**Question:**

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

$\left|\frac{2 x-1}{x-1}\right|<2, x \in \mathbf{R} .$

**Solution:**

Given:

$\left|\frac{2 x-1}{x-1}\right|<2, x \in R$

$-2<\left|\frac{2 x-1}{x-1}\right|<2$

$\frac{2 x-1}{x-1}>-2$ and $\frac{2 x-1}{x-1}<2$

When,

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

Adding 2 to both sides in the above equation

$\frac{2 x-1}{x-1}+2>-2+2$

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

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

$\frac{4 x-3}{x-1}>0$

Signs of 4x – 3:

$4 x-3=0 \rightarrow x=\frac{3}{4}$

(Adding 3 to both sides and then dividing both sides by 4)

$4 x-3>0 \rightarrow x>\frac{3}{4}$

(Adding 3 to both sides and then dividing both sides by 4)

$4 x-3<0 \rightarrow x<\frac{3}{4}$

(Adding 3 to both sides and then dividing both sides by 4)

Signs of x – 1:

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

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

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

At $x=1, \frac{4 x-3}{x-1}$ is not defined.

Intervals that satisfy the required condition: > 0

$x<\frac{3}{4}$ or $x>1$

Now, when $\frac{2 x-1}{x-1}<2$

Subtracting 2 from both the sides

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

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

$\frac{2 x-1-2 x+2)}{x-1}<0$

$\frac{1}{x-1}<0$

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)

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

Interval which satisfy the required condition: < 0

x < 1

Now, combining the intervals:

$x<\frac{3}{4}$ or $x>1$ and $x<1$

Merging the overlapping intervals:

$x<\frac{3}{4}$

Therefore,

$x \in\left(-\infty, \frac{3}{4}\right)$