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Question

Mathematics Question on linear inequalities in one variable

Solution of 2x−33x−5≥3\frac{2x - 3}{3x - 5} \ge 3 is

A

[1, 127)[1,\, \frac{12}{7})

B

(53, 127](\frac{5}{3},\, \frac{12}{7}]

C

(−∞, 53)\left(-\infty,\, \frac{5}{3}\right)

D

[127, ∞)\left[\frac{12}{7},\, \infty\right)

Answer

(53, 127](\frac{5}{3},\, \frac{12}{7}]

Explanation

Solution

We have 2x−33x−5≥3\frac{2x - 3}{3x - 5} \ge 3 or 2x−33x−5−3≥0\quad \frac{2x-3}{3x-5} - 3 \ge 0 or 7x−123x−5≤0\frac{7x - 12}{3x - 5} \le 0 ⇒\Rightarrow\quad {7x−12≤07x - 12 \le 0 and 3x−5>03x - 5 > 0} or \quad {7x−12≥07x - 12 \ge 0 and 3x−5<03x - 5 < 0} ⇒\Rightarrow\quad {x≤127x \le \frac{12}{7} and x>53x > \frac{5}{3}} or {x≥127x \ge \frac{12}{7} and x<53x < \frac{5}{3}} ⇒x∈(53, 127]\Rightarrow \quad x\in (\frac{5}{3},\, \frac{12}{7}]