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gdesaidesai
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Math: Absolute value (Modulus) 12 Feb 2021, 06:42
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VeritasKarishma
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Hi,
In the original post, in the section GMAC books, there are some prob numbers stated. Can you please explain how they relate to the post? How one will use those numbers is what I am asking.

The given problem numbers are official questions on the concept of Modulus.
You need to have those books to be able to access the given questions. The numbers give you the question numbers e.g. in OG12 in sample problem solving questions, question no 22 (on page 155) tests you on Mods.
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Math: Absolute value (Modulus) 20 Oct 2021, 18:51
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yangsta8
Hi Walker, Thanks for posting this.

You've written a property that:
|X + Y| >= |X| + |Y|
Is the same true for negative?
|X - Y| <= |X| - |Y|
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No, the comparison sign gets reversed in |X-Y|

Ex,
|X+Y| <= |X|+|Y| - Less than equals

|X-Y| >= |X|-|Y| - Greater than equals
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Math: Absolute value (Modulus) 02 Nov 2021, 02:57
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Could you please help me :cry:

As for the Example #1
Q.:|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. :shock: :dazed

SO: what did you do here to make this conclusion about those four conditions?
You equated each |...| to zero? Or what?
Sorry but you just provided those conditions as if it is so obviously, but it is not...;)
What does it mean "there are 3 key points here".. where? :-o
What is the algorithm to recieve those four conditions?
Could you please do not skip this and clarify. :please:
Thank you :tongue_opt3
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Math: Absolute value (Modulus) Updated on: 22 Jun 2023, 11:19
Originally posted by 0Lucky0 on 22 Jun 2023, 11:11.
Last edited by 0Lucky0 on 22 Jun 2023, 11:19, edited 1 time in total.
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Q.: |x+3|−|4−x|=|8+x|. How many solutions does the equation have?

KarishmaB, I have read all 6 pages on here looking for your explanations but I can't seem to find a post where you solve this question using your patented number line technique. :P I also read all 3 pages of explanations on the original question post, even there you don't actually answer this question anywhere using that technique.
You did answer a loooooot of questions on this and explained stuff using both the number line and without the number line but I couldn't find a post where you by yourself answer this question using the method you would prefer. :P
This Modulus topic is an extremely complicated topic and I really don't wanna read anyone else's explanations but yours as IMO, you are a master at this; also because reading too much of this is just frying my brain. :P So if possible, please answer this question using your patented number line method in one post. :) Thank you sooo much.. :) You truly are a genius. :)
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Math: Absolute value (Modulus) 22 Jun 2023, 11:13
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0Lucky0
Q.: |x+3|−|4−x|=|8+x|. How many solutions does the equation have?

KarishmaB, I have read all 6 pages on here looking for your explanations but I can't seem to find a post where you solve this question using your patented number line technique. :P I also read all 3 pages of explanations on the original question post, even there you don't actually answer this question anywhere using that technique.
You did answer a loooooot of questions on this and explained stuff using both the number line and without the number line but I couldn't find a post where you by yourself answer this question using the method you would prefer. :P
This Modulus topic is an extremely complicated topic and I really don't wanna read anyone else's explanations but yours as IMO, you are a master at this also because reading too much of this is just frying my brain. :P So if possible, please answer this question using your patented number line method in one post. :) Thank you sooo much.. :) You truly are a genius. :)
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You might find this discussion helpful: https://gmatclub.com/forum/x-3-4-x-8-x- ... 48996.html
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Math: Absolute value (Modulus) 22 Jun 2023, 11:17
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Bunuel
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Q.: |x+3|−|4−x|=|8+x|. How many solutions does the equation have?

KarishmaB, I have read all 6 pages on here looking for your explanations but I can't seem to find a post where you solve this question using your patented number line technique. :P I also read all 3 pages of explanations on the original question post, even there you don't actually answer this question anywhere using that technique.
You did answer a loooooot of questions on this and explained stuff using both the number line and without the number line but I couldn't find a post where you by yourself answer this question using the method you would prefer. :P
This Modulus topic is an extremely complicated topic and I really don't wanna read anyone else's explanations but yours as IMO, you are a master at this also because reading too much of this is just frying my brain. :P So if possible, please answer this question using your patented number line method in one post. :) Thank you sooo much.. :) You truly are a genius. :)

You might find this discussion helpful: https://gmatclub.com/forum/x-3-4-x-8-x- ... 48996.html
Show more
Yea, I did check this as well and this below part in my post was a reply to this topic:
"I also read all 3 pages of explanations on the original question post, even there you don't actually answer this question anywhere using that technique."
Thanks! :)
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Math: Absolute value (Modulus) 26 May 2024, 09:39
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Hi Experts,

Can you please help me understand how signs are given to each equation here? I have trouble understanding how to assigns sings to these equations. I have attached a snip for reference. ­
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Math: Absolute value (Modulus) 26 May 2024, 10:01
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Rohstar750
Hi Experts,

Can you please help me understand how signs are given to each equation here? I have trouble understanding how to assigns sings to these equations. I have attached a snip for reference. ­


 
Show more
­
Absolute Value Properties:

When \(x \le 0\), then \(|x|=-x\), or more generally, when \(\text{some expression} \le 0\), then \(|\text{some expression}| = -(\text{some expression})\).

