Question

Solve the following inequality:
$\frac{2x+3}{x^2+x-12}⩽\frac{1}{2}$

Answer 1

$\frac{2x+3}{x^2+x-12}\le \frac{1}{2}\frac{2x+3}{x^2+x-12}-\frac{1}{2}\le 0\frac{4x+6-x^2-x+12}{2(x+4)(x-3)}\le 0\frac{x^2-3x-18}{2(x+4)(x-3)}\ge 0\frac{(x-6)(x+3)}{(x+4)(x-3)}\ge 0x\in (-\infty ,-4)\cup (-3,3)\cup (6,\infty )$