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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 10 Dec 2014, 09:52
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.



So what you're saying is the order of whats inside the modulus doesnt matter as long as the sign outside changes?
I ask because i was very confused on why in gmat club book they didnt conver 4-x to -(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 10 Dec 2014, 22:15
saadis87 wrote:

So what you're saying is the order of whats inside the modulus doesnt matter as long as the sign outside changes?
I ask because i was very confused on why in gmat club book they didnt conver 4-x to -(x-4)


Note that

|x| = |-x|

If x = 5,
|5| = |-5| = 5

If x = -5,
|-5| = |-(-5)| = 5

In any case, |x| is always same as |-x|. (Note that we are not saying that |x|= x or |x| = -x because that depends on the sign of x)

So |x-4| = |-(x-4)| = |4-x|
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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 Jun 2015, 09:16
WholeLottaLove wrote:
You have |x+3| - |4-x| = |8+x|

First, look at the three values independently of their absolute value sign, in other words:
|x+3| - |4-x| = |8+x|
(x+3) - (4-x) = (8+x)

Now, you're looking at x < - 8, so x is a number less than -8. Let's pretend x = -10 here to make things a bit easier to understand.

when x=-10

I.) (x+3)
(-10+3)
(-7)


II.) (4-x)
(4-[-10]) (double negative, so it becomes positive)
(4+10)
(14)

III.) (8+x)
(8+-10)
(-2)

In other words, when x < -8, (x+3) and (8+x) are NEGATIVE. To solve problems like this, we need to check for the sign change.

Here is how I do it step by step.

I.) |x+3| - |4-x| = |8+x|

II.) IGNORE absolute value signs (for now) and find the values of x which make (x+3), (4-x) and (8+x) = to zero as follows:

(x+3)
x=-3
(-3+3) = 0

(4-x)
x=4
(4-4) = 0

(8+x)
x=-8
(8+-8) = 0

Order them from least to greatest: x=-8, x=-3, x=4 These become our ranges for x as follows:

x<-8
-8≤x<-3
-3≤x<4
x>4

So, we test values less than the smallest number, values of x between the smallest and largest number, and values of x greater than the greatest number.

So, now we test the original (x+3) - (4-x) = (8+x) with x values. This is where the sign changes in the equation become important. We need to find the number of solutions for this problem so we need to see for which values of x the problem is valid or not valid. For example:

When x < -8

(x+3) is a negative number
(4-x) is a positive number
(8+x) is a negative number

So

-(x+3) - (4-x) = -(8+x)
-x-3 -4+x = -8-x
-7=-8-x
1=-x
x=-1

Now, we are looking at values for x < -8, yet the result we got was x = -1. -1 DOES NOT fall in the range or x < -1. If you don't understand why simply draw a number line, mark down x< -8 and x=-1. Is -1 less than -8? Nope! Therefore, -1 is NOT a valid solution.

You can repeat this step for the remaining ranges of x.

I hope this helped you! :-D





Bunuel and VeritasPrepKarishma

I would like to seek some help from your end.

We do know that | x-4| = |4-x|

So in the above equation |x +3| - |4-x| = |8+x|
I have deliberately converted the middle | 4-x| = |x-4| for my convienance.

& In case I when x < -8 I checked & found that as then -8 is definately on the left hand side of ( 4-x ) or ( x-4). So definately this term should be negative. I checked with a example of -10 also.

But in both the cases the answers are different. If this is the case then I should limit my self from changing | x-4| = |4-x|
& strictly consider |4-x| in its original form??

If that were the case still by defination : | 4-x| = - (4-x) when (4-x) <=0 meaning x > 4 & similarly when x<4 I will have (4-x).

Now coming to the case I:

we have x< -8 definately less than 4. That means the bracket will open with a positive sign i.e. (4-x)

SO now adding I am getting: -(x+3) -(4-x) = -(8+x) = -1 but not in line with the initial condition.
Thus this range is not possible.

SO I am still not sure where I am getting confused.
Secondy, I am wondering why here |x-4| not equals |4-x|? Under what conditions we can do & when we cannot do such interchange?

