Question

Solve the following inequality:
$\frac{7}{(x-2)(x-3)}+\frac{9}{x-3}+1<0$

Answer 1

$\frac{7}{(x-2)(x-3)}+\frac{9}{x-3}+1<0$

$\frac{7+9(x-2)+x^2-5x+6}{(x-2)(x-3)}<0$

$\frac{x^2+4x-5}{(x-2)(x-3)}<0$

$\frac{(x+5)(x-1)}{(x-2)(x-3)}<0$

$x\in (-5,1)\cup (2,3)$