Question

Solution of $\frac{2x - 3}{3x - 5} \ge 3$ is

• A. $[1,\, \frac{12}{7})$
• B. $(\frac{5}{3},\, \frac{12}{7}]$
• C. $\left(-\infty,\, \frac{5}{3}\right)$
• D. $\left[\frac{12}{7},\, \infty\right)$

Answer 1

Answer: (B)

We have $\frac{2x - 3}{3x - 5} \ge 3 $
or $\quad \frac{2x-3}{3x-5} - 3 \ge 0$ or $\frac{7x - 12}{3x - 5} \le 0$
$\Rightarrow \quad$ {$7x - 12 \le 0$ and $3x - 5 > 0$}
or $\quad$ {$7x - 12 \ge 0$ and $3x - 5 < 0$}
$\Rightarrow \quad$ {$x \le \frac{12}{7}$ and $x > \frac{5}{3}$} or {$x \ge \frac{12}{7}$ and $x < \frac{5}{3}$}
$\Rightarrow \quad x\in (\frac{5}{3},\, \frac{12}{7}]$