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

Solution:

Given:

$\frac{1}{x-1} \leq 2, x \in R$

Subtracting 2 from both the sides in the above equation

$\frac{1}{x-1}-2 \leq 2-2$

$\frac{1-2(x-1)}{x-1} \leq 0$

$\frac{1-2 x+2}{x-1} \leq 0$

$\frac{3-2 x}{x-1} \leq 0$

Signs of 3 – 2x:

$3-2 x=0 \rightarrow x=\frac{3}{2}$

(Subtracting by 3 on both the sides, then multiplying by $-1$ on both the sides and then dividing both the sides by 2 )

$3-2 x<0 \rightarrow x>\frac{3}{2}$

(Subtracting by 3 on both the sides, then multiplying by $-1$ on both the sides and then dividing both the sides by 2)

 $3-2 x>0 \rightarrow x<\frac{3}{2}$

(Subtracting by 3 on both the sides, then multiplying by $-1$ on both the sides and then dividing both the sides by 2 )

Signs of x – 1:

$x-1=0 \rightarrow x=1$ (Adding 1 on both the sides)

$x-1<0 \rightarrow x<1$ (Adding 1 on both the sides)

$x-1>0 \rightarrow x>1$ (Adding 1 on both the sides)

Zeroes of denominator:

$x-1=0 \rightarrow x=1$

At $x=1, \frac{3-2 x}{x-1}$ is not defined

Intervals satisfying the condition: ≤ 0

$x<1$ and $x \geq \frac{3}{2}$

Therefore,

$x \in(-\infty, 1) \cup\left[\frac{3}{2}, \infty\right)$