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I just stumbled upon this particular question and cannot answer

$$\left\lvert \frac{2x+1}{x-2} \right\rvert<1$$

I know that there are some rules in order to answer this Problem

$\frac{2x+1}{x-2} $ if only the $\frac{2x+1}{x-2}≥ 0 $

and

$\frac{(-)2x+1}{x-2}$ if only the $\frac{2x+1}{x-2}< 0 $

but I just can't find the answer because everything what i've learnt and saw on YouTube mixed up, and I can't tell which one is correct.

Can anybody give me an Explanation?

*Ps, I'm sorry, I'm not really familiar with MathJax btw, I have seen another Problem on this site which is also Absolute value rational inequality, but I don't really understand.

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Just use the observation that, on the domain of the inequation ($x\ne 2$), $$\dfrac{\lvert A\rvert}{\lvert B\rvert}<1\iff\lvert A\rvert< \lvert B\rvert\iff A^2<B^2.$$ After some simplification this yields $$3x^2+8x-3<0.$$ The Rational roots theorem yields $x=-3,\dfrac13$, hence the solutions are $$-3<x<\dfrac13$$

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  • $\begingroup$ Thanks, I sort of understand with this Problem. so, by quadrating both sides this question is solved. But I don't really get it, what the rules are meaning of. $\endgroup$ – Irdiarrur Jan 5 '17 at 12:16
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    $\begingroup$ The main problem was the elimination of absolute values. This is classically done by squaring, if this answers your question. $\endgroup$ – Bernard Jan 5 '17 at 12:27
  • $\begingroup$ ok, what about a question that Limits us, so that we cannot square the Problem, for example: |x+2|<2x-1 if I square the Problem, will the answer still the same? knowing that one side has the absolute sign and the other side doesn't have. $\endgroup$ – Irdiarrur Jan 5 '17 at 13:22
  • $\begingroup$ No. For this one, you have to know that $\;\lvert A\rvert <B\iff A^2<B^2\;\textbf{and}\;B>0$. $\endgroup$ – Bernard Jan 5 '17 at 18:24
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I suggest first simplifying so you don't have $x$ in both the numerator and denominator:

$\dfrac{2x+1}{x-2}\ =\ \dfrac{2x-4+5}{x-2}\ =\ 2+\dfrac5{x-2}$

Now try and solve $$-1\ <\ 2+\dfrac5{x-2}\ <\ 1$$

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$|\frac {2x+1}{x-2}|<1$

To get rid of the absolute value:

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

When you mulitiply through by $x-2$ it is going to flip the direction of the inequalities if $x-2 < 0$

Suppose (x-2) > 0

$2-x < 2x+1 <x-2$ and $(x-2) > 0$

$2-x < 2x+1$ and $2x+1 <x-2$ and $(x-2) > 0$

$\frac 13 < x$ and $x <-3$ and $x > 2$

not compatible!

Suppose $(x-2) < 0$

$2-x > 2x+1 >x-2$ and $(x-2) < 0$

$2-x > 2x+1$ and $2x+1 > x-2$ and $(x-2) < 0$

$\frac 13 > x$ and $x > -3$ and $(x-2) < 0$

$x\in(-3 , \frac 13)$

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Here's another way:

$$\left|\frac{2x+1}{x-2}\right|<1\implies \frac{|2x+1|}{|x-2|}<1$$

$$\implies (2x+1)^2<(x-2)^2$$

$$\implies(2x+1+x-2)(2x+1-x+2)<0$$

$$\implies(3x-1)(x+3)<0$$

$$\implies -3<x<\frac 13$$

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