Question

The inequality $\dfrac{2x − 1}{3} ≥\dfrac{3x − 2}{4} −\dfrac{(2 − x)}{5}$ holds for $x$ belonging to

• A. $R$
• B. $(-∞,3]$
• C. $(-∞,-3]∪[3,∞)$
• D. $(-∞,2]$
• E. $(-∞,2]∪[4,∞)$

Answer 1

Answer: (D)

$(-∞,2]$

Answer 2

Given that:

$\dfrac{2x − 1}{3} ≥\dfrac{3x − 2}{4} −\dfrac{(2 − x)}{5}$

Then we can proceed as follows

$⇒20(2x − 1) ≥ 15(3x − 2) − 12(2 − x)$

$⇒40x-20 ≥ 45x-30-24+12x$

$⇒34 ≥ 1 > x$

$⇒x ≤ 2$

$\therefore x ∈ (−∞, 2]$