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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 Jun 2015, 09:28
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ankushbagwale wrote:
WholeLottaLove wrote:
You have |x+3| - |4-x| = |8+x|

First, look at the three values independently of their absolute value sign, in other words:
|x+3| - |4-x| = |8+x|
(x+3) - (4-x) = (8+x)

Now, you're looking at x < - 8, so x is a number less than -8. Let's pretend x = -10 here to make things a bit easier to understand.

when x=-10

I.) (x+3)
(-10+3)
(-7)


II.) (4-x)
(4-[-10]) (double negative, so it becomes positive)
(4+10)
(14)

III.) (8+x)
(8+-10)
(-2)

In other words, when x < -8, (x+3) and (8+x) are NEGATIVE. To solve problems like this, we need to check for the sign change.

Here is how I do it step by step.

I.) |x+3| - |4-x| = |8+x|

II.) IGNORE absolute value signs (for now) and find the values of x which make (x+3), (4-x) and (8+x) = to zero as follows:

(x+3)
x=-3
(-3+3) = 0

(4-x)
x=4
(4-4) = 0

(8+x)
x=-8
(8+-8) = 0

Order them from least to greatest: x=-8, x=-3, x=4 These become our ranges for x as follows:

x<-8
-8≤x<-3
-3≤x<4
x>4

So, we test values less than the smallest number, values of x between the smallest and largest number, and values of x greater than the greatest number.

So, now we test the original (x+3) - (4-x) = (8+x) with x values. This is where the sign changes in the equation become important. We need to find the number of solutions for this problem so we need to see for which values of x the problem is valid or not valid. For example:

When x < -8

(x+3) is a negative number
(4-x) is a positive number
(8+x) is a negative number

So

-(x+3) - (4-x) = -(8+x)
-x-3 -4+x = -8-x
-7=-8-x
1=-x
x=-1

Now, we are looking at values for x < -8, yet the result we got was x = -1. -1 DOES NOT fall in the range or x < -1. If you don't understand why simply draw a number line, mark down x< -8 and x=-1. Is -1 less than -8? Nope! Therefore, -1 is NOT a valid solution.

You can repeat this step for the remaining ranges of x.

I hope this helped you! :-D





Bunuel and VeritasPrepKarishma

I would like to seek some help from your end.

We do know that | x-4| = |4-x|

So in the above equation |x +3| - |4-x| = |8+x|
I have deliberately converted the middle | 4-x| = |x-4| for my convienance.

& In case I when x < -8 I checked & found that as then -8 is definately on the left hand side of ( 4-x ) or ( x-4). So definately this term should be negative. I checked with a example of -10 also.

But in both the cases the answers are different. If this is the case then I should limit my self from changing | x-4| = |4-x|
& strictly consider |4-x| in its original form??

If that were the case still by defination : | 4-x| = - (4-x) when (4-x) <=0 meaning x > 4 & similarly when x<4 I will have (4-x).

Now coming to the case I:

we have x< -8 definately less than 4. That means the bracket will open with a positive sign i.e. (4-x)

SO now adding I am getting: -(x+3) -(4-x) = -(8+x) = -1 but not in line with the initial condition.
Thus this range is not possible.

SO I am still not sure where I am getting confused.
Secondy, I am wondering why here |x-4| not equals |4-x|? Under what conditions we can do & when we cannot do such interchange?


Please read the whole thread:
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1193962
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1237206
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1238650
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1241339
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1241355
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996-20.html#p1323676
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996-20.html#p1328926
x-3-4-x-8-x-how-many-solutions-does-the-equation-148996-40.html#p1454309
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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 Jun 2015, 09:39
Thanks Bunuel for pointing out relevant thread.

I am simply amazed by the sheer depth of though process that has gone into every single question at gmat club.

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

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A better a faster method involves squaring both the sides of the equation which finally ends up with Left hand as always negative and Right hand side as always positive, which is never possible.
Hence no solution.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 05 Jul 2015, 11:29
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.


Is this the fastest way to do this problem? It took me 4 minutes to do this 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 05 Jul 2015, 12:16
TooLong150 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.


Is this the fastest way to do this problem? It took me 4 minutes to do this problem.


I did the question by a graphical method. With absolute values, it is sometimes easier to draw graphs and evaluate the questions graphically.

