Question

Solve the following inequality:
$\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$

Answer 1

$\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$

$\frac{20+10(x-3)+x^2-7x+12}{(x-3)(x-4)}>0$

$\frac{x^2+3x+12}{(x-3)(x-4)}>0$

$\frac{(x+2)(x+1)}{(x-3)(x-4)}>0$

$x\in (-\infty ,-2)\cup (-1,3)\cup (4,\infty )$