Question

solve 4÷(x+1) ≤ 3 ≤ 6 ÷ (x+1) , x>0​

Answer 1

GIVEN :–

$\\\bf \implies \dfrac{4}{x + 1} \leqslant 3 \leqslant \dfrac{6}{x + 1}$

• And –

$\\\bf \implies x > 0$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\\bf \implies \dfrac{4}{x + 1} \leqslant 3 \leqslant \dfrac{6}{x + 1}$

• First inequality –

$\\\bf \implies \dfrac{4}{x + 1} \leqslant 3$

$\\\bf \implies 4 \leqslant 3(x + 1)$

$\\\bf \implies3(x + 1) \geqslant 4$

$\\\bf \implies3x +3\geqslant 4$

$\\\bf \implies3x\geqslant 4 - 3$

$\\\bf \implies3x\geqslant1$

$\\\bf \implies x\geqslant \dfrac{1}{3} \: \: \: \: - - - -eq.(1)$

• Second inequality –

$\\\bf \implies 3 \leqslant \dfrac{6}{x + 1}$

$\\\bf \implies 3(x + 1) \leqslant6$

$\\\bf \implies 3x + 3 \leqslant6$

$\\\bf \implies 3x\leqslant6 - 3$

$\\\bf \implies 3x\leqslant3$

$\\\bf \implies x\leqslant \dfrac{3}{3}$

$\\\bf \implies x\leqslant1 \: \: \: \: \: - - - - eq.(2)$

• By eq.(1) & eq.(2) –

$\\ \large\implies \red{ \boxed{ \bf x\in \bigg[ \dfrac{1}{3},1 \bigg]}}$