[1em] \Rightarrow 8x + 6 \le 14 + 6x\\[1em] \Rightarrow 8x - 6x \le 14 - 6 \\[1em] \Rightarrow 2x \le 8 \\[1em] \Rightarrow x \le \dfrac{8}{2} \\[1em] \Rightarrow x \le 4 \text{ …….(2)}$

From (1) and (2) we get,

-3 ≤ x ≤ 4

Since, x ∈ I

Hence, solution set = {-3, -2, -1, 0, 1, 2, 3, 4}.

Solution on the number line is :