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|x+3| - |4-x| = |8+x|. How many solutions does the equation

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 28 May 2016, 14:22
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1. Might sound super stupid to you all... but when you say |4-x| = |x-4|

then why are answers different in each case if I try from both the ways,

2. Why are all not positive (|x+3| - |4-x| = |8+x|) in case of x>4?

Like they were when Karishma tried to explaing. I got it from her concept but then I read this one... and it has |4-x| ..now I dont know whats going on.

Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 28 May 2016, 14:26
Could you explain the same by solving with just one of the examples but with |4-x|

I understood the concept you have explained... but I am confused how would it be solved if I dont convert it to |x-4|..

Thanks!
VeritasPrepKarishma wrote:
MadCowMartin wrote:
I am stuck on this part in the Gmat Club book. I do understand how the conditions are set. But I can't figure out how the values for x were determined in those conditions. Ill use just a and b:

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.)

Where do the x = ' ' values come from? I have been staring at this for half an hour.
I understood the whole concept in the '3-steps approach' but the '3-steps approach for complex problems' has me stuck suddenly. There goes mij GMAT-Mojo! Anyone able to help me get it back? Thanks.


BTW: Is this 650+ level?


You solve the equation to get the x = values

First of all, you are given |x+3|-|4-x|=|8+x|
Convert this to |x+3|-|x-4|=|x+8| (since it is a mod, |4-x| is the same as |x-4|)

Now key points are -8, -3 and 4.

Case a: x< -8
When x < -8, all three expressions (x + 3), (x - 4) and (x + 8) are negative.

So |x+3| = - (x + 3) (using the definition of mod)
|x-4| = - (x - 4)
|x+8| = - (x + 8)

-(x+3) - [-(x-4)] = -(x+8)
-7 = -x - 8
x = -1
Condition not satisfied so rejected.

And no, it is 750+ level.

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 29 May 2016, 04:32
Hi Bunuel,

As you mentioned then by using any transition points, can we take the ranges as follows:

a) x≤-8
b) -8<x<-3
c) -3≤x≤4
d) x>4
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 29 May 2016, 20:51
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smritidabas wrote:
Could you explain the same by solving with just one of the examples but with |4-x|

I understood the concept you have explained... but I am confused how would it be solved if I dont convert it to |x-4|..

Thanks!
VeritasPrepKarishma wrote:
MadCowMartin wrote:
I am stuck on this part in the Gmat Club book. I do understand how the conditions are set. But I can't figure out how the values for x were determined in those conditions. Ill use just a and b:

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.)

Where do the x = ' ' values come from? I have been staring at this for half an hour.
I understood the whole concept in the '3-steps approach' but the '3-steps approach for complex problems' has me stuck suddenly. There goes mij GMAT-Mojo! Anyone able to help me get it back? Thanks.


BTW: Is this 650+ level?


You solve the equation to get the x = values

First of all, you are given |x+3|-|4-x|=|8+x|
Convert this to |x+3|-|x-4|=|x+8| (since it is a mod, |4-x| is the same as |x-4|)

Now key points are -8, -3 and 4.

Case a: x< -8
When x < -8, all three expressions (x + 3), (x - 4) and (x + 8) are negative.

So |x+3| = - (x + 3) (using the definition of mod)
|x-4| = - (x - 4)
|x+8| = - (x + 8)

-(x+3) - [-(x-4)] = -(x+8)
-7 = -x - 8
x = -1
Condition not satisfied so rejected.

And no, it is 750+ level.



It doesn't matter whether you use |x - 4| or |4 - x|. The answer will stay the same.

