Question

Solve the following systems of inequalities for all $xϵR$

$2(2x+3)-10<6(x-2)$ , $\frac{2x-3}{4}+6\ge 4+\frac{4x}{3}$

Answer 1

$2(2x+3)-10<6(x-2)$

$⟹4x+6-10<6x-12$

$⟹4x-4-6x+12<0$

$⟹-2x+8<0$

$⟹2x-8>0$

$⟹2x>8$

$⟹x>4$............(i)

$\frac{2x-3}{4}+6\ge 4+\frac{4x}{3}$

Multiplying with $12$ on both sides, we get

$⟹3(2x-3)+72\ge 48+16x$

$⟹6x-9+72\ge 48+16x$

$⟹16x+48\le 6x-9+72$

$⟹16x+48-6x+9-72\le 0$

$⟹10x-15\le 0$

$⟹x\le \frac{15}{10}$

$⟹x\le \frac{3}{2}$............................(ii)

From (i) & (ii) there is no common solution for $x$.