The given question will have 'n' solutions if the 3 lines given by the equations:

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

once you do that, it becomes apparent that there are no points that are points of intersections of 3 lines (for us to get a solution, we need to have 3 of the lines intersecting at some common points!). Attached is the graph for the same (Sets of parallel lines are: {A||B||C} and {D||E||F}). This method will be an overkill for simpler problems though.

Hope this helps
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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 Aug 2015, 15:12
I was going to reply and say that the graphical/visual way may work best for many cases but I saw that the poster above me has already covered it.

For many absolute values that are simple addition and subtraction, you might get to the answer quicker and more accurately if you just draw it out, especially if its in the form of abs(x-b) + abs(x+a) etc without a 4x-c or kx-c

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Is there any shortcut to do this question in less than 2mins?

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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 Aug 2015, 23:11
lav4 wrote:
Is there any shortcut to do this question in less than 2mins?


I have not seen such a question on GMAT but if needed, I think the graphical method mentioned in my post above would be the fastest one. Alternate method is to look at various intervals algebrically , doing so might be a bit more time consuming.
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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 2015, 11:43
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 not able to understand some part of point 2 where the range is -8<=x<=-3. What I understood is that the numbers possible are -8 to -3 as possible values of x. Now if we take -8 as value of x then |x+3| is negative but if we take -3 as value of x then it becomes 0 (-3+3=0). But we took it as negative. Same case when we take value of x is -8 for |8+x|. Where my understanding in wrong? Thanks in advance.
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 18 Mar 2016, 21:30
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, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help

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New post 18 Mar 2016, 21:44
riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help


Hi,

you have to pick a VALUE falling in the range you are taking and see what happens to the value within the MOD..
1) if the solution of the MOD is a negative number, add a -ive sign..
2)if the solution of the MOD is a positive number, add a +ive sign..


let me show with some examples --
C) -3 <= x < 4 ----- (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
take x as 0 as it falls in the given range -3 <= x < 4
see what happens to each MOD at this value
i) |x+3|.. 0+3=3 so + sign in front of MOD .. (x+3)
ii) |4-x|.. 4-0=4 so + sign in front of MOD


b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

take the value as -5..
|x+3|.. -5+3=-2 so a negative sign.. -(x+3)
|4-x|.. 4-(-5)=9.. so +sign.. (4-x)
|8+x|.. 8-5=3 so +sign... (8+x)


Hope it helps
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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 Mar 2016, 05:54
Thanks Bunuel...for a clear and step wise explaination!

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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 Mar 2016, 09:16
riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help


Let me first tell you that in this question, x < 0 has no significance.

|x| = x when x >= 0
|x| = -x when x < 0

On the same lines,
|x + 4| = x + 4 when (x + 4) >= 0
|x + 4| = -(x + 4) when (x + 4 ) < 0

In the first definition, x is just a placeholder for any expression.

|x^2 - 8| = x^2 - 8 when (x^2 - 8) >= 0
|x^2 - 8| = -(x^2 - 8) when (x^2 - 8) < 0

So, how do you get rid of |x + 3| in the original question? You take two cases: (x + 3) >= 0 or (x + 3) < 0
|x + 3| = x + 3 when (x + 3) >= 0 (i.e. when x >= -3)
|x + 3| = -(x + 3) when (x + 3) < 0 (i.e. when x < -3)
That is how you get -3 as a transition point.

Do the same for other expressions.
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New post 13 May 2016, 18:33
I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

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 !

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

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New post 15 May 2016, 23:26
ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

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 !


Note that \((a + b)^2 = a^2 + b^2 + 2ab\)

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

You missed out the last term. You would need to square it yet again and that will complicate the question further.
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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 May 2016, 17:47
ohh right...thanks.

but if there is a situation where |a-b| = |e+f| (variables or constants), then I can safely square them right? Or do I need to keep certain things in mind before doing that.

VeritasPrepKarishma wrote:
ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

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 !


Note that \((a + b)^2 = a^2 + b^2 + 2ab\)

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

You missed out the last term. You would need to square it yet again and that will complicate the question further.

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

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ameyaprabhu wrote:
ohh right...thanks.

but if there is a situation where |a-b| = |e+f| (variables or constants), then I can safely square them right? Or do I need to keep certain things in mind before doing that.



Yes you can. No problem with that. You can square without worrying, if squaring helps your case.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation   [#permalink] 17 May 2016, 00:09

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