I. Consider |x - 4|

When x >= 4, then (x - 4) is positive (try putting in x = 5) so
|x - 4| = x - 4

When x < 4, then (x - 4) is negative (try putting in x = 3) so
|x - 4| = -(x - 4) = 4 - x

II. Consider |4 - x|

When x >= 4, then (4 - x) is negative (try putting in x = 5 to see) so
|4 - x| = - (4 - x) = x - 4

When x< 4, then (4 - x) is positive (try putting in x = 3 to see) so
|4 - x| = 4 - x

Note that in both cases, when x >= 4, you get (x - 4) and when x < 4, you get (4 - x).
Just that case I is more intuitive so it's easier to flip first and then solve.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 11 Jun 2016, 07:42
Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.



Hi Bunuel,

I am getting confused at the sign changes. Please explain a bit clearly.

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Now we have four ranges:

As per property...................... |x| < 0 ... | some expression | = - some expression ---> Property 1
|x| > 0 ... | some expression | = some expression ----> Property 2.

Confirm the below:
Now case 1 : x < -8 then |x+3| = - (x+3) and |x-4| = - ( x-4) ( this has to -ve but you mentioned as positive ) and |x+8| = - (x +8 )

and confused for the other three ranges as well on how the expressions got +ve and -ve signs.
What all need to be checked to give positive and negative signs to the expressions, as you mentioned in the solution for the above four ranges.

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 12 Jun 2016, 06:59
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msk0657 wrote:
Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.



Hi Bunuel,

I am getting confused at the sign changes. Please explain a bit clearly.

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Now we have four ranges:

As per property...................... |x| < 0 ... | some expression | = - some expression ---> Property 1
|x| > 0 ... | some expression | = some expression ----> Property 2.

Confirm the below:
Now case 1 : x < -8 then |x+3| = - (x+3) and |x-4| = - ( x-4) ( this has to -ve but you mentioned as positive ) and |x+8| = - (x +8 )

and confused for the other three ranges as well on how the expressions got +ve and -ve signs.
What all need to be checked to give positive and negative signs to the expressions, as you mentioned in the solution for the above four ranges.


When x<-8:
x+3 becomes negative, thus |x+3|=-(x+3);
x+8 becomes negative, thus |x+8|=-(x+8);
4-x becomes positive, thus |4-x|=4-x. For example, 4-(-9)=13=positive.

Theory on Absolute Values: math-absolute-value-modulus-86462.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html

DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 19 Jun 2016, 21:49
VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.


Responding to a pm:
[quote]

Request you to throw some more light on this concept
I have confusion why we have taken x<-8 and why not x>-8

[/quote

The transition points are -8, -3 and 4.

Draw them on the number line:

------------------- (-8) ----------- (-3) --------------------------- (4) -------------

In different sections of this number line, the terms are going to behave differently.
When x < -8,
|x + 8| = - (x + 8)
For the other terms too, when we remove the absolute value sign, we need a negative sign.

At x = -8, the sign for |x +8| turns.
When x >= -8 but less than -3, then
|x+8| = x + 8
For the other terms, when we remove the absolute value sign, we need a negative sign.

Hence each of the four sections of the number line are considered separately.

For more on this, check: http://www.veritasprep.com/blog/2014/06 ... -the-gmat/
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 27 Jun 2016, 11:24
Definition of mod:
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0[/quote]

How is this statement correct??
|x| = -x when x < 0

Modulus is always positive!!

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 27 Jun 2016, 11:30
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kanav06 wrote:
Definition of mod:
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0


How is this statement correct??
|x| = -x when x < 0

Modulus is always positive!![/quote]

You could easily test this by picking numbers. Say x = -5 then |-5| = -(-5) = 5 = positive.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 30 Jun 2016, 23:30
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Responding to a pm:

Quote:
Will you please explain why you didn't take -(x+3) + (4-x)

Why you counted -(4-x)


It comes down to whether you understand why we take ranges and change signs. First check this post:
http://www.veritasprep.com/blog/2014/06 ... -the-gmat/

Now take a simpler example:

|x - 2| + |x - 5| = 4

What do you need to consider here?
For |x - 2|, you have to think about x < 2 and x >= 2
For |x - 5|, you have to think about x < 5 and x >= 5

To consider both together, you have
one range x < 2.
Another is x >= 2 and x < 5 so this is 2 <= x < 5.
Yet another is x >= 5

Now, will you worry about x < 2 and at the same time, x > 5? No. There will not be such a value of x. This is where |x - 2| will open as -(x - 2) and |x - 5| will open as (x - 5). This doesn't exist.

