Question

The range of values of $x$ satisfying the inequality $\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$ is

• A. $(-\infty ,-2)\cup (-1,3)\cup (4,+\infty )$
• B. $(-1,3)\cup (4,+\infty )$
• C. $(-\infty ,-2)\cup (-1,3)$
• D. $(-\infty ,-2)\cup (4,+\infty )$

Answer 1

The correct option is A $(-\infty ,-2)\cup (-1,3)\cup (4,+\infty )$

$\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$
$\frac{20+10(x-3)+(x-3)(x-4)}{(x-3)(x-4)}>0$
$\frac{x^2+7x+2}{x^2-7x+12}>0$
$\Rightarrow \frac{(x+1)(x+2)}{(x-3)(x-4)}>0$
$\Rightarrow x\in (-\infty ,-2)\cup (-1,3)\cup (4,\infty )$