Question

Solve each of the following system of equations in R.

26. $\frac{\left|x+2\right|-x}{x}<2$

Answer 1

$$\begin{gathered} \mathrm{We}have,\frac{\left|x+2\right|-x}{x}<2 \\ \Rightarrow \frac{\left|x+2\right|-x}{x}-2<0 \\ \Rightarrow \frac{\left|x+2\right|-x-2x}{x}<0 \\ \Rightarrow \frac{\left|x+2\right|-3x}{x}<0 \\ \mathrm{Now},\mathrm{two}\mathrm{cases}\mathrm{arises}. \\ \mathrm{CASE}1:\mathrm{When}x+2\ge 0,\mathrm{i}.\mathrm{e}.x\ge -2 \\ \mathrm{Then},\left|x+2\right|=x+2 \\ \mathrm{Now},\frac{\left|x+2\right|-3x}{x}<0 \\ \Rightarrow \frac{x+2-3x}{x}<0 \\ \Rightarrow \frac{-2x+2}{x}<0 \\ \Rightarrow x\in \left(-\infty ,0\right)\cup \left(1,\infty \right) \\ Butx\ge -2 \\ \therefore x\in [-2,0)\cup (1,\infty )...\left(\mathrm{i}\right) \\ \mathrm{CASE}2:\mathrm{When}x+2<0,\mathrm{i}.\mathrm{e}.x<-2 \\ \mathrm{Then},\frac{\left|x+2\right|-x}{x}<2 \\ \Rightarrow \frac{\left|x+2\right|-3x}{x}<0 \\ \Rightarrow \frac{-x-2-3x}{x}<0 \\ \Rightarrow \frac{-4x-2}{x}<0 \\ \Rightarrow \frac{2x+1}{x}>0 \\ \Rightarrow x\in \left(-\infty ,-\frac{1}{2}\right)\cup \left(0,\infty \right) \\ \mathrm{But},x<-2 \\ \therefore x\in \left(-\infty ,-\frac{1}{2}\right)...\left(\mathrm{ii}\right) \\ \Rightarrow x\in (-\infty ,0) \\ \mathrm{Therefore},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{inequation}\mathrm{is}\mathrm{union}o\mathrm{f}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right). \\ [-2,0)\cup (1,\infty )\cup \left(-\infty ,-\frac{1}{2}\right) \\ \therefore x\in (-\infty ,0)\cup (1,\infty ) \end{gathered}$$