Question

Solve : $|x+\frac{1}{3}|>\frac{8}{3}$

Answer 1

Given $|x+\frac{1}{3}|>\frac{8}{3}$

Case I : $x+\frac{1}{3}\ge 0\Rightarrow x\ge \frac{-1}{3}$ ...... $\left(1\right)$

$|x+\frac{1}{3}|=x+\frac{1}{3}>\frac{8}{3}\to x>\frac{7}{3}$..... $\left(2\right)$

From $\left(1\right),\left(2\right)x>\frac{7}{3}$

Case II: $x+\frac{1}{3}<0\Rightarrow x<\frac{-1}{3}$ ....... $\left(3\right)$

$|x+\frac{1}{3}|=-x-\frac{1}{3}>\frac{8}{3}\Rightarrow x<-3$ ..... $\left(4\right)$

from $\left(3\right),\left(4\right)x<-3$

$\Rightarrow x\in (-\infty ,-3)v(\frac{7}{3},\infty )$