|1-x|<|2x-5|

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- Nov 23rd 2012, 11:27 PMaroeHow do you solve this inequality?
|1-x|<|2x-5|

- Nov 23rd 2012, 11:35 PMProve ItRe: How do you solve this inequality?
Follow the same procedure I outlined in your last question.

- Nov 23rd 2012, 11:38 PMaroeRe: How do you solve this inequality?
- Nov 24th 2012, 12:03 AMDevenoRe: How do you solve this inequality?
using MarkFL2's clever approach, note that |a| = √(a

^{2}).

so |1-x| < |2x-5| is the same as:

√(1-x)^{2}< √(2x-5)^{2}

squaring both sides:

(1-x)^{2}< (2x-5)^{2}

1-2x+x^{2}< 4x^{2}-20x+25

0 < 3x^{2}-18x+24

0 < x^{2}-6x+8

0 < (x-2)(x-4) <---for this to be true, both factors must be either both positive, or both negative. - Nov 24th 2012, 12:07 AMaroeRe: How do you solve this inequality?
- Nov 24th 2012, 12:10 AMaroeRe: How do you solve this inequality?
I still like the other method Proveit used. If he can solve it again that way that would be great.

- Nov 24th 2012, 12:16 AMProve ItRe: How do you solve this inequality?
I'm not going to do your work for you. You know the procedure, you can try it!

- Nov 24th 2012, 12:19 AMaroeRe: How do you solve this inequality?
Sorry I miswrote the question it's actually

|1-x|>|2x-5|

And I can't find an answer for it. - Nov 24th 2012, 12:56 AMMarkFLRe: How do you solve this inequality?
Since you found:

$\displaystyle |1-x|<|2x-5|$

gives the solution $\displaystyle (-\infty,2)\,\cup\,(4,\infty)$

then you may state:

$\displaystyle |1-x|>|2x-5|$

gives the solution $\displaystyle (2,4)$