How do I solve this |2x-1|-|x|< 4 ? Do i simply treat it like an equation, meaning that it has three different solutions? or do other rules apply?

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- Feb 20th 2008, 01:15 PM #1

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- Feb 20th 2008, 02:01 PM #2
Use cases.

$\displaystyle |2x-1|-|x|<4$

There are three cases,

i. $\displaystyle x\ge\frac{1}{2}$

ii. $\displaystyle 0 < x < \frac{1}{2}$

iii. $\displaystyle x \le 0$

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i. $\displaystyle x\ge\frac{1}{2}$

The inequality becomes,

$\displaystyle 2x - 1 - x < 4$

$\displaystyle x < 5$

The interval here is, $\displaystyle \frac{1}{2}\le x < 5$

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ii. $\displaystyle 0 < x < \frac{1}{2}$

$\displaystyle -(2x-1) - x < 4$

$\displaystyle x > -1$

Interval: $\displaystyle 0 < x < \frac{1}{2}$

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iii. $\displaystyle x \le 0$

$\displaystyle -(2x -1)-(-x)<4$

Interval: $\displaystyle -3 < x \le 0$

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When we put the intervals together,

$\displaystyle \frac{1}{2}\le x < 5$

$\displaystyle 0 < x < \frac{1}{2}$

$\displaystyle -3 < x \le 0$

It makes,

$\displaystyle -3<x<5$

- Feb 20th 2008, 02:07 PM #3
the first element is positive if x > 0.5 and negative otherwise, the second element is positive if x > 0 and negative otherwise.

1. x => 0.5, in this range both elements are non negative so:

2x-1 - x < 4

x < 5

so the solution is: 0.5 <= x < 5

2. 0 <= x <= 0.5 in this range the first element is negative and the second is non negative so:

-(2x-1) - x < 4

x > -1

thus the answer is 0 <= x <= 0.5

3. x <= 0 in this range the first element is negative and the second is non positive so:

-(2x-1) + x < 4

x > -3

thus the solution is -3 < x <= 0

The final solution is the intersection of these 3 solutions thus:

-3 < x < 5

- Feb 20th 2008, 10:26 PM #4