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

Solution:

Given:

$\frac{2 x-3}{3 x-7}<0, x \in R$

Signs of $2 x-3$ :

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

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

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

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

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

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

Signs of 3x – 7:

$3 x-7=0 \rightarrow x=\frac{7}{3}$

(Adding 7 on both the sides and then dividing both sides by 3)

$3 x-7<0 \rightarrow x<\frac{7}{3}$

(Adding 7 on both the sides and then dividing both sides by 3)

$3 x-7>0 \rightarrow x>\frac{7}{3}$

(Adding 7 on both the sides and then dividing both sides by 3)

Zeroes of denominator:

3x – 7 = 0

$x=\frac{7}{3}$

(Adding 7 on both the sides and then dividing both sides by 3)

Interval that satisfies the required condition: < 0

$\frac{3}{2}