Solve the following inequation and represent the solution on the number line :

$\dfrac{3x}{5} + 2 \lt x + 4 \le \dfrac{x}{2} + 5$, x ∈ R

Given, $\dfrac{3x}{5} + 2 \lt x + 4 \le \dfrac{x}{2} + 5$

Solving L.H.S. of the inequation,

$\Rightarrow \dfrac{3x}{5} + 2 \lt x + 4 \\[1em] \Rightarrow \dfrac{3x + 10}{5} \lt x + 4 \\[1em] \Rightarrow 3x + 10 \lt 5(x + 4)\\[1em] \Rightarrow 3x + 10 \lt 5x + 20\\[1em] \Rightarrow 3x + 10 - 5x \lt 20\\[1em] \Rightarrow 10 - 2x \lt 20\\[1em] \Rightarrow -2x \lt 20 - 10\\[1em] \Rightarrow -2x \lt 10\\[1em] \Rightarrow 2x \gt -10\\[1em] \Rightarrow x \gt -\dfrac{10}{2}\\[1em] \Rightarrow x \gt -5 ……………..(1)$

Solving R.H.S. of the inequation,

$\Rightarrow x + 4 \le \dfrac{x}{2} + 5\\[1em] \Rightarrow x + 4 \le \dfrac{x + 10}{2}\\[1em] \Rightarrow 2(x + 4) \le x + 10\\[1em] \Rightarrow 2x + 8 \le x + 10\\[1em] \Rightarrow 2x + 8 - x \le 10\\[1em] \Rightarrow x + 8 \le 10\\[1em] \Rightarrow x \le 10 - 8\\[1em] \Rightarrow x \le 2 ………………………….(2)$

From (1) and (2), we get

-5 < x ≤ 2

Since, x ∈ R

The solution set of x = {x : x ∈ R, -5 < x ≤ 2}

Hence, solution set = {x : x ∈ R, -5 < x ≤ 2}.