Question

Solve the following inequality:
$\frac{2(x-3)}{x(x-6)}⩽\frac{1}{x-1}$

Answer 1

$\frac{2(x-3)}{x(x-6)}\le \frac{1}{x-1}$

$\frac{2(x-3)}{x(x-6)}-\frac{1}{x-1}\le 0$

$\frac{2(x^2-4x+3)-(x^2-6x)}{x(x-1)(x-6)}\le 0$

$\frac{x^2-2x+6}{x(x-1)(x-6)}<0$

$\frac{1}{x(x-1)(x-6)}<0$

$\therefore x\in (-\infty ,0)\cup (1,6)$