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Math: Absolute value (Modulus) 14 Apr 2014, 21:18
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Thank you Karishma. That was helpful
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Math: Absolute value (Modulus) 20 May 2014, 14:04
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Dear Honey

Here's a quick shortcut to solving this problem:

You know that 1<x<9. So, pick x=6 and substitute in each of the options. Only options C and D remain.

But, if you analyze option C, you'll find that it'll also hold true for x=0 or x=-1, which are outside the given domain of x. So, the given domain cannot be for option C.

So, the solution must be the only option remaining- D. :-D
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Math: Absolute value (Modulus) 14 Jun 2014, 19:05
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Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did.....
\(\frac{|x+5|}{|x - 5|}>3\)

==>\(\frac{x+5}{x - 5}>3\) & \(\frac{x+5}{x - 5}<-3\)
==>\(x+5>3x-15\) & \(x+5<-3x+15\)
==>-2x>-20 & 4x<10
==>x<10 & x<5/2
BUT answer is 5/2<x<10.
What am I doing wrong??
In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??
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Math: Absolute value (Modulus) 17 Jun 2014, 10:07
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In the absolute value question, I don't get why the distance between A and X is not explained. There's discussion about Y and A, Y and B, X between A and Y, distance between X and B and Y and B but not A and X , I need a reexplanation, maybe with numbers
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Math: Absolute value (Modulus) 17 Jun 2014, 20:27
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sagnik242
My questions: if we have 3 points why do we have 4 conditions? Next, for each of the letters (a,b,c,d) how do we know which equation to use and why is the inequality used? For example, for a) we do x = -8, but then how to we know the equation to use is -(x+3) - (4-x) = -(8+x), and how did the negative sign come in front of the -x+3 and why? I have the same type of questions for b,c,d so maybe an answer to a will help
Show more

Plot the 3 points: -8, -3 and 4 on the number line. Now there are 4 distinct regions on the number line:
The part that lies to the left of -8,
the part that lies between -8 and -3,
the part that lies between -3 and 4 and
the part that lies to the right of 4.

In each of these regions, each of the absolute value expressions will behave differently.

You should know the definition of absolute value:

|x| = x if x >= 0
|x| = -x if x < 0

Consider the equation:
|x+3| - |4-x| = |8+x|

What happens in the region which is at the left of -8 say x = -10
x+3 = -10+3 = -7
Since x+3 is negative, |x+3| = -(x + 3)

4-x = 4 - (-10) = 14
Since 4-x is positive, |4 - x| = 4 - x

8 + x = 8 - 10 = -2
Since x+8 is negative, |8+x| = -(8+x)

This is the logic used. Now repeat it for each region.
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Math: Absolute value (Modulus) 15 Oct 2014, 23:25
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I dont seem to understand that in the last option (d), why the sign of (4-x) is reversed to positive?
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Math: Absolute value (Modulus) 16 Oct 2014, 00:42
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mina89
I dont seem to understand that in the last option (d), why the sign of (4-x) is reversed to positive?
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When \(x \geq 4\), |4 - x|=-(4 - x), hence -|4 - x| = -(-(4 - x)) = (4 - x)
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Math: Absolute value (Modulus) 08 Dec 2014, 09:31
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VeritasPrepKarishma


If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:
Ques2.jpg
Show more

VeritasPrepKarishma

OR consider that lx+3l>3 means distance of x from -3 is more than 3.
If you go to 3 steps to right from -3, you reach 0. Anything after than is ok.
If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
Attachment:
Ques2.jpg
Show more

I didn't understand the highlighted portion, Could you please explain:
1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value?
2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

TIA!
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Math: Absolute value (Modulus) Updated on: 26 Nov 2023, 02:58
Originally posted by KarishmaB on 08 Dec 2014, 21:09.
Last edited by KarishmaB on 26 Nov 2023, 02:58, edited 1 time in total.
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appleid
VeritasPrepKarishma


If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:
Ques2.jpg

VeritasPrepKarishma

OR consider that lx+3l>3 means distance of x from -3 is more than 3.
If you go to 3 steps to right from -3, you reach 0. Anything after than is ok.
If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
Attachment:
Ques2.jpg

I didn't understand the highlighted portion, Could you please explain:
1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value?
2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

TIA!
Show more

Here is a video that explains this logic: https://youtu.be/oqVfKQBcnrs
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Math: Absolute value (Modulus) 22 Jan 2015, 08:24
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Hello,

There are a few things I don't find clear in the theory about absolute value, and particularly the 3 step method.

1) How do we decide which are the points we are interested in?
In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x?
2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest?
3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=?
4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?

