Solve the following inequation, write the solution set and represent it on the real number line.

Given,

$3x + \frac{14}{3} \gt \frac{4x}{3} - 2 \geq 3x - 4, ; x \in R$

Solving the L.H.S. of the inequation, we get:

$$\begin{aligned}\Rightarrow 3x + \frac{14}{3} \gt \frac{4x}{3} - 2 \\[1em] \Rightarrow 3x - \frac{4x}{3} \gt -2 - \frac{14}{3} \\[1em] \Rightarrow \frac{9x - 4x}{3} \gt \frac{-6 - 14}{3} \\[1em] \Rightarrow \frac{5x}{3} \gt \frac{-20}{3} \\[1em] \Rightarrow 5x \gt \frac{-20}{3} \times 3 \\[1em] \Rightarrow 5x \gt -20 \\[1em] \Rightarrow x \gt \frac{-20}{5} \\[1em] \Rightarrow x \gt -4 \text{ ......(1)}\end{aligned}$$

Solving the R.H.S. of the inequation, we get:

$$\begin{aligned}\Rightarrow \frac{4x}{3} - 2 \geq 3x - 4 \\[1em] \Rightarrow \frac{4x}{3} - 3x \ge -4 + 2 \\[1em] \Rightarrow \frac{4x - 9x}{3} \ge -2 \\[1em] \Rightarrow \frac{-5x}{3} \ge -2 \\[1em] \Rightarrow \frac{-5x}{3} \times 3 \ge -2 \times 3 \\[1em] \Rightarrow -5x \ge -6 \\[1em] \Rightarrow -5x \times -1 \le -6 \times (-1) \\[1em] \Rightarrow 5x \le 6 \\[1em] \Rightarrow x \le \frac{6}{5} \text{ ......(2)}\end{aligned}$$

From equations (1) and (2),

Solution set = $\Big\lbrace x : -4 \lt x \le \frac{6}{5}, x \in R\Big\rbrace$

Representing on the real number line :

Hence, solution set = $\bm{\Big\lbrace x : -4 \lt x \le \frac{6}{5}, x \in R\Big\rbrace}$.