If $2(2x+3) -10≤6(x-1) $ and $\frac{2x-3}{4}+6≥3+\frac{4x}{3}, x\in N $ then number of solution for 'x' are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
$2(2x+3)-10≤6(x-1)$
$4x+6-10≤6x-6$
$4x-4≤6x-6$
$2≤2x$
$x≥1$
and,
$\frac{2x-3}{4}+6≥3+\frac{4x}{3}$
$\frac{2x-3}{4}-\frac{4x}{3}≥-3$
$\frac{3(2x-3)-16x}{12}≥-3$
$\frac{-(10x+9)-16x}{12}≥-3$
$⇒10x+9≤36$
$⇒x=2.5$
∴ No. of solutions = 2 (x = 1, 2)