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

Is the only way is to multiple both sides by $(x-1)^2$?

so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:

$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$

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The simplest, as we have a homographic function, is to write it in canonical form: \begin{align*} 2<\frac{x}{x-1}\leq 3&\iff 2 < 1+\frac1{x-1}\leq 3 \iff 1< \frac1{x-1}\leq 2\\ &\iff \frac12\le x-1 <1 \iff \frac32 \le x <2 \end{align*} The third equivalence is valid because all numbers at the end of the first line have the same sign.

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  • $\begingroup$ What is the canonical form? (where can I read about homographic function?) the third equivalence you mean where you did $()^{-1}$? $\endgroup$ – gbox Jun 3 '15 at 18:24
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    $\begingroup$ A homographic function is the name for high school functions of type $\dfrac{ax+b}{cx+d}$ with $ad-bc\neq 0$. The prototype is simply $f(x)=\dfrac cx, \enspace c\neq 0$. The canonical form is $C+\dfrac K{x- A}$ (you can always get one dividing numerator by denominator). As for the third equivalence, yes: you can invert an inequality only if you're sure all numbers have the same sign. $\endgroup$ – Bernard Jun 3 '15 at 18:33
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Why quadratic? Linear is sufficient.

If $x > 1$, then $(x-1)$ is positive so we can multiply by that to get: $2(x-1) \lt x \le 3(x-1)$. Left-hand gives us $x < 2$ and right-hand gives us $x \ge \frac32$. Thus: $\frac32 \le x < 2$.

If $x < 1$, then we have to flip the signs and get $2(x-1) > x \ge 3(x-1)$. But there, the left-hand gives us $x > 2$, so there's no solution here.

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Multiplying $(x-1)^2\gt 0$ is a good idea, but you made mistakes after that. $$2(x-1)^2\lt x(x-1)\iff x^2-3x+2\lt 0$$$$\iff (x-2)(x-1)\lt 0\iff 1\lt x\lt 2$$ and $$x(x-1)\le 3(x-1)^2\iff 2x^2-5x+3\ge 0$$$$\iff (x-1)(2x-3)\ge 0\iff x\le 1\ \text{or}\ x\ge\frac 32.$$ Hence, the answer is $$\frac 32\le x\lt 2.$$

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you can rewrite $$2 < \frac x{x-1} = 1 + \frac 1{x-1} \le 3 \to 1 < \frac1 {x-1} \le 2 \tag 1$$ from the graph of $y = \frac1{x-1},$ we see that there are no solutions $(-\infty, 1)$ and $x - 1 > 0$ is necessary so $(1)$ is equivalent to $$\frac 12 \le x - 1< 1\to \frac32 \le x < 2. $$

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You should just multiply the inequality by (x-1) instead of $(x-1)^2$.

Also, the solution would then be $\frac{3}{2}\le x\lt 2$.

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  • $\begingroup$ Multiplying an inequality by a negative number reverses the signs, forcing us to consider the cases where $x-1$ is positive and where it is negative separately. Multiplying by $(x-1)^2 > 0$ avoids that issue. $\endgroup$ – epimorphic Jun 5 '15 at 20:39

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