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

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

$$\begin{aligned}\Rightarrow -6 + \frac{12x}{5} \le \frac{7x}{5} - 3 \\[1em] \Rightarrow \frac{12x}{5} - \frac{7x}{5} \le 6 - 3 \\[1em] \Rightarrow \frac{12x - 7x}{5} \le 3 \\[1em] \Rightarrow \frac{5x}{5} \le 3 \\[1em] \Rightarrow x \le 3 \text{ ......(1)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{7x}{5} - 3 \lt \frac{2}{5} + 2x \\[1em] \Rightarrow \frac{7x}{5} - 2x \lt \frac{2}{5} + 3 \\[1em] \Rightarrow \frac{7x - 10x}{5} \lt \frac{2 + 15}{5} \\[1em] \Rightarrow \frac{-3x}{5} \lt \frac{17}{5} \\[1em] \Rightarrow -3x \lt 17 \\[1em] \Rightarrow x \gt \frac{-17}{3} \\[1em] \Rightarrow x \gt -5\frac{2}{3} \text{ ......(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = ${x : -5\frac{2}{3} \lt x \le 3, x \in R}$

Hence, solution set = {$\bm{x} : \mathbf{-5\frac{2}{3}} \lt \bm{x} \le 3, \bm{x}$ ∈ R}.