Mathematics

Solve the following inequation and write the solution and represent it on the real number line.
3 - 2x ≥ x + $\dfrac{1 - x}{3}$ > $\dfrac{2x}{5}$ , x ∈ R.

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To prove:

3 - 2x ≥ x + $\dfrac{1 - x}{3}$ > $\dfrac{2x}{5}$

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

⇒ 3 - 2x ≥ x + $\dfrac{1 - x}{3}$

⇒ 3 - 2x ≥ $\dfrac{3x + 1 - x}{3}$

⇒ 3(3 - 2x) ≥ 2x + 1

⇒ 9 - 6x ≥ 2x + 1

⇒ 2x + 6x ≤ 9 - 1

⇒ 8x ≤ 8

⇒ x ≤ $\dfrac{8}{8}$

⇒ x ≤ 1 …………(1)

Solving R.H.S. of the above equation, we get :

⇒ x + $\dfrac{1 - x}{3} \gt \dfrac{2x}{5}$

⇒ $\dfrac{3x + 1 - x}{3} \gt \dfrac{2x}{5}$

⇒ 5(2x + 1) > 3 × 2x

⇒ 10x + 5 > 6x

⇒ 10x - 6x > -5

⇒ 4x > -5

⇒ x > $-\dfrac{5}{4}$ ………..(2)

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

Solution set = {x : $-\dfrac{5}{4}$ < x ≤ 1, x ∈ R}

Representation of solution set on real number line is :

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