Solve each of the following in equations and represent the solution set on

Solution:

Given:

$|3 x-7|>4, x \in R .$

$3 x-7<-4$ or $3 x-7>4$

(Because $|x|>a, a>0$ then $x<-a$ and $x>a$ )

$3 x-7<-4$

Now, adding 7 to both the sides in the above equation

$3 x-7+7<-4+7$

$3 x<3$

Now, dividing by 3 on both the sides of above equation

$\frac{3 x}{3}<\frac{3}{3}$

$x<1$

Now,

$3 x-7>4$

Adding 7 on both the sides in above equation

$3 x-7+7>4+7$

$3 x>11$

Now, dividing by 3 on both the sides in the above equation

$\frac{3 x}{3}>\frac{11}{3}$

$x>\frac{11}{3}$

Therefore,

$x \in(-\infty, 1) \cup\left(\frac{11}{3}, \infty\right)$