Question

Find the value of $x$ in $\frac{32}{x}-\frac{12}{x}+\frac{11}{x}\le -3$, if $x$ is a negative number.

• A. $x\le \frac{23}{3}$ and $x>0$
• B. $x\le \frac{-23}{3}$ and $x<0$
  
  
• C. $x\le \frac{3}{23}$ and $x<0$
  
• D. $x\ge \frac{-23}{3}$ and $x<0$
  

Answer 1

The correct option is D $x\ge \frac{-23}{3}$ and $x<0$


Given, $\frac{32}{x}-\frac{12}{x}+\frac{11}{x}\le -3$.

$\frac{32}{x}-\frac{12}{x}+\frac{11}{x}\le -3$
Since the denominator is same (i.e. $x$), the LCM would be $x$.

$\Rightarrow \frac{32-12+11}{x}\le -3$

$\Rightarrow \frac{23}{x}\le -3$

$\therefore$ Multiplying both sides with $x$ will change the sign of inequality because $x$ is a negative number.

$\frac{23}{x}\times x\ge -3\times x$

$23\ge -3x$

$\to$Transfer -3 to L.H.S and reverse the inequality sign due to negative integer -3.
$\frac{-23}{3}\le x$

$\frac{-23}{3}\le x$ which can also be written as $x\ge \frac{-23}{3}$
As $x$ is a negative number, $x<0$
Hence, $x\ge \frac{-23}{3}$ and $x<0$

Hence, option (d.) is the correct choice.