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Math: Absolute value (Modulus)

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 11 Jul 2015, 19:08
Engr2012 wrote:
ranaazad wrote:
walker wrote:
I've added a new example:


Example #2
Q.: \(|x^2-4| = 1\). What is x?
Solution: There are 2 conditions:

a) \((x^2-4)\geq0\) --> \(x \leq -2\) or \(x\geq2\). \(x^2-4=1\) --> \(x^2 = 5\). x e {\(-\sqrt{5}\), \(\sqrt{5}\)} and both solutions satisfy the condition.

b) \((x^2-4)<0\) --> \(-2 < x < 2\). \(-(x^2-4) = 1\) --> \(x^2 = 3\). x e {\(-\sqrt{3}\), \(\sqrt{3}\)} and both solutions satisfy the condition.

Answer: \(-\sqrt{5}\), \(-\sqrt{3}\), \(\sqrt{3}\), \(\sqrt{5}\)



Hello,
Nice post. I can't understand why the first condition equals to 0 as well when the second condition is only less than zero but NOT equal to zero? Could you please explain?

Thanks! :D


For modulus or absolute value questions, you need to take the equality with the '>' sign. That is the convention and follows the definition of an absolute value. Additionally, you only need to account for equality once in your question.

Also, think of modulus or absolute value this way:

|x| = x for x=0,1,2,3,4,5... or true for all NON-NEGATIVE numbers. Remember the 'non-negative' part. This includes 0 as well.

But |x| = -x for x<0, x=-1,-2,-0.25 ...

This is the reason why we put equality with the '>' sign to account for all non-negative numbers. The 'nature' of 'x' does not change if it is 0 and above but it does change (multiply x by -1) if x is <0.

Hope this helps.



Thanks. Now I'm observing the => convention |x| has. The equal sign is included because we are guessing x could be 0 as well. But if 0 is a non-negative number, it can be a non-positive number too (In that non positive number scenario, we would have followed the <= convention for |x|). In fact 0 is a neutral number. So, in this case, why is 0 implied as a non negative number?
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Re: Math: Absolute value (Modulus) [#permalink]

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Engr2012 wrote:

For modulus or absolute value questions, you need to take the equality with the '>' sign. That is the convention and follows the definition of an absolute value. Additionally, you only need to account for equality once in your question.

Also, think of modulus or absolute value this way:

|x| = x for x=0,1,2,3,4,5... or true for all NON-NEGATIVE numbers. Remember the 'non-negative' part. This includes 0 as well.

But |x| = -x for x<0, x=-1,-2,-0.25 ...

This is the reason why we put equality with the '>' sign to account for all non-negative numbers. The 'nature' of 'x' does not change if it is 0 and above but it does change (multiply x by -1) if x is <0.

Hope this helps.


Thanks. Now I'm observing the => convention |x| has. The equal sign is included because we are guessing x could be 0 as well. But if 0 is a non-negative number, it can be a non-positive number too (In that non positive number scenario, we would have followed the <= convention for |x|). In fact 0 is a neutral number. So, in this case, why is 0 implied as a non negative number?


Please refer to the text in red above. The nature of |x| does not change when x = 0 or >0 while it changes when x <0. This is the reason why we club =0 with >0 as they have similar "nature".

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 30 Jul 2015, 22:29
II. Converting inequalities with modulus into a range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).

X should be equal to (-4,4) ??

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Re: Math: Absolute value (Modulus) [#permalink]

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gargatul wrote:
II. Converting inequalities with modulus into a range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).

X should be equal to (-4,4) ??


|x| < 5 means \(-5<x<5\) ---> x can be \(-4.99999999\leq x \leq4.99999999\)

x can be equal to (-4,4) ONLY IF you are given that x MUST be an integer. If not, then x can be -4.99 or 4.99 as well.

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 04 Oct 2015, 16:01
This is most likely a basic question, so I'm missing some foundational knowledge here ...

Why is it true that when X<-8, the equation turns into: −(x+3)−(4−x)=−(8+x) ? Thanks!
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Re: Math: Absolute value (Modulus) [#permalink]

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happyface101 wrote:
This is most likely a basic question, so I'm missing some foundational knowledge here ...

Why is it true that when X<-8, the equation turns into: −(x+3)−(4−x)=−(8+x) ? Thanks!