So in a question like this: |x - 2| + |x - 5| = 4, you don't have 4 cases ((x - 2) positive or negative and (x - 5) positive or negative). You have only 3 cases. The 4th case will not exist.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 04 Jul 2016, 08:41
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


A faster way to solve this is to notice that:
|x+3|-|4-x|=|8+x|>=0

this give us 2 equations:
1. |x+3|-|4-x|>=0 - > solving 4 cases here (which 2 cases are not possible) give us x>=1/2 and x<=1/2 -> 1/2 is the only value for all the 4 options
2. |8+x|>=0 -> give us x>=-8 or x<=-8 -> which give us 1 value x=1/2

Those 2 equations should have a shared area on the number line, and since they do no have it, there is no solution for this euqation -> 0 values.

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 29 Aug 2016, 05:45
johnwesley wrote:
You do it in the proper way. This method is called the "critical values" method. And once you have the critical values (by doing each absolute term equal to zero), you have to place them in the Real numbers line to make all the possible intervals. Then you just do the intervals as follows: x<lowest number in your real line, and then you take the intervals from each critical value to before the next one: i.e. x<-8, -8<=x<-3, -3<=x<4, x<=4. Therefore, you are getting all the possible intervals in the real line, and splitting the intervals from one critical value (including it) to before the next critical value (not including it).

Then, as you have done, you just set the predominant sign for each term under each condition, you solve the equation, and finally you check if the result saqtisfies the condition.

--------------------------------------------------------------------------------

I find the above technique easier for these problems.

Can someone please confirm if my below approach is correct :

The roots/criticial values for the given equations are : -8,-3,4.

Now placing these values on the no. line we have four ranges & checking for each ranges :

1. For x greater than 4

Randomly selecting any value greater than 4 , lets say x = 10

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

|10+3| - |4-10| is not equal to |8+10|

So any value greater than 4 doesnot satisfy the equation.

2. For x between -3 & 4

Let x = 2
|x+3| - |4-x| = |8+x|

|2+3| - |4-2| is not equal to |8+2|

So any value between -3 & 4 doesnot satisfy the equation.

3. For x between -8 & -3,

Let x = -2

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

|-2+3| - |4-(-2)| is not equal to |8+(-2)|

So any value between -8 & -3 doesnot satisfy the equation.

4. For x lesser than -8,

Let x = -9

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

|-9+3| - |4-(-9)| is not equal to |8+(-9)|

So any value less than -8 doesnot satisfy the equation.

So the total solution is 0.

Please confirm if my above approach is correct .


Thanks
Kshitij

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 31 Aug 2016, 00:54
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kshitij89 wrote:
johnwesley wrote:
You do it in the proper way. This method is called the "critical values" method. And once you have the critical values (by doing each absolute term equal to zero), you have to place them in the Real numbers line to make all the possible intervals. Then you just do the intervals as follows: x<lowest number in your real line, and then you take the intervals from each critical value to before the next one: i.e. x<-8, -8<=x<-3, -3<=x<4, x<=4. Therefore, you are getting all the possible intervals in the real line, and splitting the intervals from one critical value (including it) to before the next critical value (not including it).

Then, as you have done, you just set the predominant sign for each term under each condition, you solve the equation, and finally you check if the result saqtisfies the condition.

--------------------------------------------------------------------------------

I find the above technique easier for these problems.

Can someone please confirm if my below approach is correct :

The roots/criticial values for the given equations are : -8,-3,4.

