Question

Solve the following inequality:
$\frac{4}{x+2}>3-x$

Answer 1

$\frac{4}{x+2}>3-x\frac{4}{x+2}-(3-x)>0\frac{4-(3x+6-x^2-2x)}{x+2}>0\frac{(x-2)(x+1)}{x+2}>0x\in (-2,-1)\cup (2,\infty )$