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Re: Math: Absolute value (Modulus) [#permalink]
 17 Jan 2011, 01:25
VeritasPrepKarishma wrote: Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0
|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)
Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|
You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3
Now, x can be either greater than 2, between -3 and 2 or less than -3.
So you solve for these 3 cases:
Case 1: x >= 2
(x - 2) = (x + 3)
-2 = 3
No solution
Case 2: -3 <= x < 2
-(x - 2) = (x + 3)
x = -1/2 which lies between -3 and 2
So this is a solution to the equation
Case 3: x <= -3
-(x - 2) = -(x + 3)
2 = -3
No solution
Similarly you can solve for as many terms as you want. Thanks Karishma! Much clearer now. +1
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Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 04:41
i have a question related to Example 1:
|x+3| - |4-x| = |8+x|
-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks
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Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 05:47
tinki wrote: i have a question related to Example 1:
|x+3| - |4-x| = |8+x|
-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example: Question: |4-x| > 5 Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1 Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9 Answer: x is either less than -1 or greater than 9. Now switch: Question: |x-4| > 5 Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9 Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1 Answer: x is either less than -1 or greater than 9. The same two cases in both the questions.. same answer in both...
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Re: Math: Absolute value (Modulus) [#permalink]
09 Feb 2011, 11:54
VeritasPrepKarishma wrote: tinki wrote: i have a question related to Example 1:
|x+3| - |4-x| = |8+x|
-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example: Question: |4-x| > 5 Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1 Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9 Answer: x is either less than -1 or greater than 9. Now switch: Question: |x-4| > 5 Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9 Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1 Answer: x is either less than -1 or greater than 9. The same two cases in both the questions.. same answer in both... THANKS 
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Re: Math: Absolute value (Modulus) [#permalink]
13 Feb 2011, 17:16
Thanks for this Walker!
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Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 08:52
Hi one small question can we write |x|>=0 when 0 is neither positive nor negative but neutral.
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Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 10:25
Yup, this is the first property - absolute value of x is always non-negative.
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Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 13:27
Very Helpful notes.Thanks But can you pls explain the following third Trick III. Thinking about absolute values as distance between points at the number line. For example, Problem: A<X<Y<B. Is |A-X| <|X-B|? 1) |Y-A|<|B-Y| Solution: Image We can think about absolute values here as distance between points. Statement 1 means than distance between Y and A is less than Y and B. Because X is between A and Y, distance between |X-A| < |Y-A| and at the same time distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.
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Re: Math: Absolute value (Modulus) [#permalink]
14 Feb 2011, 19:03
GMATD11 wrote: Hi
one small question can we write |x|>=0
when 0 is neither positive nor negative but neutral. I am not sure I got your question.. |x|>=0 implies |x| is positive(when x is positive or negative) or 0 (when x = 0). |x| is of course never negative.
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Re: Math: Absolute value (Modulus) [#permalink]
19 Feb 2011, 23:40
Hi,
In the original post, in the section GMAC books, there are some prob numbers stated. Can you please explain how they relate to the post? How one will use those numbers is what I am asking.
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Re: Math: Absolute value (Modulus) [#permalink]
19 Mar 2011, 21:53
thank you, very thorough explanation!
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Re: Math: Absolute value (Modulus) [#permalink]
26 May 2011, 14:34
walker wrote:
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 \leq 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 \leq 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 \geq 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)
This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios. As we are looking at x+3 >=0 we get x>=-3 Similarly when x+3 <0 we have x<-3 Based on these the ranges I have for the 4 cases 1. x<-8 2. -8<= x <-3 3. -3<= x <= 4 4. x> 4 the last two cases are whats different . This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance
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Re: Math: Absolute value (Modulus) [#permalink]
27 May 2011, 06:44
VeritasPrepKarishma wrote: someonear wrote: walker wrote:
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 \leq 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 \leq 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 \geq 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)
This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios. As we are looking at x+3 >=0 we get x>=-3 Similarly when x+3 <0 we have x<-3 Based on these the ranges I have for the 4 cases 1. x<-8 2. -8<= x <-3 3. -3<= x <= 4 4. x> 4 the last two cases are whats different . This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance You don't need to solve anything to get these four ranges. You see that the points where the signs will vary are -8, 4 and -3. To cover all the numbers on the number line, the ranges are: x is less than -8, then between -8 and -3, then between -3 and 4 and then greater than 4. It doesn't matter where you put the '='. In each range the sign of the terms will be different. After assigning the proper signs, you will get a value for x and you have to check if the value lies in the range you were considering. If it does, it is a solution, else it is not. I am worried about values of x that are on the border of the ranges Say hypothetically for a particular set of equations we end with the identical 4 cases we have here. Now if say I have x=4 then the way I had come up with the ranges I will get a solution between -3 and 4 as x=4 exists in -3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4 Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
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Re: Math: Absolute value (Modulus) [#permalink]
27 May 2011, 07:15
someonear wrote:
I am worried about values of x that are on the border of the ranges
Say hypothetically for a particular set of equations we end with the identical 4 cases we have here.
Now if say I have x=4 then the way I had come up with the ranges I will get a solution between -3 and 4 as x=4 exists in -3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4
Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
Your aim is to get the solution. You create the ranges to help yourself solve the problem. It doesn't matter at all in which range you consider the border value to lie. Say, when x = 4, (4 - x) = 0. You solve saying that in the range -3<= x<4, (4 - x) is positive and in the range x>= 4, (4 - x) is negative. At the border value i.e. x = 4, (4 - x) = 0. There is no negative or positive at this point. Hence it doesn't matter where you include the '='. Put it wherever you like. I just like to go in a regular fashion like walker did above. Include the first point in the first range, the second one in the second range (but not the third one i.e. -3 <= x < 4) and so on.
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Re: Math: Absolute value (Modulus) [#permalink]
30 May 2011, 03:26
wow thank you!
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Re: Math: Absolute value (Modulus) [#permalink]
04 Jun 2011, 17:04
Really helpful...thanks Walker
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Re: Math: Absolute value (Modulus) [#permalink]
21 Jun 2011, 00:01
Hi Walker,
Can you please explain this?
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:
How do we get 3 key points and 4 conditions?
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Re: Math: Absolute value (Modulus) [#permalink]
21 Jun 2011, 20:43
walker wrote: There are 3 points where one of the modules is zero:
1)x+3=0 --> x = -3
2)4-x=0 --> x = 4
3)8+x=0 --> x = -8
Those 3 points divide the number line by 4 pieces:
1) -inf, -8
2) -8,-3
3) -3, 4
4) 4, +inf
and for each condition we are solving the equation separately. Hi Walker, Since |x-3| is is modulus why did we not do x-3<0 or x-3>0 therefore, x = 3 or x =-3 and similarly for 4-x and 8+x as well?
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Re: Math: Absolute value (Modulus) [#permalink]
20 Jul 2011, 00:23
Thanks Walker.. great post.. very well explained..
I finally understood, i hope, most of what is there to absolute values.
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Re: Math: Absolute value (Modulus) [#permalink]
20 Jul 2011, 05:34
Found this Q on the Manhattan site but was unable to comprehend the solution. I esp did not understand how to use algebra to solve it. Please help. If y = /x + 7/ + /2 - x/, is y = 9? (1) x < 2 (2) x > -7 IMO
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Re: Math: Absolute value (Modulus)
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20 Jul 2011, 05:34
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