Now placing these values on the no. line we have four ranges & checking for each ranges :

1. For x greater than 4

Randomly selecting any value greater than 4 , lets say x = 10

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

|10+3| - |4-10| is not equal to |8+10|

So any value greater than 4 doesnot satisfy the equation.

2. For x between -3 & 4

Let x = 2
|x+3| - |4-x| = |8+x|

|2+3| - |4-2| is not equal to |8+2|

So any value between -3 & 4 doesnot satisfy the equation.

3. For x between -8 & -3,

Let x = -2

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

|-2+3| - |4-(-2)| is not equal to |8+(-2)|

So any value between -8 & -3 doesnot satisfy the equation.

4. For x lesser than -8,

Let x = -9

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

|-9+3| - |4-(-9)| is not equal to |8+(-9)|

So any value less than -8 doesnot satisfy the equation.

So the total solution is 0.

Please confirm if my above approach is correct .


Thanks
Kshitij


Note that in this case, all values in the range will not satisfy the equation.
Say when you take x < -8, you could have got x = -9. But what if you had actually tried x = -10 only and decided that no value in the range satisfies?
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 10 Sep 2016, 07:51
Hi Bunuel,

Is it possible to solve the equation this way?

We see that (x+8)-(x+3)=5
|4-x|=|x-4| is positive. Therefore, -(x-4)=5 and x=-1. With x=-1 two sides of the equation are not equal, thus there are no solutions.

Thank you in advance.


Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.

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New post 09 Oct 2016, 17:05
Thanks everyone, but, would you solve this problem or just take a guess? Even though we know how to solve it, it would take way more than 2 minutes to do such.

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 26 Mar 2017, 05:30
previous solutions are confusing ! but finally understood sign problem

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 09 Apr 2017, 06:34
WholeLottaLove wrote:
Can someone tell me if this approach is correct?

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

So we have:

x=-3
x=4
x=-8

x<-8
-(x+3) - (4+x) = -(8+x)
-x-3 - 4 - x = -8-x
-2x-7=-8-x
1=x (fails, as x is 1 when it must be less than -8)

-8<x<-3
-(x+3) - (4-x) = (8+x)
-x-3 -4+x=8+x
-7=8+x
-15=x (fails, as x is -15 when it must be between -8 and -3)

-3<x<4
(x+3)-(4+x)=8+x
-1=8+x
-9=x (fails, as x is -9 when it must be between -3 and 4)

x>4
(x+3) - -(4-x) = (8+x)
x+3 - (-4+x) = (8+x)
x+3 +4-x=8+x
7=8+x
x=-1 (fails, as x=-1 when it must be greater than 4)

Is this correct?

Thanks!




I understood how to calculate key point (or range)
but plz tell me how do we decide the sign (x+3), (4-x) & (8+x) for different ranges

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 16 Apr 2017, 18:46
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this problem has a very simple solution, give me your kudos after reading it, thanks

so, first move the -[4-x] to the right,
2nd step, square both sides of the equation
3rd, take all variables to one side, simplify the result, (right now, one side should have 0)
4th, you can see one side always is greater than 0, then the equation cannot hold, then there is no value of x

pls, give me kudos, thanks

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 13 Jul 2017, 09:55
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4


[Reveal] Spoiler:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


Can we use number line to solve this?
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 26 Jul 2017, 09:58
VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?


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 <= 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)

thanks !


|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.


hi mam

in "a" when assumed that x is less than -8, the RHS modulus (8+x) is correctly negated. I am okay with this operation. Side by side, however, the LHS modulus (x + 3) is also negated...

what is the main logic underlying this negation ..? is this such that to make the LHS negative, some number should be subtracted from some bigger negatives..? if so, how can we conclude that (x+3) is bigger in value than is (4-x) ..?

thanks in advance ...

Kudos [?]: 0 [0], given: 176

|x+3| - |4-x| = |8+x|. How many solutions does the equation   [#permalink] 26 Jul 2017, 09:58

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