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