# Solve the following inequation: (2x-3)/4 + 8 ≥ 2 + 4x/3; x ∈ R.

4.5k views

closed

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

+1 vote
by (48.4k points)
selected by

$\frac{2x-3}{4}$ + 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]