# Absolute Value Inequalities - Math Question

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Does anyone know how to solve this?

|x+2k| > |x-k|

I need to find what k is

and k is a positive constant

|x+2k| > |x-k|

I need to find what k is

and k is a positive constant

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(Original post by

Yes, but I don't believe we can just provide solutions.

What have you attempted?

**Maths Fan**)Yes, but I don't believe we can just provide solutions.

What have you attempted?

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#4

It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/

https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

https://www.stitz-zeager.com/

https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

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**Maths Fan**)

Yes, but I don't believe we can just provide solutions.

What have you attempted?

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reply

(Original post by

It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/

https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

**0le**)It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/

https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

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reply

**0le**)

It may help you to also have a read through Chapter 0, Section 0.5 of Stitz and Zeager, which goes through inequalities of absolute terms:

https://www.stitz-zeager.com/

https://www.stitz-zeager.com/ch_0_links.pdf

Page 55 has some worked examples.

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#8

In red: The first line is correct, the second isn't. You have to multiply BOTH sides by -1 when flipping the inequality.

Note: you could divide through by k at this point as it's a positive quantity.

Note: you could divide through by k at this point as it's a positive quantity.

Last edited by ghostwalker; 4 weeks ago

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(Original post by

In red: The first line is correct, the second isn't. You have to multiply BOTH sides by -1 when flipping the inequality.

Note: you could divide through by k at this point as it's non-zero.

**ghostwalker**)In red: The first line is correct, the second isn't. You have to multiply BOTH sides by -1 when flipping the inequality.

Note: you could divide through by k at this point as it's non-zero.

I was finally left with k>-2x

What do I do next?

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#10

(Original post by

Thank you so much for pointing that out!!

I was finally left with k>-2x

What do I do next?

**Kundana Amudala**)Thank you so much for pointing that out!!

I was finally left with k>-2x

What do I do next?

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#11

**Kundana Amudala**)

Thank you so much for pointing that out!!

I was finally left with k>-2x

What do I do next?

*positive*quantity. Just being non-zero isn't good enough with inequalities.

Well you can't go any further, other than rearrange to x> -k/2 perhaps.

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(Original post by

What does the original question actually say? Can you upload an image?

**mqb2766**)What does the original question actually say? Can you upload an image?

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#13

So they want you to find a range of x (in terms of k) which you have done?

Without squaring, you could have noted one inequality is

x+2k > x-k

which is trivially satisfied, as well as the other combiations.

x+2k > -(x-k)

etc ...

If you thought about it on the number line, they want the xs which are closer to the point "k" than to the point "-2k". The midpoint -k/2 represents the boundary (equal distance from these two points). Just do a quick sketch.

Without squaring, you could have noted one inequality is

x+2k > x-k

which is trivially satisfied, as well as the other combiations.

x+2k > -(x-k)

etc ...

If you thought about it on the number line, they want the xs which are closer to the point "k" than to the point "-2k". The midpoint -k/2 represents the boundary (equal distance from these two points). Just do a quick sketch.

Last edited by mqb2766; 4 weeks ago

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#14

(Original post by

There's unfortunately nothing for absolute value inequalities WITH constants...

**Kundana Amudala**)There's unfortunately nothing for absolute value inequalities WITH constants...

If , then the solution is either:

or

By doing it this way, you avoid the need to work with squared terms. Sometimes, as in the case here, when you work through one of those inequalities it becomes apparent that it leads to a "dead-end" so to speak. So in your question, doing it this way leads to the two inequalities:

and

Obviously the first one does not provide any useful information since we know that k was positive anyway.

EDIT: Beaten to it above!

Last edited by 0le; 4 weeks ago

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#15

(Original post by

It does not matter. The constant is not really the issue here but rather the logic. The solution using the methods described in that book is as follows:

If , then the solution is either:

or

By doing it this way, you avoid the need to work with squared terms. Sometimes, as in the case here, when you work through one of those inequalities it becomes apparent that it leads to a "dead-end" so to speak. So in your question, doing it this way leads to the two inequalities:

and

Obviously the first one does not provide any useful information since we know that k was positive anyway.

EDIT: Beaten to it above!

**0le**)It does not matter. The constant is not really the issue here but rather the logic. The solution using the methods described in that book is as follows:

If , then the solution is either:

or

By doing it this way, you avoid the need to work with squared terms. Sometimes, as in the case here, when you work through one of those inequalities it becomes apparent that it leads to a "dead-end" so to speak. So in your question, doing it this way leads to the two inequalities:

and

Obviously the first one does not provide any useful information since we know that k was positive anyway.

EDIT: Beaten to it above!

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