Question

Solve the following inequality:
$\frac{2(x-4)}{(x-1)(x-7)}⩾\frac{1}{x-2}$

Answer 1

$=>\frac{2(x-4)(x-2)}{(x-1)(x-7)}\ge 1$

$=>\frac{2(x-4)(x-2)}{(x-1)(x-7)}-1\ge 0$

$=>\frac{2(x^2-6x+8)-(x^2-8x-7)}{(x-1)(x-7)}\ge 0$

$=>\frac{x^2-4x+9}{(x-1)(x-7)}\ge 0$

$\because x^2-4x+9$ is always positive for all value of x as $\infty =13-36<0$

$xϵ(-\infty ,1)\cup (7,\infty )$.