Question

Solve the following inequations simultaneously. $\frac{x}{2x+1}\ge \frac{1}{4};\frac{6x}{4x-1}<\frac{1}{2}$.

Answer 1

$\frac{x}{2x+1}\ge \frac{1}{4}⟹\frac{x}{2x+1}-\frac{1}{4}\ge 0⟹\frac{4x-2x-1}{8x+4}\ge 0⟹\frac{2x-1}{8x+4}\ge 0$

Hence, either both numerator and denominator should be positive or both should be negative.

$⟹\left\{\right\{2x-1\ge 0\}\cap \{8x+4\ge 0\left\}\right\}\cup \left\{\right\{2x-1\le 0\}\cap \{8x+4\le 0\left\}\right\}$

$⟹\{x\ge \frac{1}{2}\}\cap \{x<-\frac{1}{2}\}$ (as $x=-0.5$ gives undefined value)

$⟹x\in (-\infty ,-\frac{1}{2})\cup [\frac{1}{2},\infty )$

Also,

$\frac{6x}{4x-1}<\frac{1}{2}⟹\frac{6x}{4x-1}-\frac{1}{2}<0⟹\frac{12x-4x+1}{8x-2}<0⟹\frac{8x+1}{8x-2}<0$

Hence, numerator and denominator should be of opposite sign.

$⟹\left\{\right\{8x+1<0\}\cap \{8x-2>0\left\}\right\}\cup \left\{\right\{8x+1>0\}\cap \{8x-2<0\left\}\right\}$

$⟹\left\{\varphi \right\}\cup \{\{x>-\frac{1}{8}\}\cap \{x<\frac{1}{4}\}\}$

$⟹x\in (-\frac{1}{8},\frac{1}{4})$

Now, the required solution is $x\in \{(-\infty ,-\frac{1}{2})\cup [\frac{1}{2},\infty )\}\cap (-\frac{1}{8},\frac{1}{4})$

$⟹x\in [\frac{1}{2},\frac{1}{4})$