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

2

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)