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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 29 Nov 2013, 08:33
Dear Karishma,
I have been trying to follow your concept from this page
http://www.veritasprep.com/blog/2011/01/quarter-wit-quarter-wisdom-the-holistic-approach-to-mods-part-ii/#comment-5122

,however its still unclear.Would you mind applying the concept to this equation.I tried and got stuck multiple places.
I got upto here.
total distance is 15 when x=-3 equation becomes 0=5+7.not suff , we need to move a step to the left of -3 .one step is equal to 3 units so lets try -5.new equation is 2=3+9.The thing is i don't see any value that would satisfy this equation.so my answer is 0.Could i be misunderstanding the concept?

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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 Dec 2013, 13:09
hi,
I understood how the values of x have taken, also the case a,d but nowhere the options b,c are discussed. I have doubts in it.

b) -8<= x < -3 => -(x+3)-(4-x)= (8+x) (as per the pdf)

i.e. x>=-8 and x<-3 - is this correct?
for x>=-8, x can be -ve( -8 to -1)or +ve (0 to inf) = more value for +ve , so |8+x|= +(8+x)?
and for x<3 , x can be -ve (inf to -1) and +ve (0 to 2)= more values for -ve, so |x+3|= -(x+3)? and sign for |4-x| will remain same?

Is this understanding right?

same for option c?

Please help me to understand this.

Thanks,
M.

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

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New post 14 Jan 2014, 22:11
Hi
Is there any shorter way to solve this question? It took me more than two and a half minutes to solve it..
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 14 Jan 2014, 22:17
akankshasoneja wrote:
Hi
Is there any shorter way to solve this question? It took me more than two and a half minutes to solve it..
Thanks


Different methods to solve this question have been discussed above. Pick whichever you are most comfortable with. Also, this is a 750+ level question. It will certainly take some time to 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 29 Jan 2014, 00:29
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VeritasPrepKarishma wrote:
|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.



Quote:
In GMATCLUB Maths book it is given that to solve any Modulus Questions we need to follow 3 step process. First is by opening the Modulus and exposing to signs. In the problem given below I am not able to deduce, as how to proceed for opening the mod and exposing to signs (both +ve and -ve)
------------------------------
Q: Mod (x+3) - Mod (4-x)= Mod (8+x). How many solutions does the equation have?
------------------------------
As a general approach, the terms in MOD are put in brackets and then solved by putting positive and then negative signs.


The process given in the book is perfectly fine but don't expect it to be 100% mechanical. You will need to evaluate the question every time. You CANNOT just make equations taking +ve and -ve value for every mod and get 2*2*2 = 8 equations. Understand the reason for this.

By definition,
|x+8| = x+8 when x+8 >= 0 i.e. when x >= -8
|x+8| = -(x+8) when x+8 < 0 i.e. when x < -8

Similarly,
|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

|x-4| = x-4 when x-4 >= 0 i.e. when x >= 4
|x-4| = -(x-4) when x-4 < 0 i.e. when x < 4

Now think, is it possible that |x+8| = -(x+8) but |x+3| = x+3? No because this will happen only when x < -8 AND x >= -3. There is no such value of x.

As mentioned in the post above - you will have 3 transition points: -8, -3 and 4. In between these ranges, the signs of the 3 mods change.
x < -8 ------ Here, |x+8| = -(x+8), |x+3| = -(x+3), |x-4| = -(x-4). So you make an equation using these.
-8 <= x < -3 ------- Here, |x+8| = (x+8), |x+3| = -(x+3), |x-4| = -(x-4). So you make an equation using these.
-3 <= x < 4 ------- |x+8| = (x+8), |x+3| = (x+3), |x-4| = -(x-4). So you make an equation using these.
x >=4 -------- |x+8| = (x+8), |x+3| = (x+3), |x-4| = (x-4). So you make an equation using these.

The transition points decide the number of equations you will get. If you have 3 transition points, you will get 4 equations. If you have 4 transition points, you will get 5 equations and so on...
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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, 02:28
[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


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

Now when |x+ 3| is lesser than |8+x| and |4-x| will always be positive or zero hence this equation will not have any solution.
When |x+3| is greater than |8 + x| which will be when x is lesser than -5.5 then |4 - x| will be greater than |x+3|.

Hence answer is A.
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Last edited by PerfectScores on 07 Feb 2014, 04:01, edited 1 time in total.

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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, 03:53
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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 07 Feb 2014, 04:01
VeritasPrepKarishma wrote:
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|


Thanks Karishma, completely missed one part. Edited the post.
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Need explanation of Absolute values. [#permalink]

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New post 13 Feb 2014, 08:24
Dear All,

i need explanation of how to solve Absolute Values. I have read GMAT club book and have confusion in the following example

3-steps approach for complex problems
Let’s consider following examples,
Example #1
Q.: |X+3| - |4 – X| = |8 + X| How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) .X< -8. –(X + 3) – (4 – X)= -(8 + X)--> X= -1 . We reject the solution because our condition is
not satisfied (-1 is not less than -8)

b) -8 ≤ 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)


i understand how to calculate the key points and value of x and to make inequalities?
but i dont understand the signs they use to calculate the values of x since they are changing in each options.

