Solve the following
Question:

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

Solution:

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

$\Rightarrow\left(x+\frac{1}{3}\right)<\frac{-8}{3}$ or $\left(x+\frac{1}{3}\right)>\frac{8}{3} \quad($ As, $|x|>a \Rightarrow x<-a$ or $x>a)$

$\Rightarrow x<-\frac{1}{3}-\frac{8}{3}$ or $x>\frac{8}{3}-\frac{1}{3}$

$\Rightarrow x<-\frac{9}{3}$ or $x>\frac{7}{3}$

$\Rightarrow x<-3$ or $x>\frac{7}{3}$

$\therefore x \in(-\infty,-3) \cup\left(\frac{7}{3}, \infty\right)$

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