Question

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

Solution

## $$\dfrac{x}{2x+1}\geq \dfrac{1}{4}\implies \dfrac{x}{2x+1}-\dfrac{1}{4}\geq 0 \implies \dfrac{4x-2x-1}{8x+4}\geq 0 \implies \dfrac{2x-1}{8x+4}\geq 0$$Hence, either both numerator and denominator should be positive or both should be negative.$$\implies \{\{2x-1\geq0\}\cap \{8x+4\geq0\}\}\cup \{\{2x-1\leq0\}\cap \{8x+4\leq0\}\}$$$$\implies \left\lbrace x\geq\dfrac{1}{2}\right\rbrace\cap \left\lbrace x<-\dfrac{1}{2}\right\rbrace$$ (as $$x=-0.5$$ gives undefined value)$$\implies x\in\left(-\infty,-\dfrac{1}{2}\right)\cup\left[\dfrac{1}{2},\infty\right)$$Also, $$\dfrac{6x}{4x-1}<\dfrac{1}{2}\implies \dfrac{6x}{4x-1}-\dfrac{1}{2}<0\implies \dfrac{12x-4x+1}{8x-2}<0\implies \dfrac{8x+1}{8x-2}<0$$Hence, numerator and denominator should be of opposite sign.$$\implies \{\{8x+1<0\}\cap \{8x-2>0\}\}\cup \{\{8x+1>0\}\cap \{8x-2<0\}\}$$$$\implies \{\phi\}\cup \left\{\left\{x>-\dfrac{1}{8}\right\}\cap \left\{x<\dfrac{1}{4}\right\}\right\}$$$$\implies x\in\left(-\dfrac{1}{8},\dfrac{1}{4}\right)$$Now, the required solution is $$x\in\left\{\left(-\infty,-\dfrac{1}{2}\right)\cup\left[\dfrac{1}{2},\infty\right)\right\}\cap\left(-\dfrac{1}{8},\dfrac{1}{4}\right)$$$$\implies x\in\left[\dfrac{1}{2},\dfrac{1}{4}\right)$$Maths

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