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

$2x - 1 \ge x + \dfrac{(7 - x)}{3} \gt 2$, x ∈ R

Given,

$\Rightarrow 2x - 1 \ge x + \dfrac{(7 - x)}{3} \gt 2$

Solving L.H.S. of the inequation,

$\Rightarrow 2x - 1 \ge x +\dfrac{(7 - x)}{3} \\[1em] \Rightarrow 2x - x \ge \dfrac{(7 - x)}{3} + 1 \\[1em] \Rightarrow x \ge \dfrac{7 - x + 3}{3} \\[1em] \Rightarrow x \ge \dfrac{10 - x}{3} \\[1em] \Rightarrow 3x \ge 10 - x \\[1em] \Rightarrow 3x + x \ge 10 \\[1em] \Rightarrow 4x \ge 10 \\[1em] \Rightarrow x \ge \dfrac{10}{4} \\[1em] \Rightarrow x \ge \dfrac{5}{2} \text{ ………(1)}$

Solving R.H.S. of the inequation,

$\Rightarrow x + \dfrac{(7 - x)}{3} \gt 2 \\[1em] \Rightarrow \dfrac{3x + 7 - x}{3} \gt 2 \\[1em] \Rightarrow \dfrac{2x + 7}{3} \gt 2 \\[1em] \Rightarrow 2x + 7 \gt 6 \\[1em] \Rightarrow 2x \gt 6 - 7 \\[1em] \Rightarrow 2x \gt -1 \\[1em] \Rightarrow x \gt -\dfrac{1}{2} \\[1em] \Rightarrow x \gt -0.5 \text{ ………..(2)}$

From (1) and (2) we get,

x ≥ $\dfrac{5}{2}$

Hence, solution set = {x : x ≥ $\dfrac{5}{2}$, x ∈ R}.

Solution on the number line is :