Also please tell me the good book and guide to solve absolute values with practice questions.

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Re: Need explanation of Absolute values. [#permalink]

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New post 13 Feb 2014, 08:50
Hi heyhadi,

The most important thing to understand when dealing with absolute values is that |x| is either equal to x or –x whether x is positive or negative. For example if x=5 then |x|= x = 5 and if x=-3 then |x| = -x = -(-3) = 3 !

This is why the most efficient way to solve an exercise with absolute value is very often to divide the exercise in two different cases : the case when what is inside the absolute value is positive and then |x| = x and the case where it is negative and |x|=-x. This way you can drop the absolute value and just solve a standard (nice) equation.

In your example we have the following cases :
a) What is inside the first absolute value is negative, what is inside the second is positive, and what is inside the last is negative. Therefore |x+3| = -(x+3), |4-x|=4-x and |8 + x|=-(8+x)
b) What is inside the first absolute value is negative, what is inside the second is positive, and what is inside the last is positive. Therefore |x+3| = -(x+3), |4-x|=4-x and |8 + x|=+(8+x)
c) What is inside the first absolute value is positive, what is inside the second is positive, and what is inside the last is positive. Therefore |x+3| = +(x+3), |4-x|=4-x and |8 + x|=+(8+x)
d) What is inside the first absolute value is positive, what is inside the second is negative, and what is inside the last is positive. Therefore |x+3| = +(x+3), |4-x|=-(4-x) and |8 + x|=+(8+x)

After these distinction you just have the solve the remaining equation !

Hope this helps,
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Re: Need explanation of Absolute values. [#permalink]

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New post 13 Feb 2014, 09:13
GMarc wrote:
Hi heyhadi,

The most important thing to understand when dealing with absolute values is that |x| is either equal to x or –x whether x is positive or negative. For example if x=5 then |x|= x = 5 and if x=-3 then |x| = -x = -(-3) = 3 !

This is why the most efficient way to solve an exercise with absolute value is very often to divide the exercise in two different cases : the case when what is inside the absolute value is positive and then |x| = x and the case where it is negative and |x|=-x. This way you can drop the absolute value and just solve a standard (nice) equation.

In your example we have the following cases :
a) What is inside the first absolute value is negative, what is inside the second is positive, and what is inside the last is negative. Therefore |x+3| = -(x+3), |4-x|=4-x and |8 + x|=-(8+x)
b) What is inside the first absolute value is negative, what is inside the second is positive, and what is inside the last is positive. Therefore |x+3| = -(x+3), |4-x|=4-x and |8 + x|=+(8+x)
c) What is inside the first absolute value is positive, what is inside the second is positive, and what is inside the last is positive. Therefore |x+3| = +(x+3), |4-x|=4-x and |8 + x|=+(8+x)
d) What is inside the first absolute value is positive, what is inside the second is negative, and what is inside the last is positive. Therefore |x+3| = +(x+3), |4-x|=-(4-x) and |8 + x|=+(8+x)

After these distinction you just have the solve the remaining equation !

Hope this helps,



Thanks for your help..

but the confusion is how would i know that which should i take negative or positive.

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Re: Need explanation of Absolute values. [#permalink]

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New post 13 Feb 2014, 10:46
heyhadi wrote:
Dear All,

i need explanation of how to solve Absolute Values. I have read GMAT club book and have confusion in the following example

3-steps approach for complex problems
Let’s consider following examples,
Example #1
Q.: |X+3| - |4 – X| = |8 + X| How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) .X< -8. –(X + 3) – (4 – X)= -(8 + X)--> X= -1 . We reject the solution because our condition is
not satisfied (-1 is not less than -8)

b) -8 ≤ 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)


i understand how to calculate the key points and value of x and to make inequalities?
but i dont understand the signs they use to calculate the values of x since they are changing in each options.

Also please tell me the good book and guide to solve absolute values with practice questions.


Merging similar topics. Please read the discussion, I think it answers the questions you have. For a detailed solution check this: x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html#p1241355

Theory on Abolute Values: math-absolute-value-modulus-86462.html

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

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


P.S. Please name topics properly. Check rule 3 here: rules-for-posting-please-read-this-before-posting-133935.html Thank you.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 17 May 2014, 03:46
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?
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation [#permalink]

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New post 17 May 2014, 05:55
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?


Yes, you can apply this approach to all similar questions.
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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, 21:49
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 2016 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 2010 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, 04:39

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 1947 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, 03:08
[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, 03:11
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, 07:40
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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|x+3| - |4-x| = |8+x|. How many solutions does the equation   [#permalink] 03 Sep 2014, 07:40

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