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

-3(x - 7) ≥ 15 - 7x > $\dfrac{x + 1}{3}$, x ∈ R

Given,

-3(x - 7) ≥ 15 - 7x > $\dfrac{x + 1}{3}$

Solving L.H.S. of the inequation,

⇒ -3(x - 7) ≥ 15 - 7x

⇒ -3x + 21 ≥ 15 - 7x

⇒ -3x + 7x ≥ 15 - 21

⇒ 4x ≥ -6

⇒ x ≥ -$\dfrac{6}{4}$

⇒ x ≥ -$\dfrac{3}{2}$

⇒ x ≥ -1.5 …..(i)

Solving L.H.S. of the inequation,

$\Rightarrow 15 - 7x \gt \dfrac{x + 1}{3} \\[1em] \Rightarrow 3(15 - 7x) \gt x + 1 \\[1em] \Rightarrow x + 1 \lt 45 - 21x \\[1em] \Rightarrow x + 21x \lt 45 - 1 \\[1em] \Rightarrow 22x \lt 44 \\[1em] \Rightarrow x \lt 2 \space …..(\text{ii})$

From (i) and (ii) we get,

-1.5 ≤ x < 2

Solution set = {x : x ∈ R and -1.5 ≤ x < 2}.

Solution on the number line is :