Question

If $\frac{5x+8}{4-x}<2$, then $x\in$

• A. $(0,4)$
• B. $(-\infty ,0)$
• C. $(4,\infty )$
• D. $(-\infty ,0)\cup (4,\infty )$

Answer 1

The correct option is D $(-\infty ,0)\cup (4,\infty )$

$\frac{5x+8}{4-x}<2$
$\Rightarrow$ $\frac{5x+8}{4-x}-2<0$
$\Rightarrow$ $\frac{5x+8-8+2x}{4-x}<0$
$\Rightarrow$ $\frac{7x}{x-4}>0$
$x\in (-\infty ,0)\cup (4,\infty )$
$\therefore$ (4) is correct