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Solve the following inequation: \(\frac{2x-3}{4}\) + 8 ≥ 2 + \(\frac{4x}{3}\); x ∈ R.

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\(\frac{2x-3}{4}\) + 8 ≥ 2 + \(\frac{4x}{3}\) ⇒ 12 \(\big(\frac{2x-3}{4}\big)\) + 12 x 8 ≥ 12 x 2 + \(\frac{4x}{3}\) x 12 (Multiplying each term by LCM = 12) 

⇒ 3(2x – 3) + 96 > 24 + 16x ⇒ 6x – 9 + 96 > 24 + 16x 

⇒ 6x + 87 > 24 + 16x ⇒ 87 – 24 > 16x – 6x ⇒ 63 > 10x 

⇒ 10x < 63 ⇒ x < 6.3 

∴ x ∈ (– ∞, 6.3]

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