When \(x \ge 0\), then \(|x|=x\), or more generally, when \(\text{some expression} \ge 0\), then \(|\text{some expression}| = \text{some expression}\).­

So, for example, for |x + 3|, when x + 3 <= 0, or x <= -3, then |x + 3| = -(x + 3) and when x + 3 >= 0, or x >= -3, then |x + 3| = x + 3.
 ­
This is applied to the above problem when expanding absolute values in different ranges. 
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Math: Absolute value (Modulus) 07 Jun 2025, 05:47
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Could you please explain the curve method to find the range of a modulus function?

Bunuel

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread

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Math: Absolute value (Modulus) 24 Oct 2025, 15:16
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Hi Bunuel, in the first example, in the last interval, if i take x=4 then |4-x| should open positively correct? then why did we open it with a -?
I get that if x>4, then the interval will be <0 but not with x=4?
PLEASE help me
walker
ABSOLUTE VALUE
(Modulus)

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel



Definition

The absolute value (or modulus) \(|x|\) of a real number x is x's numerical value without regard to its sign.

For example, \(|3| = 3\); \(|-12| = 12\); \(|-1.3|=1.3\)



Graph:



Important properties:

\(|x|\geq0\)

\(|0|=0\)

\(|-x|=|x|\)

\(|x|+|y|\geq|x+y|\)


How to approach equations with moduli

It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression.

1) \(|x| = \sqrt{x^2}\). This approach might be helpful if an equation has × and /.

2) |x| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below).

3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.
To solve/open a modulus, you need to consider 2 situations to find all roots:
  • Positive (or rather non-negative)
  • Negative

For example, \(|x-1|=4\)
a) Positive: if \((x-1)\geq0\), we can rewrite the equation as: \(x-1=4\)
b) Negative: if \((x-1)<0\), we can rewrite the equation as: \(-(x-1)=4\)
We can also think about conditions like graphics. \(x=1\) is a key point in which the expression under modulus equals zero. All points right are the first condition \((x>1)\) and all points left are second condition \((x<1)\).


2. Solve new equations:
a) \(x-1=4\) --> x=5
b) \(-x+1=4\) --> x=-3

3. Check conditions for each solution:
a) \(x=5\) has to satisfy initial condition \(x-1>=0\). \(5-1=4>0\). It satisfies. Otherwise, we would have to reject x=5.
b) \(x=-3\) has to satisfy initial condition \(x-1<0\). \(-3-1=-4<0\). It satisfies. Otherwise, we would have to reject x=-3.


3-steps approach for complex problems

Let’s consider following examples,

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

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

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

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

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

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Answer: 0

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.

(Optional) The following illustration may help you understand how to open modulus at different conditions.

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


Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,
Problem: 1<x<9. What inequality represents this condition?

A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. 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.

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+3|>3 is equal to x e (-inf,-6)&(0,+inf)

III. Thinking about absolute values as the distance between points at the number line.

For example,
Problem: A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|
Solution:

We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, |X-A| < |Y-A| and at the same time the distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.


Pitfalls

The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions.




Practice Questions

Easy:

Medium:


Hard:



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Math: Absolute value (Modulus) Updated on: 25 Oct 2025, 01:01
Originally posted by Bunuel on 25 Oct 2025, 01:01.
Last edited by Bunuel on 25 Oct 2025, 01:01, edited 1 time in total.
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Saun2511
Hi Bunuel, in the first example, in the last interval, if i take x=4 then |4-x| should open positively correct? then why did we open it with a -?
I get that if x>4, then the interval will be <0 but not with x=4?
PLEASE help me

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At x = 4, |4 - x| = 0, so it doesn’t affect the sign: |0| = -0 = 0.

Recall:

When x ≥ 0, |x| = x
When x ≤ 0, |x| = -x

Given |x + 3| - |4 - x| = |8 + x|

For x ≥ 4, (4 - x) ≤ 0, so |4 - x| = -(4 - x). Substituting:

(x + 3) - [-(4 - x)] = (8 + x)
(x + 3) + (4 - x) = (8 + x)
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Math: Absolute value (Modulus) 25 Oct 2025, 03:24
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When x ≥ 0, |x| = x
When x ≤ 0, |x| = -x

Here, it was x<0 right? not equal to? because then x=0 is both + and - Bunuel
Bunuel


At x = 4, |4 - x| = 0, so it doesn’t affect the sign: |0| = -0 = 0.

Recall:

When x ≥ 0, |x| = x
When x ≤ 0, |x| = -x

Given |x + 3| - |4 - x| = |8 + x|

For x ≥ 4, (4 - x) ≤ 0, so |4 - x| = -(4 - x). Substituting:

(x + 3) - [-(4 - x)] = (8 + x)
(x + 3) + (4 - x) = (8 + x)
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