Question

The following system of inequalities
$\frac{x}{2x+1}\ge \frac{1}{4}$, $\frac{6x}{4x-1}<\frac{1}{2}$ have

• A. infinitely many solution
• B. unique solution
• C. no solution
• D. None of these

Answer 1

Answer: (C)

From the first inequality, we have $\frac{x}{2x+1}-\frac{1}{4}\ge 0$
$\Rightarrow \frac{2x-1}{2x+1}\ge 0$
$\Rightarrow$ ($2x-1\ge 0$ and $2x+1>0$)
or ($2x-1\le 0$ and $2x+1<0$) [Since $2x+1\ne 0$]
$\Rightarrow$ ($x\ge \frac{1}{2}$ and $x>-\frac{1}{2}$) or ($x\le \frac{1}{2}$ and $x<-\frac{1}{2}$)
$\Rightarrow x\ge \frac{1}{2}$ or $x<-\frac{1}{2}\Rightarrow x\in \left(-\infty ,-\frac{1}{2}\right)\cup \left[\frac{1}{2},\infty \right]...\left(1\right)$
From the second inequality, we have $\frac{6x}{4x-1}-\frac{1}{2}<0$
$\Rightarrow \frac{8x+1}{4x-1}<0$
$\Rightarrow$ ($8x+1<0$ and $4x-1>0$)
or ($8x+1>0$ and $4x-1<0$)
$\Rightarrow$ ($x<-\frac{1}{8}$ and $x>\frac{1}{4}$, it is not possible)
or ($x>-\frac{1}{8}$ and $x<\frac{1}{4}$)
$\Rightarrow x\in \left(-\frac{1}{8},\frac{1}{4}\right)...\left(2\right)$
Note that the common solution of $(1)$ and $(2)$ is null set. Hence, the given system of inequalities has no solution.