Solve absolute value inequality : |3x−8|>7

Step 1
the given absolute value inequality is:
${\displaystyle \left|3x-8\right|>7}$
we have to solve the given absolute value inequality.
Step 2
the given absolute value inequality is ${\displaystyle \left|3x-8\right|>7}$
${\displaystyle \left|3x-8\right|>7}$
as we know that if ${\displaystyle \left|x\right|>a}$ then ${\displaystyle x\in (-\mathrm{\infty },-a)\cup (a,\mathrm{\infty })}$
that implies if ${\displaystyle \left|x\right|>a}$ then ${\displaystyle x<-a\text{or}x>a}$
therefore,
if ${\displaystyle \left|3x-8\right|>7}$ then
${\displaystyle 3x-8<-7\text{or}3x-8>7}$
Therefore,
for ${\displaystyle 3x-8<-7}$
${\displaystyle 3x-8+8<-7+8}$
${\displaystyle 3x<1}$
${\displaystyle x<\frac{1}{3}}$
Step 3
for ${\displaystyle 3x-8>7}$
${\displaystyle 3x-8+8>7+8}$
${\displaystyle 3x>15}$
${\displaystyle x>\frac{15}{3}}$
${\displaystyle x>5}$
therefore, if ${\displaystyle \left|3x-8\right|>7}$ then ${\displaystyle x<13\text{or}x>5}$
therefore the solution of the given absolute value inequality ${\displaystyle \left|3x-8\right|>7}$ is ${\displaystyle x<13\text{or}x>5}$
therefore the solution of the given absolute value inequality ${\displaystyle \left|3x-8\right|>7}$ in interval notation is
${\displaystyle x\in (-\mathrm{\infty },\frac{1}{3})\cup (5,\mathrm{\infty })}$

Answer is given below (on video)