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

Given:

$\frac{3}{x-2}<2, x \in R$

Subtracting 2 from both the sides in the above equation

$\frac{3}{x-2}-2<2-2$

$\frac{3-2(x-2)}{x-2}<0$

$\frac{3-2 x+4}{x-2}<0$

$\frac{7-2 x}{x-2}<0$

Signs of $7-2 x$ :

$7-2 x=0 \rightarrow x=\frac{7}{2}$

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

$7-2 x<0 \rightarrow x>\frac{7}{2}$

(Subtracting by 7 on both the sides, then multiplying by $-1$ on both the sides and then dividing both the sides by 2)

$7-2 x>0 \rightarrow x<\frac{7}{2}$

(Subtracting by 7 on both the sides, then multiplying by $-1$ on both the sides and then dividing both the sides by 2 )

Signs of $x-2$ :

$x-2=0 \rightarrow x=2$ (Adding 2 on both the sides)

$x-2<0 \rightarrow x<2$ (Adding 2 on both the sides)

$x-2>0 \rightarrow x>2$ (Adding 2 on both the sides)

Zeroes of denominator:

$x-2=0 \rightarrow x=2$

At $x=2, \frac{7-2 x}{x-2}$ is not defined

Intervals satisfying the condition: < 0

$x<2$ and $x>\frac{7}{2}$

Therefore

$x \in(-\infty, 2) \cup\left(\frac{7}{2}, \infty\right)$