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chinmay96
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Math: Absolute value (Modulus) 04 Jun 2018, 21:35
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Can someone please explain how we change signs in the steps below?

3-steps approach for complex problems



Let’s consider following examples,

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)

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: 0
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Math: Absolute value (Modulus) 04 Jun 2018, 21:49
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chinmay96
Can someone please explain how we change signs in the steps below?

3-steps approach for complex problems



Let’s consider following examples,

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)

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: 0
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Explained here: https://gmatclub.com/forum/x-3-4-x-8-x- ... l#p1241355 Hope it helps.
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chinmay96
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Math: Absolute value (Modulus) 04 Jun 2018, 22:47
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Got it! Thanks a ton! :)

Bunuel
chinmay96
Can someone please explain how we change signs in the steps below?

3-steps approach for complex problems



Let’s consider following examples,

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)

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: 0
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Math: Absolute value (Modulus) 08 Jun 2018, 08:39
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walker
ABSOLUTE VALUE


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.

....

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 (-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)

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

Answer: 0

Show more

Hey abhimahna, for step "3. Check conditions for each solution:", shouldn't we always have to plug in the value of x in the original equation with absolute value || or it's fine to directly evaluate if the value of x satisfies by just comparing each value of x to its inequality that we got from the points? I believe it's way faster if we just compare it to the inequality, as the example 1 above is doing.

For example: If |x-2|=|2x-3|, what are the possible values for x?

If you see the uploaded photo, you'll see that we get 2 same values of x (i.e. x=1) out of which one satisfies its inequality (x<3/2) and the other doesn't satisfy its inequality (x>=2). So we would count that as one solution. What the same value satisfied 2 inequalities, then would be consider that as 2 solutions or just 1?

Range 1: \(x<3/2\)
-(x-2) = -(2x-3)
x=1 (Satisfies)

Range 2: \(\frac{3}{2}\leq{x} < 2\)
- (x-2) = (2x-3)
x=5/3 (Satisfies)

Range 3: \(x\geq2\)
(x-2)=(2x-3)
x=1 (Doesn't satisfy the range)
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Math: Absolute value (Modulus) 06 Jul 2018, 09:20
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Thanks Sir for this amazing tutorial. Could you please explain how you solved this question because i cant get your solution clearly.

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

For example,
Problem: 1<x<9. What inequality represents this condition?
Image
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.
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Math: Absolute value (Modulus) 25 Jul 2018, 06:15
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hi walker thanks for explainig the concept
can anyone please tell me the solution of |2x+1|=|5x-2| please?
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Math: Absolute value (Modulus) 23 Aug 2018, 02:37
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How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
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Math: Absolute value (Modulus) 01 Sep 2018, 15:16
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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.
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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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Chitralekhakar
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Math: Absolute value (Modulus) 03 Sep 2018, 21:55
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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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Math: Absolute value (Modulus) 22 Sep 2018, 03:50
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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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Math: Absolute value (Modulus) 29 Sep 2018, 06:26
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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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GMAT Focus 1: 735 Q90 V89 DI81
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Math: Absolute value (Modulus) 29 Sep 2018, 10:02
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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?
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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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Math: Absolute value (Modulus) 29 Sep 2018, 10:24
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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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Math: Absolute value (Modulus) 20 Nov 2018, 13:54
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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!!!!!
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I didn't understand how the sign for each expression was determined.
I'd appreciate any help.
Thanks!!
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Math: Absolute value (Modulus) 26 Nov 2018, 16:07
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Bunuel

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread

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When are we supposed to use the 3-step method? I mean, on which kind of problems?
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Math: Absolute value (Modulus) 15 May 2019, 20:55
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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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Math: Absolute value (Modulus) 05 Jul 2019, 05:44
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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.
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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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Math: Absolute value (Modulus) 05 Jul 2020, 13:06
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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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Math: Absolute value (Modulus) 08 Jul 2020, 22:42
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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
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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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Math: Absolute value (Modulus) 31 Jul 2020, 00:08
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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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Fixed. Thank you.
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