You need to brush up on your absolute value theory which states that \(|x| = x\) when \(x\geq 0\) but \(|x|=-x\) when \(x<0\)

Now when x<-8, assume x=-9, (x+3) <0, (4-x) >0 and (8+x) < 0

Thus, for all the terms <0, |x|=-x --> |x+3| = -(x+3) and |8+x| = -(8+x)

Also, as (4-x)>0 for x <-8 and as |x|=x for x >0, you get |4-x| = (4-x)

Thus, \(|x+3|−|4−x|=|8+x|\) ---> −(x+3)−(4−x)=−(8+x)

Hope this helps.

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Re: Math: Absolute value (Modulus) [#permalink]

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Engr2012 wrote:
happyface101 wrote:
This is most likely a basic question, so I'm missing some foundational knowledge here ...

Why is it true that when X<-8, the equation turns into: −(x+3)−(4−x)=−(8+x) ? Thanks!


You need to brush up on your absolute value theory which states that \(|x| = x\) when \(x\geq 0\) but \(|x|=-x\) when \(x<0\)

Now when x<-8, assume x=-9, (x+3) <0, (4-x) >0 and (8+x) < 0

Thus, for all the terms <0, |x|=-x --> |x+3| = -(x+3) and |8+x| = -(8+x)

Also, as (4-x)>0 for x <-8 and as |x|=x for x >0, you get |4-x| = (4-x)

Thus, \(|x+3|−|4−x|=|8+x|\) ---> −(x+3)−(4−x)=−(8+x)

Hope this helps.


+1 Kudo!

Thank you for the quick and clear reply - this makes so much sense now! Basically |x| = -x when |x| < 0; |x| = x when |x| > or = 0, so I need to test each of the |x+3|, |4-x| and |8+x| to see if they are > 0 or <0 and translate them accordingly to + or - (x+3), (4-x) etc. Is this the correct understanding?
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Re: Math: Absolute value (Modulus) [#permalink]

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New post 04 Oct 2015, 16:49
happyface101 wrote:
Engr2012 wrote:
happyface101 wrote:
This is most likely a basic question, so I'm missing some foundational knowledge here ...

Why is it true that when X<-8, the equation turns into: −(x+3)−(4−x)=−(8+x) ? Thanks!


You need to brush up on your absolute value theory which states that \(|x| = x\) when \(x\geq 0\) but \(|x|=-x\) when \(x<0\)

Now when x<-8, assume x=-9, (x+3) <0, (4-x) >0 and (8+x) < 0

Thus, for all the terms <0, |x|=-x --> |x+3| = -(x+3) and |8+x| = -(8+x)

Also, as (4-x)>0 for x <-8 and as |x|=x for x >0, you get |4-x| = (4-x)

Thus, \(|x+3|−|4−x|=|8+x|\) ---> −(x+3)−(4−x)=−(8+x)

Hope this helps.


+1 Kudo!

Thank you for the quick and clear reply - this makes so much sense now! Basically |x| = -x when |x| < 0; |x| = x when |x| > or = 0, so I need to test each of the |x+3|, |4-x| and |8+x| to see if they are > 0 or <0 and translate them accordingly to + or - (x+3), (4-x) etc. Is this the correct understanding?


Yes, that is correct.

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Math: Absolute value (Modulus) [#permalink]

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New post 29 Oct 2015, 22:35
a) x<−8. −(x+3)−(4−x)=−(8+x) --> x=−1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) −8≤x<−3. −(x+3)−(4−x)=(8+x) --> x=−15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) −3≤x<4. (x+3)−(4−x)=(8+x) --> x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x≥4. (x+3)+(4−x)=(8+x) --> x=−1. We reject the solution because our condition is not satisfied (-1 is not more than 4)



I have a question about the above. Why is there a negative (-) sign in front of the x+3 term in part a), and not in front of the 4-x term? I think this gets at some basics of absolute values, but am not sure. Any insight is much appreciated. Thanks.


Edit: see question asked right before me, never mind.

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 12 Nov 2015, 14:16
Hi, Walker. many thanks for your great explanation but I think I need a further explanation for stupid person like me. how did you decide the sign of the three expressions in the four cases. please note that I may miss some advanced concepts of absolute value. I need like a map of what are the kind of absolute value question and what are the best and fastest approaches to solve each kind. please help. I need to master my self in absolute value questions especially I have like 2 month left to take the test.

