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Re: Math: Absolute value (Modulus) [#permalink]
17 Jun 2014, 08:16

My questions: if we have 3 points why do we have 4 conditions? Next, for each of the letters (a,b,c,d) how do we know which equation to use and why is the inequality used? For example, for a) we do x = -8, but then how to we know the equation to use is -(x+3) - (4-x) = -(8+x), and how did the negative sign come in front of the -x+3 and why? I have the same type of questions for b,c,d so maybe an answer to a will help

Re: Math: Absolute value (Modulus) [#permalink]
17 Jun 2014, 09:07

In the absolute value question, I don't get why the distance between A and X is not explained. There's discussion about Y and A, Y and B, X between A and Y, distance between X and B and Y and B but not A and X , I need a reexplanation, maybe with numbers

Re: Math: Absolute value (Modulus) [#permalink]
17 Jun 2014, 19:27

Expert's post

sagnik242 wrote:

My questions: if we have 3 points why do we have 4 conditions? Next, for each of the letters (a,b,c,d) how do we know which equation to use and why is the inequality used? For example, for a) we do x = -8, but then how to we know the equation to use is -(x+3) - (4-x) = -(8+x), and how did the negative sign come in front of the -x+3 and why? I have the same type of questions for b,c,d so maybe an answer to a will help

Plot the 3 points: -8, -3 and 4 on the number line. Now there are 4 distinct regions on the number line: The part that lies to the left of -8, the part that lies between -8 and -3, the part that lies between -3 and 4 and the part that lies to the right of 4.

In each of these regions, each of the absolute value expressions will behave differently.

You should know the definition of absolute value:

|x| = x if x >= 0 |x| = -x if x < 0

Consider the equation: |x+3| - |4-x| = |8+x|

What happens in the region which is at the left of -8 say x = -10 x+3 = -10+3 = -7 Since x+3 is negative, |x+3| = -(x + 3)

4-x = 4 - (-10) = 14 Since 4-x is positive, |4 - x| = 4 - x

8 + x = 8 - 10 = -2 Since x+8 is negative, |8+x| = -(8+x)

This is the logic used. Now repeat it for each region. _________________

Re: Math: Absolute value (Modulus) [#permalink]
31 Jul 2014, 10:21

i am really confused in case of double/triple modulus. Someone please help me! --> how to open modulus in case of double/triple mod. |x+3| - |4-x| = |8+x| Here origins will be -8, -3 and 4 (I am crystal clear till this point) So we have 4 regions ==> x>4, -3<x<4 , -8<x<-3, and x<-8 I do not understand how to open modulus in these 4 regions? (when to consider positive/negative signs while opening modulus)

Re: Math: Absolute value (Modulus) [#permalink]
31 Jul 2014, 10:25

Expert's post

AkshayDavid wrote:

i am really confused in case of double/triple modulus. Someone please help me! --> how to open modulus in case of double/triple mod. |x+3| - |4-x| = |8+x| Here origins will be -8, -3 and 4 (I am crystal clear till this point) So we have 4 regions ==> x>4, -3<x<4 , -8<x<-3, and x<-8 I do not understand how to open modulus in these 4 regions? (when to consider positive/negative signs while opening modulus)

Please go through the thread. This question is extensively discussed there. _________________

Re: Math: Absolute value (Modulus) [#permalink]
09 Aug 2014, 11:04

can someone help me with this, |x-1|+|x-3|+|x-5|>=9, the solution that I came up for this was completely different than in the book I am working with. Can someone please tell how to proceed with this inequality. The solution I got was (-infinity,1)U(5,infinity). TIA

Re: Math: Absolute value (Modulus) [#permalink]
08 Dec 2014, 08:31

VeritasPrepKarishma wrote:

If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.

Attachment:

Ques2.jpg

VeritasPrepKarishma wrote:

OR consider that lx+3l>3 means distance of x from -3 is more than 3. If you go to 3 steps to right from -3, you reach 0. Anything after than is ok. If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.

Attachment:

Ques2.jpg

I didn't understand the highlighted portion, Could you please explain: 1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value? 2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

Re: Math: Absolute value (Modulus) [#permalink]
08 Dec 2014, 20:09

Expert's post

appleid wrote:

VeritasPrepKarishma wrote:

If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.

Attachment:

Ques2.jpg

VeritasPrepKarishma wrote:

OR consider that lx+3l>3 means distance of x from -3 is more than 3. If you go to 3 steps to right from -3, you reach 0. Anything after than is ok. If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.

Attachment:

Ques2.jpg

I didn't understand the highlighted portion, Could you please explain: 1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value? 2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

Re: Math: Absolute value (Modulus) [#permalink]
22 Jan 2015, 07:24

Hello,

There are a few things I don't find clear in the theory about absolute value, and particularly the 3 step method.

1) How do we decide which are the points we are interested in? In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x? 2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest? 3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=? 4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?

Re: Math: Absolute value (Modulus) [#permalink]
29 Jan 2015, 09:32

Expert's post

pacifist85 wrote:

1) How do we decide which are the points we are interested in? In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x? 2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest? 3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=? 4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?

