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Math: Absolute value (Modulus)

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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Feb 2011, 04:19
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worldogvictor wrote:
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.


The given problem numbers are official questions on the concept of Modulus.
You need to have those books to be able to access the given questions. The numbers give you the question numbers e.g. in OG12 in sample problem solving questions, question no 22 (on page 155) tests you on Mods.
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Re: Math: Absolute value (Modulus) [#permalink] New post 19 Mar 2011, 20:53
thank you, very thorough explanation!
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Re: Math: Absolute value (Modulus) [#permalink] New post 26 May 2011, 13:34
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] New post 27 May 2011, 04:00
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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.
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 May 2011, 05:44
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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] New post 27 May 2011, 06:15
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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] New post 30 May 2011, 02:26
wow thank you! :)
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Re: Math: Absolute value (Modulus) [#permalink] New post 04 Jun 2011, 16:04
Really helpful...thanks Walker
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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jun 2011, 23:01
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] New post 21 Jun 2011, 02:28
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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.
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Re: Math: Absolute value (Modulus) [#permalink] New post 21 Jun 2011, 19:43
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] New post 19 Jul 2011, 23:23
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] New post 20 Jul 2011, 04:34
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
[Reveal] Spoiler:
C

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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jul 2011, 04:47
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y can be 9 only if both moduli are open with positive sign. -- > x+7 +2 -x = 9

We can open both moduli with positive sign if x>-7 for first modulus and x < 2 for second modulus. In other words, x should be (-7,2)
Now, let's look at our options... we need both of them. So, C
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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jul 2011, 06:28
Hi Walker,

Quote:
We can open both moduli with positive sign if x>-7 for first modulus and x < 2 for second modulus.


I may sound really dumb, but could you please explain how the second modulus is positive at x<2.

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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jul 2011, 08:53
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2-x > 0
2 > x
x < 2
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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jul 2011, 11:45
Cool lesson for students in grade 10, 11 and higher. It contains all the basics to advance level of absolute value.
I tweeted this link to my twitter.
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Re: Math: Absolute value (Modulus) [#permalink] New post 20 Jul 2011, 20:39
Thanks a ton Walker.. Gosh! i guess my doubt was really silly, i appreciate the time u took to reply
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Re: Math: Absolute value (Modulus) [#permalink] New post 21 Jul 2011, 08:15
This is of intrinsic value!
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Re: Math: Absolute value (Modulus) [#permalink] New post 13 Nov 2011, 19:53
Very helpful module! I understand it a little bit better now!
Re: Math: Absolute value (Modulus)   [#permalink] 13 Nov 2011, 19:53
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