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 12 Nov 2015, 17:55
hatemnag wrote:
Hi, Walker. many thanks for your great explanation but I think I need a further explanation for stupid person like me. how did you decide the sign of the three expressions in the four cases. please note that I may miss some advanced concepts of absolute value. I need like a map of what are the kind of absolute value question and what are the best and fastest approaches to solve each kind. please help. I need to master my self in absolute value questions especially I have like 2 month left to take the test.


In order for anyone to answer your question to the best of their abilities, make sure to quote the question and the step that you have a doubt about. A general question such as the one above will not serve you any good. Please re post your question quoting a particular question from the 1st page of this topic and then we can look into your specific question.

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 23 Nov 2015, 09:09
Hi Karishma

Can u please help? Is that correct that I could solve the 3-mod equation/inequality by opening mod one by one and/or in different combinations (negative-positive) and then test the resulting values OR by setting the intervals and testing values out of those intervals to see whether they suffice the requirements set by the intervals AND get the same result?

The reason for intervals is that they save time, while dealing with 3-mod questions can become a time-consuming exercise) and prone to errors.

THANK YOU
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Re: Math: Absolute value (Modulus) [#permalink]

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New post 13 Aug 2016, 07:07
Hi all,

I have been reading this from GMAT Club Math Book and have been trying to understand the 3-step process to solving absolute value inequalities/equalities and vice versa. I cannot seem to understand how to use this approach.

If anyone can explain it in more elaborate detail I would be very greatful.

Thank you

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 25 Sep 2016, 05:03
Hi Walker,

Very nice and informative post! Thanks a lot.

I need a small help in explaining a few bits to me in a little detail (I am somehow missing something here to get it all right). Can anyone please help. TIA

1) In the tricks example I please explain this bit in a little detail- "Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D."

2) In the trick no II, please explain the second example- "|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)"

sorry if its a very basic query, just want to get the basics right :)

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 25 Sep 2016, 20:14
naren01 wrote:

2) In the trick no II, please explain the second example- "|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)"

sorry if its a very basic query, just want to get the basics right :)


http://www.veritasprep.com/blog/2011/01 ... edore-did/
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Re: Math: Absolute value (Modulus) [#permalink]

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New post 07 Jan 2017, 04:24
Hi everyone,

I'm having problems understanding the example 1 ( as written below). Why are they key points -8, -3 and 4, instead of 3, 4, 8?

Many thanks

Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) --> x=−15x=−15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) --> x=9x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 07 Jan 2017, 05:08
lauramo wrote:
Hi everyone,

I'm having problems understanding the example 1 ( as written below). Why are they key points -8, -3 and 4, instead of 3, 4, 8?

Many thanks

Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) --> x=−15x=−15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) --> x=9x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not more than 4)


|x - a| = b
means x is b units away from a.

|x + a| = b
|x - (-a)| = b
means x is b units away from -a.

Hence, if we have |x + 3|, it means the transition point is -3.

For more on this, check:
https://www.veritasprep.com/blog/2011/0 ... edore-did/
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Re: Math: Absolute value (Modulus) [#permalink]

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New post 09 Jan 2017, 02:43
Got it!
Thanks heaps!

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 20 Jan 2017, 09:15
If x is an integer, what is the value of x?
(1) –2(x + 5) < –1
(2) –3x > 9


from S1 it should be x>4.5...........but in explanation its given x>-4.5.
pls explain the logic

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Re: Math: Absolute value (Modulus) [#permalink]

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New post 21 Jan 2017, 03:33
gupta87 wrote:
If x is an integer, what is the value of x?
(1) –2(x + 5) < –1
(2) –3x > 9


from S1 it should be x>4.5...........but in explanation its given x>-4.5.
pls explain the logic


–2(x + 5) < –1

Divide by -2 and flip the sign: x + 5 > 0.5;

Deduct 5: x > 0.5 - 5

x > -4.5

Discussed here: if-x-is-an-integer-what-is-the-value-of-x-109983.html

Hope it helps.
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Re: Math: Absolute value (Modulus)   [#permalink] 21 Jan 2017, 03:33

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