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Re: Math: Absolute value (Modulus) [#permalink]
20 Feb 2011, 04:19

1

This post received KUDOS

Expert's post

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. _________________

Re: Math: Absolute value (Modulus) [#permalink]
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

Re: Math: Absolute value (Modulus) [#permalink]
27 May 2011, 04:00

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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. _________________

Re: Math: Absolute value (Modulus) [#permalink]
27 May 2011, 05:44

1

This post was BOOKMARKED

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

Re: Math: Absolute value (Modulus) [#permalink]
27 May 2011, 06:15

Expert's post

1

This post was BOOKMARKED

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. _________________

Re: Math: Absolute value (Modulus) [#permalink]
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:

Re: Math: Absolute value (Modulus) [#permalink]
20 Jul 2011, 04:47

Expert's post

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 _________________

Re: Math: Absolute value (Modulus) [#permalink]
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. _________________

The Cambridge open day wasn’t quite what I was used to; no sample lectures, no hard and heavy approach; and it even started with a sandwich lunch. Overall...

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...