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

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

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New post 10 Mar 2013, 12:55
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|x+3| – |4-x| = |8+x| How many solutions will this equation have?

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


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 10 Mar 2013, 23:32
15
21
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.
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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 Jun 2013, 09:54
10
5
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



rrsnathan wrote:
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 Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?

Thanks in advance,
RRSNATHAN.
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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 Mar 2013, 15:15
3
4
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.
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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 Jun 2013, 14:17
4
4
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!
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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 Jun 2013, 20:12
1
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!


You need to consider the boundary value points somewhere in the range, as sometimes the nature of the equation might behave differently after and before, and ON the transition point itself.
While in this particular question, this was not an issue as with or without considering it, you could get the right answer.
However, as a rule of thumb we should always involve the = part in one of the ranges to make sure the solution is consistent and not missing on any boundary value conditions.
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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 Jun 2013, 09:42
After reviewing the material in the link, I can't help but think that it is irrelevant whether it is ≤ or as long as there is only one ≤ or ≥ sign in each number less than x less than number i.e.

-3≤x<4
OR
-3<x≤4

Is that correct?
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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 Jun 2013, 09:46
WholeLottaLove wrote:
After reviewing the material in the link, I can't help but think that it is irrelevant whether it is ≤ or as long as there is only one ≤ or ≥ sign in each number less than x less than number i.e.

-3≤x<4
OR
-3<x≤4

Is that correct?


Yes, that' correct. The point is to include the transition points but it really doesn't matter in which 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 18 Jun 2013, 08:50
2
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 Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?

Thanks in advance,
RRSNATHAN.
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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 Jun 2013, 09:43
4
rrsnathan wrote:
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 Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?

Thanks in advance,
RRSNATHAN.


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

When x <= -8, all three expressions (x + 3), (x - 4) and (x + 8) are negative when x <= -8.
So |x+3| = - (x + 3) (using the definition of mod)
|x-4| = - (x - 4)
|x+8| = - (x + 8)

Definition of mod:
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0
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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 Jun 2013, 22:42
rrsnathan wrote:
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 Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?

Thanks in advance,
RRSNATHAN.


It just follows a very simple logic:
If we have been given say |x|
And now we need to get rid of the mod in order to evaluate its ranges.

So, If x<0? The in that case x will take any negative value, for eg take x= -7
|-7| = 7 which is equal to -x
and if x>0 then , suppose x = 7
|7| = 7 which is equal to x.

Hence, if the value inside Mod is resulting in negative value, then we need to put a - sign before it, to get its actual value like we saw the case with -7 over here.
similarly if the value inside mod is positive, in that case we do not need to negate it, and we can write it as it is.

Now in the main question you can go ahead checking, which term results in positive or negative, and put a - sign accordingly.

Hope this helped.
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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 Jun 2013, 00:30
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Dear all,

I have understood the concept of critical points and applied it successfully when solving this question. My problem is time. It took me well over 3min to solve this question.
Can someone check my method and tell me where I am losing the time or on which part I might be able to speed up/take a shortcut?

What I did:
1. Find special points at first glance --> -8, -3, 4
2. Set up equation for x < -8 and solve --> x=9 which is not in defined range --> stop
3. Set up equation for -8 <= x < -3 and solve --> x=-7/3 which is not in defined range --> stop
4. Set up equation for -3 <= x < 4 and solve --> x=-1 which IS in defined range --> check in original equation --> -3 = 7 --> no solution
5. Set up equation for x => 4 and solve --> x=9 which IS in defined range --> check in original equation --> 7 = 17 --> no solution
6. Answer is zero solutions --> A

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

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New post 22 Jun 2013, 20:07
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?
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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New post 22 Jun 2013, 21:12
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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?


Well look at the question again, its :

|x+3| – |4-x| = |8+x|
Lets try to make all the terms positive first,

|x+3| = |x+8| + |x-4| ( since |x-a| = |a-x|)

Now you can see that for |x+3|, it will be have differently for

x>-3 and x<-3

since, refer to the property |x| = x for x>=0, and |x| = -x for x<0

hence, same way for |x+8|, will behave differently for x>-8 and x<-8

and for |x-4| will behave differently for x>4 and x<4

therefore, we get our conditions : (put it on the number line for clarity)

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 22 Jun 2013, 23:08
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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 23 Jun 2013, 03:28
Thank you both for making it all much more clearer!
I did try to solve the equation yesterday, but I messed up the distribution of the minus signs I guess. As it didn't seem to solve to the numbers given, I lost my way.

Following your steps I'm able to recreate the answer.

Good to know this is 750+ level. My target goal in 600 (Well personally 650, but 600 gets me into the program) and my test is in 10 days.
I first thought that the GMAT Club book was an overview of the basic fundamentals, but I should watch out for spending too much time on 750+ issues so I can spend more time on grasping the basics needed for 600.
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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 2013, 09:57
Responding to pm:
smartmanav wrote:
Hi Bunuel

Need your help for the concept of modulus.

Can we take minus one common out of a modulus number ?

For ex: I have an equation |x+3| - |4-x| = |8+x|

Can i write it as |x+3| + |x-4| = |8+x| ??

Thanks
Aakash


No. You cannot do that.

Modulus (|expression|) is not the same as parentheses ((expression)).

Also, |4-x|=|x-4|, so |x+3| - |4-x| = |8+x| is the same as |x+3| - |x-4| = |8+x|

Hope it's clear.
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|x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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New post 30 Jun 2013, 11:07
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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


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 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 Updated on: 07 Feb 2014, 04:01
[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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Originally posted by PerfectScores on 07 Feb 2014, 02:28.
Last edited by PerfectScores on 07 Feb 2014, 04:01, edited 1 time in total.
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation &nbs [#permalink] 07 Feb 2014, 02:28

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