Question

Solve the following inequality:
$2+\frac{3}{x+1}>\frac{2}{x}$

Answer 1

$2+\frac{3}{x+1}>\frac{2}{x}$

$2+\frac{3}{x+1}-\frac{2}{x}>0$

$\frac{2\left(x\right(x+1\left)\right)+3x-2(x+1)}{x(x+1)}>0$

$\frac{2{(x}^2+x)+3x-2x-2)}{x(x+1)}>0$

$\frac{2x^2+2x+3x-2x-2}{x(x+1)}>0$

$\frac{2x^2+3x-2}{x(x+1)}>0$

$\frac{2x^2+4x-x-2}{x(x+1)}>0$

$\frac{2x(x+2)-1(x+2)}{x(x+1)}>0$

$\frac{(2x-1)(x+2)}{(x-0)(x+1)}>0(x=1,\frac{(2-1)\left(3\right)}{1\times 2}>0)$

$x\epsilon (-\infty ,-2)\bigcup (-1,0)\bigcup (\frac{1}{2},\infty )$