Solve absolute value inequality: |3−(23)x|>5.

Step 1
According to the question, we have to solve the inequality ${\displaystyle |3-\left(\frac{2}{3}\right)x|>5}$.
As the above inequality is modulus type inequality and we know that a modulus function is a function which gives the absolute value of a number or variable. It produces the magnitude of the number of variables.
So, we have to take two condition, one for postive values and another for negative values and solve accordingly.
Step 2
Rewrite the given expression,
${\displaystyle |3-\left(\frac{2}{3}\right)x|>5}$
Firstly taking for postive values,
${\displaystyle (3-\left(\frac{2}{3}\right)x)>5}$
${\displaystyle 3-\frac{2}{3}x>5}$
${\displaystyle -\frac{2}{3}x>5-3}$
${\displaystyle -\frac{2}{3}x>2}$
Further solving by multiplying negative sign both sides, the inequality sign gets reverse
${\displaystyle \frac{2}{3}x<2}$
${\displaystyle x<2\cdot \frac{3}{2}}$
x<3
For this condition the solution set is ${\displaystyle (-\mathrm{\infty },3)}$.
Step 3
Now for taking negative values,
${\displaystyle -(3-\left(\frac{2}{3}\right)x)>5}$
${\displaystyle (3-\left(\frac{2}{3}\right)x)<-5}$
${\displaystyle -\left(\frac{2}{3}\right)x<-5-3}$
${\displaystyle -\left(\frac{2}{3}\right)x<-8}$
${\displaystyle \left(\frac{2}{3}\right)x>8}$
${\displaystyle x>8\cdot \frac{3}{2}}$
x>4*3
x>12
From this condition, the solution set is ${\displaystyle (12,\mathrm{\infty })}$.
Hence, combining the both condition the solution set for the inequality ${\displaystyle |3-\left(\frac{2}{3}\right)x|>5is(-\mathrm{\infty },3)\cup (12,\mathrm{\infty })}$.