Thank you.
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Math: Absolute value (Modulus) 29 Jan 2015, 10:32
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pacifist85

1) How do we decide which are the points we are interested in?
In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x?
2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest?
3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=?
4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?
Show more

1) When the expression under modulus equals 0. The expression may or may not change its sign in those special points.
2) Yes, because the number line goes from -infinity all the way to +infinity. -4 is not a special point (see #1; 4 - (-4) = 8 ≠ 0)
3) All conditions cover the number line and break at the special points.
4) Using the definition of modulus: for y = |x|, y = x if x >=0 and y = -x if x < 0.
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Math: Absolute value (Modulus) 12 Jun 2015, 02:29
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I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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)
Why are the all the signs here negative?
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.)
Why are the signs in the first part here negative and not the second?
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.)
Why is this similar to the original equation?
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)
This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Thank you!
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Math: Absolute value (Modulus) 07 Jan 2017, 04:24
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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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Math: Absolute value (Modulus) Updated on: 26 Nov 2023, 02:58
Originally posted by KarishmaB on 07 Jan 2017, 05:08.
Last edited by KarishmaB on 26 Nov 2023, 02:58, edited 1 time in total.
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lauramo
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)
Show more

|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://youtu.be/oqVfKQBcnrs
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Math: Absolute value (Modulus) 20 Jan 2017, 09:15
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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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Math: Absolute value (Modulus) 21 Jan 2017, 03:33
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gupta87
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
Show more

–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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Math: Absolute value (Modulus) 21 Jan 2017, 07:55
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gupta87
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.
Show more

understood.but please correct me where am i going wrong??
-2x-10<-1 => -2x<9 =>2x>9 => x>4.5
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regarding this |x - 3| well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? :? \(x\) could be -5 no :? ?

can you advice in which case can i square both sides od modulus ? can i do it |x+4| + |x - 3| = 10


How about this question \(|x-2| + |x+5| - |x-3| > |x+13|\) woukd i be able to square both sides ? if not please explain why ? :? :)


thank you in advance for your kind explanation and have an awesome day :)
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Math: Absolute value (Modulus) 09 Feb 2018, 12:32
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VeritasPrepKarishma
dave13


regarding this |x - 3| well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? :? \(x\) could be -5 no :? ?

can you advice in which case can i square both sides od modulus ? can i do it |x+4| + |x - 3| = 10


How about this question \(|x-2| + |x+5| - |x-3| > |x+13|\) woukd i be able to square both sides ? if not please explain why ? :? :)


thank you in advance for your kind explanation and have an awesome day :)

Hey Dave,

I am not sure I understand your question but here is something that might help you:

Why do we take the range as 'x - 3 > 0' ? or 'x + 5 > 0' ?

The definition of absolute value will help you understand this. It is discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/

Can you square |x+4| + |x - 3| = 10?
Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?
Show more



Hi VeritasPrepKarishma, many thanks for you reply and the article :)

i wish i could answer your question, :) i dont know how i would take care off it but there is one math wizzard called pushpitkc (perhaps he learnt from you :) ) who actually applied techique of squaring modulus in this question :) https://gmatclub.com/forum/how-many-val ... l#p2011543

i extracted his solution here (in wine red :-) with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened :) otherwise your laptop risks to be frozen like mine :)

so see below this magician squares modulus :) so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ?

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5


If |x-7| = a, the equation will become |a - 9| = 11 ( how did he break this equation into two btw ? :)

Squaring on both sides,
\(a^2\)−18a+81=121
\(a^2\)−18a+81=121

\(a^2\)−18a−40=0
\(a^2\)−18a−40=0

Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)
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Math: Absolute value (Modulus) 18 Feb 2018, 08:26
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Hi VeritasPrepKarishma

many thanks for taking to explain ! :)

i still have some questions :-)

you say: "
Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign."

what does "within the sign" mean ?

Another question: ok so we have \(|x+4| + |x - 3| = 10\)

tha formula we use: \((a + b)^2 = a^2 + b^2 + 2ab\)

Do you mean this formula wont work because

\(a = |x + 4|\) and \(b = |x - 3|\) whereas is in other equations we only one modulus on each side like this one \(|2x + 3| = |x - 4|\) or this one \(|a - 9| = 11\)

but on the other hand \(a = |x + 4|\) and \(b = |x - 3|\) i could write / divide left side into two modulus |x + 4| and |x - 3|

so \(|x + 4|\) --> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get ---> \(x^2+8x+16 = 0\)

\(|x - 3|\)---> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get ---> \(x^2 - 9x +9 = 0\)


Another point: you write this -->

"So\((|x+4| + |x - 3| )^2\)= \(|x + 4|^2 + |x - 3|^2 + 2*|x + 4|*|x - 3|\)

\(= (x + 4)^2 + (x - 3)^2 + 2*|x + 4|*|x - 3|\)"

i dont understand why after this (|x+4| + |x - 3| )^2 you still keep modulus sign :? we square it right

for example here everything is simple and clear ---> \(|x + 2|^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left) <--- you square and modulus sign is gone:)

i would appreciate your explanation :)

best,

D.
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Bunuel
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