<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] Make $0$ one side of the inequality:
$\displaystyle\frac{x}{3x-5}-\frac{2}{x-1}\le 0$
Write the expression in $x$ as a single fraction:
$\displaystyle\frac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\le 0$
Expand the numerator:
$\displaystyle\frac{x^2-7x+10}{(3x-5)(x-1)}\le 0$
Factorise the numerator:
$\displaystyle\frac{(x-2)(x-5)}{(3x-5)(x-1)}\le 0$
Multiply throughout by the square of the denominator which is positive, thus preserving the inequality, while noting the values $x$ cannot take for the initial fractions to be defined:
$\displaystyle(x-1)(x-\frac{5}{3})(x-2)(x-5)\le 0$ and $x\ne 1$ and $x\ne \frac{5}{3}$
By considering regions of positivity and negativity, we have:
$\displaystyle 1<x<\frac{5}{3}$ or $2\le x\le 5$
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