Solve the inequation, write down the solution set and represent it on a real number line:

Given,

$3x - 16 \lt \frac{2x}{5} - 3 \le -\frac{3}{5} + 2x$

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow 3x - 16 \lt \frac{2x}{5} - 3 \\[1em] \Rightarrow 3x - \frac{2x}{5} \lt 16 - 3 \\[1em] \Rightarrow \frac{15x - 2x}{5} \lt 13 \\[1em] \Rightarrow 15x - 2x \lt 13 \times 5 \\[1em] \Rightarrow 13x \lt 65 \\[1em] \Rightarrow x \lt \frac{65}{13} \\[1em] \Rightarrow x \lt 5\text{ ........(1)}\end{aligned}$$

Solving R.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{2x}{5} - 3 \le -\frac{3}{5} + 2x \\[1em] \Rightarrow \frac{2x}{5} - 2x \le -\frac{3}{5} + 3\\[1em] \Rightarrow \frac{2x - 10x}{5} \le \frac{-3 + 15}{5}\\[1em] \Rightarrow \frac{-8x}{5} \le \frac{12}{5} \\[1em] \Rightarrow -8x \le 12 \\[1em] \Rightarrow 8x \ge -12 \\[1em] \Rightarrow x \ge -\frac{12}{8} \\[1em] \Rightarrow x \ge -\frac{3}{2} \text{ ........(2)}\end{aligned}$$

From equation (1) and (2), we get :

$-\frac{3}{2} \le x \lt 5$

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