Question

If $\frac{x-3}{x+4}<0,$ then

• A. $x\in (-4,3)$
• B. $x\in (-\infty ,3)$
• C. $x\in (-\infty ,-4)$
• D. $x\in (-4,-3)$

Answer 1

The correct option is A $x\in (-4,3)$

We know that, $\frac{a}{b}<0$ when $a>0$ and $b<0$ or $a<0$ and $b>0$.
$\therefore$ Either $(x-3>0$ and $x+4<0)$ or $(x-3<0$ and $x+4>0)$

Case $1:$ When $x-3>0$ and $x+4<0$
$\Rightarrow x>3$ and $x<-4$
This is not possible, as we can never find a real number which is greater than $3$ and less than $-4$.

Case $2:$ When $x-3<0$ and $x+4>0$
$\Rightarrow x<3$ and $x>-4$
$\Rightarrow -4<x<3$
$\Rightarrow x\in (-4,3)$

$\therefore$ Solution set $=(-4,3)$.