Question:
$\frac{5 x+8}{4-x}<2$
Solution:
We have, $\frac{5 x+8}{4-x}<2$
$\Rightarrow \frac{5 x+8}{4-x}-2<0$
$\Rightarrow \frac{5 x+8-2(4-x)}{4-x}<0$
$\Rightarrow \frac{5 x+8-8+2 x}{4-x}<0$
$\Rightarrow \frac{7 x}{4-x}<0$
$\Rightarrow \frac{7 x}{x-4}>0$
$\therefore x \in(-\infty, 0) \cup(4, \infty)$
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