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In the book I'm using an example is given as follow:

$\frac{2x - 5}{x-2}< 1$

then it proceeds to say that we could multiple both sides by $x-2$ to get rid of the denominator in the left hand side (I understand that). But then it goes on to say that this method would require to consider the following cases,

$x-2 > 0$ and $x-2<0$ separately.

How did the author of the book reached to the conclusion that we will need to evaluate such cases? and why did the orientation of the inequalites change?

I tried doing some algebraic manipulation myself but I could reach a conclusion. I tried:

multiplying both sides by $x-2$

$2x-5<x-2$

subtract x from both sides

$x-5 < -2 $

add 2 to both sides

$x-3 < 0$

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  • $\begingroup$ Multiplying by a negative number flips the inequality $\endgroup$
    – Riquelme
    Commented Feb 29, 2020 at 14:07
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    $\begingroup$ A different possibly better approach could be to look at $\frac{2x-5}{x-2}-1<0$ so $\frac{x-3}{x-2}<0$ now it's clear that quotient of two numbers is negative if they are of opposite signs. $\endgroup$
    – kingW3
    Commented Feb 29, 2020 at 14:10
  • $\begingroup$ @Riquelme because its implied by the inequality that the denominator must be a negative value? $\endgroup$
    – KetDog
    Commented Feb 29, 2020 at 14:16
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    $\begingroup$ You can get $2x-5\lt x-2$ from $\frac{2x-5}{x-2}\lt 1$ only when $x-2$ is positive. You are implicitly assuming that $x-2$ is positive. $\endgroup$
    – mathlove
    Commented Feb 29, 2020 at 14:55
  • $\begingroup$ @mathlove thanks now I think I understand. $\endgroup$
    – KetDog
    Commented Feb 29, 2020 at 15:01

3 Answers 3

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How did the author of the book reached to the conclusion that we will need to evaluate such cases?

Because we don't know the sign of $x-2.$ All we know is that it may either be positive ($>0$) or negative ($<0$); the case $x-2=0$ not arising since then the fraction is not a real number.

why did the orientation of the inequalites change?

I don't know which of the inequalities you're talking about, but when you're considering the case when $x-2<0,$ you have to change the original inequality from $\text{LHS}<\text{RHS}$ to $\text{LHS}>\text{RHS}$ because whenever you multiply both sides of an inequality by a negative number, the order is reversed.

A way to proceed by not evaluating cases is simply to multiply both sides by $(x-2)^2$ without bothering about signs, since a square can never be negative.

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$$\frac{2x-5}{x-2}<1$$ it is alwys better that you have 0 on the RHS. Then $$\frac{2x-5}{x-2}-1<0 \implies \frac{x-3}{x-2} <0$$ $A/B <0 \implies AB<0 \implies (1):$ $A<0$ and $B>0$ or (2): $A>0$ and $B<0$ So we get (1): $x-3<0$ and $x-2>0 \implies 2<x<3.$ or $(ii) x>3$ and $x<2 \implies$ no solution (as the oprverlap is null).

So the the solution is $2<x<3$

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Because $$\frac{a}{b}<1\iff\begin{cases}a<b\\ b>0\end{cases}\quad \text{or}\quad \begin{cases}a>b\\ b<0\end{cases}.$$

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