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

3. Check conditions for each solution:
a) x=5x=5 has to satisfy initial condition x−1>=0x−1>=0. 5−1=4>05−1=4>0. It satisfies. Otherwise, we would have to reject x=5.
b) x=−3x=−3 has to satisfy initial condition x−1<0x−1<0. −3−1=−4<0−3−1=−4<0. It satisfies. Otherwise, we would have to reject x=-3.

So...just to clarify...the answer can either be 5 or -3?

Just need the clarification on what this is saying
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In the following example of 3-steps approach for complex problems


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) Can any one please explain me how x<-8 because whenever I am doing it I am getting x>=-8 for both |8+x| and -|8+x|

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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IN Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
C condition has typo error
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.)

its not -15 but 9
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chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand :P
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?
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hibobotamuss
chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand :P
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?

Please mention the portion you are talking about..
No problems. Pl keep asking questions and I would reply to each whenever I get time :)
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"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."

This is the "trick" mentioned in the post, don't know what it means chetan2u
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gettinit
Let’s consider following examples,

Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
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) \(x < -8\). \(-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)\) --> \(x = -1\). 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 (-15 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)

I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!


I didn't understand how the sign for each expression was determined.
I'd appreciate any help.
Thanks!!
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OP or Bunuel

There is a typo in the first post.

It says "(-15 is not within (-3,4) interval." for the range in (c) when it should actually say "9 is not within the -3<x<4 range"
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dcummins
OP or Bunuel

There is a typo in the first post.

It says "(-15 is not within (-3,4) interval." for the range in (c) when it should actually say "9 is not within the -3<x<4 range"
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Edited. Thank you.
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VeritasKarishma
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hi,

i have a doubt in the example

Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0



how the range for this example is (X^2-4)>=0 and not (x^2-4)>0?


regards,
RRSNATHAN

You can ignore "=0" part. (x^2 - 4) cannot be 0 since its mod is 1. So you can certainly write the range as (x^2 - 4) > 0. In the overall scheme, it doesn't change anything. People usually write the ranges are ">=0" and "<0" to cover everything. Here "=0" is not needed.

Posted from my mobile device


I have a doubt for this question

Does not this expression mean that x^2 is at a distance of 1 from 4. Then should not the correct answers be only sq root 5 and sq root 3 instead of -sq root 3 and -sq root 5 as these don't have a distance of 1 from 4 ?

Regards
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You said in each range signs of the terms will be different . I am unable to assign proper signs to terms in ranges VeritasKarishma please help

Posted from my mobile device
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Vini07
VeritasKarishma
rrsnathan
hi,

i have a doubt in the example

Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0



how the range for this example is (X^2-4)>=0 and not (x^2-4)>0?


regards,
RRSNATHAN

You can ignore "=0" part. (x^2 - 4) cannot be 0 since its mod is 1. So you can certainly write the range as (x^2 - 4) > 0. In the overall scheme, it doesn't change anything. People usually write the ranges are ">=0" and "<0" to cover everything. Here "=0" is not needed.

Posted from my mobile device


I have a doubt for this question

Does not this expression mean that x^2 is at a distance of 1 from 4. Then should not the correct answers be only sq root 5 and sq root 3 instead of -sq root 3 and -sq root 5 as these don't have a distance of 1 from 4 ?

Regards

Vini07,

It means "x^2" is at a distance 1 from 4, not x.
So x^2 is 5 or 3. Then x can be -sqrt(5) and also sqrt(5) and -sqrt(3) and also sqrt(3)
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Syeedarimshah
You said in each range signs of the terms will be different . I am unable to assign proper signs to terms in ranges VeritasKarishma please help

Posted from my mobile device


The rightmost region is positive and you alternate from there (assuming the factors are in (ax + b) or (ax - b) form). Check out the logic explained in this video:

https://youtu.be/PWsUOe77__E
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Genoa2000
walker
ABSOLUTE VALUE

\(|x|\geq0\)

\(|0|=0\)

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

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

\(|x|\geq0\)


Little typo walker Bunuel. Hope it helps :)
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VeritasKarishma
worldogvictor
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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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|

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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