Question

# Solve each of the following system of equations in R. 27. $\left|\frac{2x-1}{x-1}\right|>2$

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Solution

## Clearly, LHS of the inequality does not hold true for x = 1. $\left|\frac{2x-1}{x-1}\right|>2\phantom{\rule{0ex}{0ex}}\mathrm{Clearly},\mathrm{LHS}\mathrm{of}\mathrm{the}\mathrm{inequality}\mathrm{is}\mathrm{meaningful}\mathrm{for}x\ne \frac{1}{2}.\phantom{\rule{0ex}{0ex}}⇒\left|\frac{2x-1}{x-1}\right|-2>0\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{two}\mathrm{cases}\mathrm{arise}.\phantom{\rule{0ex}{0ex}}\mathrm{CASE}1:\frac{2x-1}{x-1}>0\mathrm{i}.\mathrm{e}.x\in \left(-\infty ,\frac{1}{2}\right)\cup \left(1,\infty \right)\phantom{\rule{0ex}{0ex}}\mathrm{Then},\left|\frac{2x-1}{x-1}\right|=\frac{2x-1}{x-1}\phantom{\rule{0ex}{0ex}}\therefore \left|\frac{2x-1}{x-1}\right|-2>0\phantom{\rule{0ex}{0ex}}⇒\frac{2x-1}{x-1}-2>0\phantom{\rule{0ex}{0ex}}⇒\frac{2x-1-2x+2}{x-1}>0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x-1}>0\phantom{\rule{0ex}{0ex}}⇒x\in \left(1,\infty \right)...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{CASE}2:\mathrm{When},\frac{2x-1}{x-1}<0,\mathrm{i}.\mathrm{e}.\mathrm{x}\in \left(\frac{1}{2},1\right)\phantom{\rule{0ex}{0ex}}⇒\left|\frac{2x-1}{x-1}\right|=\frac{-\left(2x-1\right)}{x-1}\phantom{\rule{0ex}{0ex}}\therefore \left|\frac{2x-1}{x-1}\right|-2>0\phantom{\rule{0ex}{0ex}}⇒\frac{-\left(2x-1\right)}{x-1}-2>0\phantom{\rule{0ex}{0ex}}⇒\frac{-2x+1-2x+2}{x-1}>0\phantom{\rule{0ex}{0ex}}⇒\frac{-4x+3}{x-1}>0\phantom{\rule{0ex}{0ex}}⇒x\in \left(\frac{3}{4},1\right)...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{set}\mathrm{of}\mathrm{inequalities}\mathrm{is}\mathrm{the}\mathrm{union}\mathrm{of}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right).\phantom{\rule{0ex}{0ex}}x\in \left(\frac{3}{4},1\right)\cup \left(1,\infty \right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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