Question

The solution set of the inequation $\frac{x+11}{x-3}>0$is

• A. $\left(-\infty ,-11\right)\cup (3,\infty )$
• B. $\left(-\infty ,-10\right)\cup (2,\infty )$
• C. $\left(-100,-11\right)\cup (1,\infty )$
• D. $\left(-5,0\right)\cup (3,7)$
• E. $\left(0,5\right)\cup (-1,0)$

Answer 1

The correct option is A

$\left(-\infty ,-11\right)\cup (3,\infty )$

Explanation for the correct option

$\frac{x+11}{x-3}>0$

$\Rightarrow x+11>0$ and $(x-3)>0$ or $x+11<0$ and $x-3<0$

$\Rightarrow x>-11$and $x>3$ or $x<-11$and $x<3$

$\Rightarrow x<-11$and $x>3$

$\Rightarrow x\in (-\infty ,-11)\cup (3,\infty )$

Hence the correct option is option(a) i.e. $\left(-\infty ,-11\right)\cup (3,\infty )$