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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 07 Feb 2014, 02:53
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PerfectScores wrote:
Now |x+ 3| will always be lesser than |8+x|


Not necessary.
Put x = -10

|x+3| = |-10 + 3| = 7
|8+x| = |8-10| = 2

It is not necessary that |x+3| will be less than |8+x|.

Note the meaning of absolute value.

|x+3| is the distance of x from -3.
|x+8| is the distance of x from -8.
It is not essential that every point on the number line will be closer to -3 than to -8. All points to the left of -5.5 will be closer to -8 while all points to the right of -5.5 will be closer to -3.
So when x > -5.5, |x+3| < |x+8|
When x < -5.5, |x+3| > |x+8|
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 18 May 2014, 20:49
NGGMAT wrote:
Bunuel wrote:
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. \(x\geq{4}\)




IS THE RED HIGHLIGHTED PORTION A RULE TYPE??? AS IN WHEN WE HAVE SUCH EQUATIONS... WE FIRST FIND THE CRITICAL POINTS AND THE RANGE CHECKS ARE ALWAYS IN THE FOLLOWING PATTERN?

X<= LOWEST VALUE
LOWEST VALUE <X< 2ND LOWEST VALUE
2ND LOWEST VALUE <= X<= HIGHEST VALUE
X> HIGHEST VALUE


I UNDERSTOOD THE FACT THAT TRANSITION POINTS ARE INCLUDED ONLY ONCE...
SO SHOULD I TAKE IT AS A RULE AND IF ANY SUCH QS COMES UP... APPLY THIS RULE?


In addition, check out this post: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
It explains this question type in detail.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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To All who get confused about what sign to put when opening the modulus



This little tip will help you a lot.

After you've determined the critical points, plot them on the number line.

Attachment:
Critical Points on Number Line.PNG
Critical Points on Number Line.PNG [ 2.04 KiB | Viewed 2668 times ]


Now, Check the coefficient of x in each modulus.

For Example, let's take |8+x|. The coefficient of x is +1. Positive. So, on the number line, put a + sign on the Right of the critical point -8, and a - sign on the Left of -8.

Similarly, for |x+3| too, the coefficient of x is positive. Again, we'll put a + sign on the Right of critical point -3, and a - sign on Left of -3.

For |4-x|, the coefficient of x is -1. It's negative. So here the direction of signs will be reversed, that is, we'll put a + sign on Left of critical point 4 and a - sign on Right of 4.

This is what you'll get:

Attachment:
Opening the Modulus.PNG
Opening the Modulus.PNG [ 2.73 KiB | Viewed 2663 times ]


So, now, while opening the modulus, just refer to this plot.

Example:

For -8<=x<=-3
|x+8| will be opened as x+8 [since the graph shows that (x+8) is positive for all x>-8]
|4-x| will be opened as 4-x [since the graph shows that (4-x) is positive for all x<4]
|x+3| will be opened as -(x+3) [as the graph shows that (x+3) is negative for all x<-3]

This plot takes only a second and leaves no chance of confusion about the signs in opening the Modulus.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 21 May 2014, 03:39

A question to practice the above tip:



Plot on a number line the signs for each modulus in the following equation:

\(-|2x-3| + |5-x| + |x-10| = |3-x|\)



[Reveal] Spoiler:
Attachment:
Solution.PNG
Solution.PNG [ 3.09 KiB | Viewed 2565 times ]

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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 Jun 2014, 02:08
[quote="guerrero25"]|x+3| – |4-x| = |8+x| How many solutions will this equation have?

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

Logically and imaginary it means that the distance from X to -3 minus distance from X to 4 must be equal to distance from X to -8. Whereever you place X in number line you never get such outcome

________-8__________-3________________4______


But what if we make change like that:

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

I found -15 as only solution with image use
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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 Jun 2014, 02:11
Temurkhon 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

Logically and imaginary it means that the distance from X to -3 minus distance from X to 4 must be equal to distance from X to -8. Whereever you place X in number line you never get such outcome

________-8__________-3________________4______


But what if we make change like that:

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

I found -15 as only solution with image use


|4-x| - |x+3| = |8+x| has TWO solutions -15 and -7/3.
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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 03 Sep 2014, 06:40
I'm not the first one to say this but the way I solved this problem was by thinking about the equation. The equation says the distance from -3 minus the distance from 4 equals the distance from -8. I drew a picture and the closest I got was -2 which is 1-6=|-5| (but -2 is 6 away from -8). If you think this way you can see that x can't be less than -8 or greater than 4... but for nonintegers, I think the 4 step method is much more precise.
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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, 08:52
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, 21:15
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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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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ankushbagwale wrote:
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:
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1193962
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1237206
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1238650
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1241339
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1241355
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1323676
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#p1328926
http://gmatclub.com/forum/x-3-4-x-8-x-h ... l#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, 08:39
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, 11:16
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, 14:12
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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|x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 18 Mar 2016, 20:30
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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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 18 Mar 2016, 20:44
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 29 Mar 2016, 08:16
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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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 13 May 2016, 17:33
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, 22:26
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, 16:47
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.
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation   [#permalink] 16 May 2016, 16:47

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