Question

Solve each of the following system of equations in R. 25. $\frac{1}{\left|x\right|-3}<\frac{1}{2}$

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Solution

$\mathrm{We}have,\frac{1}{\left|x\right|-3}<\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\left|x\right|-3}-\frac{1}{2}<0\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{two}\mathrm{cases}\mathrm{arises}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{CASE}1:x>0\phantom{\rule{0ex}{0ex}}\mathrm{Then},\left|x\right|=x\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{\left|x\right|-3}-\frac{1}{2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x-3}-\frac{1}{2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{2-x+3}{2x-6}<0\phantom{\rule{0ex}{0ex}}⇒\frac{-x+5}{2x-6}<0\phantom{\rule{0ex}{0ex}}⇒x\in \left(-\infty ,3\right)\cup \left(5,\infty \right)\phantom{\rule{0ex}{0ex}}\mathrm{But}\mathrm{in}\mathrm{this}\mathrm{case}x\mathrm{is}\mathrm{positive}\phantom{\rule{0ex}{0ex}}\therefore x\in \left(0,3\right)\cup \left(5,\infty \right)....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{CASE}2:\mathrm{When}x<0\phantom{\rule{0ex}{0ex}}\left|x\right|=-x\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{\left|x\right|-3}-\frac{1}{2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{-x-3}-\frac{1}{2}<0\phantom{\rule{0ex}{0ex}}⇒\frac{2+x+3}{-2x-6}<0\phantom{\rule{0ex}{0ex}}⇒\frac{x+5}{-2x-6}<0\phantom{\rule{0ex}{0ex}}⇒x\in \left(-\infty ,-5\right)\cup \left(-3,\infty \right)\phantom{\rule{0ex}{0ex}}\mathrm{But}\mathrm{in}\mathrm{this}\mathrm{case}x\mathrm{is}\mathrm{negative}\phantom{\rule{0ex}{0ex}}\therefore x\in \left(-\infty ,-5\right)\cup \left(-3,0\right)....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{inequation}\mathrm{is}\mathrm{the}\mathrm{union}\mathrm{of}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right).\phantom{\rule{0ex}{0ex}}\therefore x\in \left(-\infty ,-5\right)\cup \left(-3,3\right)\cup \left(5,\infty \right)$

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