<s>[INST] Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
 [/INST] Note that $3x - 5 = 0$ when $x = \frac{5}{3}$. When $x < \frac{5}{3}$, $3x - 5 < 0$, and when $x > \frac{5}{3}$, $3x - 5 > 0$.
Also, $x - 1 = 0$ when $x = 1$. When $x < 1$, $x - 1 < 0$, and when $x > 1$, $x - 1 > 0$.
Now you have three cases:


*

*$x < 1$,

*$1 < x < \frac{5}{3}$, and

*$x > \frac{5}{3}$.


In each case, multiply by $(3x - 5)(x - 1)$ which will can be arranged to give you a quadratic inequality. Keep in mind, you need to consider the sign of $(3x - 5)(x - 1)$. If you multiply an inequality by a positive number, the inequality sign remains the same. However, if you multiply by a negative number, the inequality sign reverses.
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