1) When the expression under modulus equals 0. The expression may or may not change its sign in those special points. 2) Yes, because the number line goes from -infinity all the way to +infinity. -4 is not a special point (see #1; 4 - (-4) = 8 ≠ 0) 3) All conditions cover the number line and break at the special points. 4) Using the definition of modulus: for y = |x|, y = x if x >=0 and y = -x if x < 0. _________________

Math: Absolute value (Modulus) [#permalink]
23 Feb 2015, 00:29

gettinit wrote:

Let’s consider following examples,

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)

I understand (a) and (d)... but i don't really understand (b) and (c). Firstly for (b), since x<0, shouldn't the (-)sign be applied to all factors instead of just (x+3) and (4-x)? Why is the (-)sign not applied to (8+x)? as for (c), x can be >0 or <0, so what should be done here?

Re: Math: Absolute value (Modulus) [#permalink]
12 Jun 2015, 01:29

I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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) Why are the all the signs here negative? 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.) Why are the signs in the first part here negative and not the second? 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.) Why is this similar to the original equation? 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) This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Re: Math: Absolute value (Modulus) [#permalink]
13 Jun 2015, 23:44

1

This post received KUDOS

pacifist85 wrote:

I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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) Why are the all the signs here negative? 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.) Why are the signs in the first part here negative and not the second? 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.) Why is this similar to the original equation? 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) This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Thank you!

hi pacifist85, 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) Why are the all the signs here negative? we have three mod here .. 1)|x+3|.. this term will be positive because of mod sign but if x<-8, say -9.. x+3=-9+3=-6... so the value of x+3 is actually negative but l-6l=6.. therefore to get the value of x we put a negative sign when we open modulus sign.. 2)|4−x|.. here 4-(-9)=13 so the values do not change whether with mod sign or without.. so the existing -ive sign remains.. 3)|8+x|... when x=-9, 8=(-9)=-1.. same as 1 above ..

you can similarly check for other three parts.. hope it helped

Re: Math: Absolute value (Modulus) [#permalink]
14 Jun 2015, 21:14

1

This post received KUDOS

Expert's post

pacifist85 wrote:

I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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) Why are the all the signs here negative? 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.) Why are the signs in the first part here negative and not the second? 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.) Why is this similar to the original equation? 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) This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Thank you!

IT is derives from the definition of absolute value:

|x| = x if x is positive and -x if x is negative.

You have three transition points: -8, -3 and 4.

When x < -8, all three expressions (x + 8), (x + 3) and (x - 4) will be negative (check for say x = -10 to understand). So when you remove the absolute value from |x + 8|, you will get -(x + 8) (because x + 8 is negative).Similarly, when you remove absolute value from |x + 3|, you will get - (x + 3) because x+3 is negative and so on...

Similarly, when -8 < x < -3, (x is to the right of -8) (x+8) will be positive but other two expressions will be negative (check for say x = -5 to understand).

Re: Math: Absolute value (Modulus) [#permalink]
11 Jul 2015, 16:57

walker wrote:

I've added a new example:

Example #2 Q.: \(|x^2-4| = 1\). What is x? Solution: There are 2 conditions:

a) \((x^2-4)\geq0\) --> \(x \leq -2\) or \(x\geq2\). \(x^2-4=1\) --> \(x^2 = 5\). x e {\(-\sqrt{5}\), \(\sqrt{5}\)} and both solutions satisfy the condition.

b) \((x^2-4)<0\) --> \(-2 < x < 2\). \(-(x^2-4) = 1\) --> \(x^2 = 3\). x e {\(-\sqrt{3}\), \(\sqrt{3}\)} and both solutions satisfy the condition.

Hello, Nice post. I can't understand why the first condition equals to 0 as well when the second condition is only less than zero but NOT equal to zero? Could you please explain?

Math: Absolute value (Modulus) [#permalink]
11 Jul 2015, 17:22

1

This post received KUDOS

1

This post was BOOKMARKED

ranaazad wrote:

walker wrote:

I've added a new example:

Example #2 Q.: \(|x^2-4| = 1\). What is x? Solution: There are 2 conditions:

a) \((x^2-4)\geq0\) --> \(x \leq -2\) or \(x\geq2\). \(x^2-4=1\) --> \(x^2 = 5\). x e {\(-\sqrt{5}\), \(\sqrt{5}\)} and both solutions satisfy the condition.

b) \((x^2-4)<0\) --> \(-2 < x < 2\). \(-(x^2-4) = 1\) --> \(x^2 = 3\). x e {\(-\sqrt{3}\), \(\sqrt{3}\)} and both solutions satisfy the condition.

Hello, Nice post. I can't understand why the first condition equals to 0 as well when the second condition is only less than zero but NOT equal to zero? Could you please explain?

Thanks! :D

For modulus or absolute value questions, you need to take the equality with the '>' sign. That is the convention and follows the definition of an absolute value. Additionally, you only need to account for equality once in your question.

Also, think of modulus or absolute value this way:

|x| = x for x=0,1,2,3,4,5... or true for all NON-NEGATIVE numbers. Remember the 'non-negative' part. This includes 0 as well.

But |x| = -x for x<0, x=-1,-2,-0.25 ...

This is the reason why we put equality with the '>' sign to account for all non-negative numbers. The 'nature' of 'x' does not change if it is 0 and above but it does change (multiply x by -1) if x is <0.

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