Question

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

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

Answer 1

Answer: (A)

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

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