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 Post subject: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 6:00 am 
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What is the solution set for |3x-2|\leq|2x-5|

One way to solve is to square both the terms of course , but what is other way of solving it.


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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 6:34 am 
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GMATMadeeasy wrote:
What is the solution set for |3x-2|\leq|2x-5|

One way to solve is to square both the terms of course , but what is other way of solving it.


First you should determine the check points (key points): \frac{2}{3} and \frac{5}{2}. Hence we'll have three ranges to check:

A. x<\frac{2}{3} --> -3x+2\leq-2x+5 --> -3\leq{x}, as x<\frac{2}{3}, then -3\leq{x}<\frac{2}{3};

B. \frac{2}{3}\leq{x}\leq\frac{5}{2} --> 3x-2\leq-2x+5 --> -x\leq\frac{7}{5}, as \frac{2}{3}\leq{x}\leq\frac{5}{2} , then \frac{2}{3}\leq{x}\leq\frac{7}{5};

C. x>\frac{5}{2} --> 3x-2\leq2x-5 --> x\leq{-3}, as x>\frac{5}{7}, then in this range we have no solution;

Ranges from A and B give us the solution as: -3\leq{x}\leq\frac{7}{5}.

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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities


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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 8:15 am 
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Thank you , it helps greatly.

Question : What was normal way of doing it back in school? I am wondering how I used to solve them ?

Working out questions from your post and other guys notes on the site in the mean time.

Thanks a lot Bunuel.


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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 9:49 am 
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@ Bunuel:

I didn't pick your strategy.

My way of doing is simple but it is giving different answer.

|3x-2| <= |2x-5|
We will consider two scenarios
3x-2 <= 2x-5 & -(3x-2) <= 2x-5
x<= -3 & x >= 7/5

So the range will be x<= -3 & x >= 7/5

Kindly let me know where did I go wrong???

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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 11:04 am 
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Hussain15 wrote:
@ Bunuel:

I didn't pick your strategy.

My way of doing is simple but it is giving different answer.

|3x-2| <= |2x-5|
We will consider two scenarios
3x-2 <= 2x-5 & -(3x-2) <= 2x-5
x<= -3 & x >= 7/5

So the range will be x<= -3 & x >= 7/5

Kindly let me know where did I go wrong???


If you plug the numbers from the ranges you got, you'll see that the inequality doesn't hold true.

As for the solution: we have two absolute values |3x-2| and |2x-5|. |3x-2| changes sign at \frac{2}{3} and |2x-5| changes sign at \frac{5}{2}.

---(I)---\frac{2}{3}---(II)---\frac{5}{2}---(III)---

We got three ranges as above. We should expand given inequality in these ranges and see what we'll get.

Hope it's clear.

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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities


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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 11:39 am 
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Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??

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 Post subject: Re: Inequalities - Challenging and tricky One [#permalink]
PostPosted: Sat Jan 16, 2010 11:54 am 
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Hussain15 wrote:
Thanks for the reply Bunue!

But there is one more issue.

In range A, you have changed the signs of both modules, in range B you have done it only for 2x+5 and in third case you haven't changed the signs. What's the logic behind this one??


I got your point. I'm not "changing" the signs, I'm expanding the absolute values in each range.

In range A, when x<\frac{2}{3}: |3x-2|=-3x+2 and |2x-5|=-2x+5, so we get -3x+2\leq-2x+5.

In range B, when \frac{2}{3}\leq{x}\leq\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=-2x+5, so we get 3x-2\leq-2x+5.

In range C, when x>\frac{5}{2}: |3x-2|=3x-2 and |2x-5|=2x-5, so we get 3x-2\leq2x-5.

_________________
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities


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