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?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
 A: Hint: right side - left side is of the form $\dfrac{\text{quadratic}}{(x-1)(3x-5)}$, and the quadratic can be factored.
A: 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 $2\le x\le 5$
A: 
$$\frac{x}{3x-5}\leq \frac{2}{x-1}$$

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 $$\dfrac{(x-2)(x-5)}{(x-1)(3x-5)} \leq 0 \;\;\iff\;\; 1 \lt x <\dfrac 53\;\;\text{OR}\;\;2 \leq x \leq 5$$
A: 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.
