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

Given:

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

Intervals of $|x-3|$

$|x-3|=-(x-3)$ or $(x-3)$

When $|x-3|=x-3$

$x-3 \geq 0$

Therefore, $x \geq 3$

When $|x-3|=-(x-3)$

$(x-3)<0$

Therefore, $x<3$

Intervals: $x \geq 3$ or $x<3$

Domain of $\frac{|x-3|-x}{x}<2$ :

$\frac{|x-3|-x}{x}$ is not defined for $x=0$

Therefore, x > 0 or x < 0

Now, combining intervals and domain:

x < 0 or 0 < x < 3 or x ≥ 3

For x = 0

$\frac{|x-3|-x}{x}<2 \underset{\rightarrow}{\frac{-(x-3)-x}{x}<2}$

Now, subtracting 2 from both the sides

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

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

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

Signs of $3-4 x:$

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

(Subtracting 3 from both the sides and then dividing both sides by -1)

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

(Subtracting 3 from both the sides and then multiplying both sides by -1)

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

(Subtracting 3 from both the sides and then multiplying both sides by $-1$ )

Signs of x:

$x=0$

$x<0$

$x>0$

Intervals satisfying the required condition: < 0

$x<0$ or $x>\frac{3}{4}$

Combining the intervals:

$x<0$ or $x>\frac{3}{4}$ and $x<0$

Merging the overlapping intervals:

x < 0

Similarly, for $0

$x<0$ or $x>\frac{3}{4}$ and $0

Merging the overlapping intervals:

$\frac{3}{4}

For, $x \geq 3$

$\frac{|x-3|-x}{x}<2 \underset{\rightarrow}{\frac{(x-3)-x}{x}<2}$

Now, subtracting 2 from both the sides

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

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

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

Signs of $-3-2 x$ :

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

(Adding 3 to both the sides and then dividing both sides by -2)

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

(Adding 3 to both the sides and then multiplying both sides by -1)

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

(Adding 3 to both the sides and then multiplying both sides by -1)

Signs of x:

$x=0$

$x<0$

$x>0$

Intervals satisfying the required condition: < 0

$\mathrm{X}<\frac{-3}{2}$ or $\mathrm{x}>0$

Combining the intervals:

$x<\frac{-3}{2}$ or $x>0$ and $x \geq 3$

Merging the overlapping intervals

x ≥ 3

Combining all the intervals:

$x<0$ or $\frac{3}{4}

Merging overlapping intervals:

$x<0$ and $x>\frac{3}{4}$

Therefore

$x \in(-\infty, 0) \cup\left(\frac{3}{4}, \infty\right)$