$\frac{2x-3}{4}+8≥2+\frac{4x}{3}$
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : $(-∞, 6.3]$
$\frac{2x-3}{4}+8≥2+\frac{4x}{3}$
Multiplying by 12 (LCM of 4, 3) both the sides
$3(2x-3)+96≥24+16x$
$6x-16x≥24-87$
$-10x≥-63$
$x≤6.3$
∴ Solution set is $(-∞, 6.3]$