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Absolute Value Question

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Absolute Value Question [#permalink] New post 20 Jan 2012, 13:01
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When is abs(x-4)=4-x?

Can someone explain to me how to solve this algebraically (i.e. without using plug and chug and/or educated guessing)?

Going the normal route would lead to the following:

x-4=4-x -> x=4 (after getting to this point I understand I can guess above and below 4 to see which one works and then arrive at the correct solution, but I don't want to do this as it is not a rigorous method)

or

-(x-4)=4-x -> -x+4=4-x -> 0=0

So how do we go from these equalities to the inequality that correctly characterizes the solution to this problems:

x<=4

Thanks for your input.
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Re: Absolute Value Question [#permalink] New post 20 Jan 2012, 13:48
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adavydov7 wrote:
When is abs(x-4)=4-x?

Can someone explain to me how to solve this algebraically (i.e. without using plug and chug and/or educated guessing)?

Going the normal route would lead to the following:

x-4=4-x -> x=4 (after getting to this point I understand I can guess above and below 4 to see which one works and then arrive at the correct solution, but I don't want to do this as it is not a rigorous method)

or

-(x-4)=4-x -> -x+4=4-x -> 0=0

So how do we go from these equalities to the inequality that correctly characterizes the solution to this problems:

x<=4

Thanks for your input.


We have |x-4|=4-x. LHS=|x-4| to be equal to RHS=4-x it should expand with minus sign (in this case -(x-4)=4-x), or which is the same, the expression in absolute value must not be positive: x-4\leq{0} --> x\leq{4}.

Theory: check Absolute Value chapter of Math Book (math-absolute-value-modulus-86462.html)

DS questions on absolute values: search.php?search_id=tag&tag_id=37
PS questions on absolute values: search.php?search_id=tag&tag_id=58
Hardest DS questions on absolute values/inequalities with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Re: Absolute Value Question [#permalink] New post 20 Jan 2012, 19:19
Bunuel, thanks for the explanation and the links. Between the two I think I pieced together the complete correct answer (I think the one you provide is correct but not entirely complete). I will outline the complete answer below (as I understand) and highlight what was missing from your explanation in bold so others can see how to arrive at the answer rigorously. If I go wrong anywhere or my understanding is incorrect please be sure to correct me.

abs(x-4)=4-x

Step 1: Set conditions (consequently the step I was skipping when I first did this problem)
a) x-4>=0
b) x-4<0 (you state that this should be x-4<=0 and state that this is where the answer comes from, in actuality this condition is strictly < not <=, in order to get the = part we need to solve (a) as I do below in Step 2 and ensure that it satisfies the condition set out above in Step 1a as I do in Step 3a)

Step 2: Solve the original equations
a) x-4=4-x solves to x=4
b) -(x-4)=4-x solves to 0=0

Step 3: Check the solutions to the equations in Step 2 against their respective conditions in Step 1
a) plug x=4 into x-4>=0 which yields 0>=0 which is TRUE hence 4 is a solution to the problem, i.e. x=4
b) because the equation in Step 2b completely cancels out we are left the with the condition from Step 1b as the solution, i.e. x-4<0 or x<4

Combining the possible solutions from Step 3 for both (a) (x=4) and (b) (x<4) (since we have verified that both ARE solutions to the original problem by checking them against the conditions set in Step 1) we come up with the answer to the problem, or x<=4.

I hope you see what I mean when I say that your method wasn't entirely complete. I just want to be very rigorous in solving these is all. However, if my approach is incorrect please do let me know as I will have to go back and re-read the rules on solving these things to make sure I have it down.

Thanks again!
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Re: Absolute Value Question [#permalink] New post 21 Jan 2012, 01:20
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adavydov7 wrote:
Bunuel, thanks for the explanation and the links. Between the two I think I pieced together the complete correct answer (I think the one you provide is correct but not entirely complete). I will outline the complete answer below (as I understand) and highlight what was missing from your explanation in bold so others can see how to arrive at the answer rigorously. If I go wrong anywhere or my understanding is incorrect please be sure to correct me.

abs(x-4)=4-x

Step 1: Set conditions (consequently the step I was skipping when I first did this problem)
a) x-4>=0
b) x-4<0 (you state that this should be x-4<=0 and state that this is where the answer comes from, in actuality this condition is strictly < not <=, in order to get the = part we need to solve (a) as I do below in Step 2 and ensure that it satisfies the condition set out above in Step 1a as I do in Step 3a)

Step 2: Solve the original equations
a) x-4=4-x solves to x=4
b) -(x-4)=4-x solves to 0=0

Step 3: Check the solutions to the equations in Step 2 against their respective conditions in Step 1
a) plug x=4 into x-4>=0 which yields 0>=0 which is TRUE hence 4 is a solution to the problem, i.e. x=4
b) because the equation in Step 2b completely cancels out we are left the with the condition from Step 1b as the solution, i.e. x-4<0 or x<4

Combining the possible solutions from Step 3 for both (a) (x=4) and (b) (x<4) (since we have verified that both ARE solutions to the original problem by checking them against the conditions set in Step 1) we come up with the answer to the problem, or x<=4.

I hope you see what I mean when I say that your method wasn't entirely complete. I just want to be very rigorous in solving these is all. However, if my approach is incorrect please do let me know as I will have to go back and re-read the rules on solving these things to make sure I have it down.

Thanks again!


It seems that you need to brush up fundamentals on absolute value. So, I do advice to follow the links in my previous posts for that.

As for the solution: it's correct. |x-4|=4-x to hold true LHS must not be positive which means: x-4\leq{0} --> x\leq{4}. Exactly as I wrote.

Usually when you expand absolute value you should check when the expression in it is <=0 and >0 (you can put = sign for either of case). But x-4>0 can not hold true as in this case RHS=4-x<0 and we would have that LHS=|x-4|<0 which is cannot possibly be correct as absolute value is always non-negative.

Hope it's clear.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Absolute Value Question [#permalink] New post 21 Jan 2012, 21:02
Thanks so much, I hadn't realized that it doesn't matter which condition in Step 1 the = sign was attached to, i.e. as you say it can be to either the > or the < (in the links you provided the = sign always went with the > so I just assumed that this was by rule). This makes your solution completely correct! And everything else follows. Again thank you!
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Re: Absolute Value Question [#permalink] New post 22 Jan 2012, 00:08
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Let me add the definition of mod here:

|x|= x when x >= 0
|x|= -x when x < 0 (the '=' works here as well)

Therefore, |x-4| = x-4 when (x-4) >= 0
|x-4| = -(x-4) = 4-x when (x-4) <= 0 i.e. when x <= 4
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Re: Absolute Value Question [#permalink] New post 22 Jan 2012, 07:57
Karishma, that is also a very good and simple explanation. Thank you.
Re: Absolute Value Question   [#permalink] 22 Jan 2012, 07:57
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