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?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$


$$\frac{x}{3x-5}\leq \frac{2}{x-1}$$

Subtracting the right-hand side from each side, then finding the common denominator and subtracting resulting numerators gives us:$$\begin{align}  
 \dfrac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\leq 0 & \iff \dfrac{x^2 - 7x + 10}{(x-1)(3x-5)} \leq 0 \\ \\ 
& \iff \dfrac{(x-2)(x-5)}{(x-1)(3x-5)} \leq 0\quad x \neq 1, \;x \neq \frac 53\\ \\
\end{align}$$
This inequality will be satisfied when exactly one or exactly three of the factors is/are negative or when $x = 2$ or $x = 5$, regardless of the signs of the other factors. 

A nice way to approach this sort of problem is through the use of a sign-chart, ordering the roots (or points at which one factor switches from positive to negative):
$x \quad = \qquad 1 \qquad \frac 53 \qquad 2\;\qquad 5$
$\qquad\quad+\qquad - \qquad+\;\;\,0\;\; -\;\;0\quad +$
This gives us $$\dfrac{(x-2)(x-5)}{(x-1)(3x-5)} \leq 0 \;\;\iff\;\; 1 \lt x <\dfrac 53\;\;\text{OR}\;\;2 \leq x \leq 5$$
