# Thread: |2x-1|-|x|< 4

1. ## |2x-1|-|x|< 4

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?

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,
$\displaystyle -3<x<5$

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

4. Originally Posted by weasley74
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?
If you like - or allowed to - you can solve it graphically.