Question

Solve the following system of equations in R.

|x+1|+|x|>3

Answer 1

$|x+1|+|x|>3Case1:When-\infty <x<-1|x+1|=-(x+1)and|x|=-x\therefore |x+1|+|x|>3\Rightarrow -(x+1)-x>3\Rightarrow -2x>4\Rightarrow x<-2But,-\infty <x<-1\therefore \text{The solution set of the given inequation is}(-\infty ,-2).Case2:When-1\le x<0|x+1|=(x+1)and|x|=-x\therefore |x+1|+|x|>3\Rightarrow (x+1)-x>3\Rightarrow 1>3\text{Which is not true}Case3:When0<x<\infty |x+1|=(x+1)and|x|=x\therefore |x+1|+|x|>3\Rightarrow (x+1)+x>3\Rightarrow 2x>2\Rightarrow x>1But,0<x<\infty \therefore Theso{l}^n\text{set of the given inequation is}(1,\infty ).\text{Combining case (i),case (ii) and (iii) we obtain that the solution set of given inequality is}(-\infty ,-2)\cup (1,\infty )$