Question

Solution of $\frac{x-7}{x+3}>2$ is

• A. $\left(-3,\infty \right)$
• B. $\left(-\infty ,-13\right)$
• C. $(-13,-3)$
• D. $(-13,3)$

Answer 1

Answer: (C)

We have, $\frac{x-7}{x+3}>2$
or $\frac{x-7}{x+3}-2>0$ or $\frac{x+13}{x+3}<0$
$\Rightarrow$ {$x+13>0$ and $x+3<0$} or {$x+13<0$ and $x+3>0$}
$\Rightarrow$ {$x>-13$ and $x<-3$} or {$x<-13$ and $x>-3$}
it is not possible
$\Rightarrow x\in \left(-13,-3\right)$