|x+3| - |4-x| = |8+x|. How many solutions does the equation : Problem Solving (PS)

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

18 Jun 2013, 09:54

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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]

15 Jun 2013, 14:17

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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]

10 Mar 2013, 15:15

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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.

L

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

18 Jun 2013, 09:43

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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]

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]

21 May 2014, 03:55

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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 [ 2.04 KiB | Viewed 17249 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 [ 2.73 KiB | Viewed 17245 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.

L

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

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...

L

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

12 Jun 2016, 06:59

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msk0657 wrote:

Bunuel wrote:

guerrero25 wrote:

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

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

Show Spoiler

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

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

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

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

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

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

Responding to pm.

Absolute value properties:

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

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

STEP BY STEP SOLUTION:

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

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

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

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

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

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

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

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

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

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

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

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

Answer: A.

Hope it's clear.

Hi Bunuel,

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

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

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

Now we have four ranges:

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

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

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

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

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

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

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

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

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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KarishmaB

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

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]

18 Jun 2013, 08:50

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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]

Updated on: 07 Feb 2014, 04:01

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[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.

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.

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KarishmaB

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

26 Jul 2017, 20:54

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ssislam 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 mam

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

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

thanks in advance ...

Consider this:
If x < -8 (i.e. x can be -9, -10, -11, -11.4, -20 and so on...),
what can you say about the sign of (x + 3) ?
Can we say that when x is less than -3, (x + 3) is negative? Sure. So if we know that x is less than -8, then obviously, it is less than -3 too. So in this case, (x+3) will certainly be negative.
Hence |x + 3| will also translate into -(x+3).

To see it easily, drawing a number line helps.

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

For (x + 3), all values of x to the left of -3 will give |x + 3| = -(x + 3). So if x < -8, it is to the left of -3 and hence will lead to |x+3| = -(x + 3).
To the right of -3, all values of x will give |x + 3| = (x + 3).

For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/

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KarishmaB

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

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]

18 Mar 2016, 21:44

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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]

29 May 2016, 20:51

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smritidabas wrote:

Could you explain the same by solving with just one of the examples but with |4-x|

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

Thanks!

VeritasPrepKarishma wrote:

MadCowMartin wrote:

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

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

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

BTW: Is this 650+ level?

You solve the equation to get the x = values

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

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

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

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

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

And no, it is 750+ level.

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

I. Consider |x - 4|

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

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

II. Consider |4 - x|

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

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

Note that in both cases, when x >= 4, you get (x - 4) and when x < 4, you get (4 - x).
Just that case I is more intuitive so it's easier to flip first and then solve.

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

30 Jun 2016, 23:30

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Responding to a pm:

Quote:

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

Why you counted -(4-x)

It comes down to whether you understand why we take ranges and change signs. First check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/

Now take a simpler example:

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

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

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

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

So in a question like this: |x - 2| + |x - 5| = 4, you don't have 4 cases ((x - 2) positive or negative and (x - 5) positive or negative). You have only 3 cases. The 4th case will not exist.

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

30 Jun 2016, 23:30

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