Question

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

Answer 1

$$\begin{gathered} \mathrm{As},\left|x+\frac{1}{3}\right|>\frac{8}{3} \\ \Rightarrow \left(x+\frac{1}{3}\right)<\frac{-8}{3}\mathrm{or}\left(x+\frac{1}{3}\right)>\frac{8}{3}\left(\mathrm{As},\left|x\right|>a\Rightarrow x<-a\mathrm{or}x>a\right) \\ \Rightarrow x<-\frac{1}{3}-\frac{8}{3}\mathrm{or}x>\frac{8}{3}-\frac{1}{3} \\ \Rightarrow x<-\frac{9}{3}\mathrm{or}x>\frac{7}{3} \\ \Rightarrow x<-3\mathrm{or}x>\frac{7}{3} \\ \therefore x\in \left(-\infty ,-3\right)\cup \left(\frac{7}{3},\infty \right) \end{gathered}$$