Given :
\(\frac{3}{{\text{x}}-2}<2\), x ∈ R
Subtracting 2 from both the sides in the above equation,
\(\frac{3}{{\text{x}}-2}-2\)<2-2
\(\frac{3-2({\text{x}}-2)}{{\text{x}}-2}<0\)
\(\frac{3-2{\text{x}}+4}{{\text{x}}-2}\) <0
\(\frac{7-2{\text{x}}}{{\text{x}}-2} <0\)
Signs of 7 – 2x:
7 - 2x = 0 → 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 - 2x < 0 → 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 – 2x > 0 → 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 → x = 2 (Adding 2 on both the sides)
x – 2 < 0 → x < 2 (Adding 2 on both the sides)
x – 2 > 0 → x > 2 (Adding 2 on both the sides)
Zeroes of denominator:
x – 2 = 0 → x = 2
At x = 2, \(\frac{7-2{\text{x}}}{{\text{x}}-2} \) is not defined
intervals satisfying the condition: <0
x<2 and x > \(\frac{7}{2}\)
Therefore,
x ∈ (-∞,2) U \(\big(\frac{7}{2}, ∞\big)\)
