Question

Find the value of $x$ in $\frac{3}{x}+\frac{9}{\frac{x}{2}\times 2}\ge \frac{9}{3}$, if $x$ is a positive number.

• A. $4\le x$
• B. $4<x$
• C. $4\ge x$
• D. $\frac{1}{4}\le x$

Answer 1

The correct option is C $4\ge x$

Given, $\frac{3}{x}+\frac{9}{\frac{x}{2}\times 2}\ge \frac{9}{3}$.

Simplify the equation to it's full capacity.
$\frac{3}{x}+\frac{9}{x}\ge 3$.[$\because \frac{9}{\frac{x}{2}\times 2}=\frac{9}{\frac{x}{1}\times 1}=\frac{9}{x}$]

Since the denominator is same (i.e. $x$), the LCM would be $x$.

$\Rightarrow \frac{3+9}{x}\ge 3$

$\Rightarrow \frac{12}{x}\ge 3$

$\therefore$ Multiplying both sides with $x$ will not change the sign of inequality because $x$ is a positive number.

$\frac{12}{x}\times x\ge 3\times x$

$12\ge 3x$

$\to$Transfer 3 to L.H.S

$\frac{12}{3}\ge x$

$4\ge x$

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