What is the solution set for |3x-2|<=|2x-5| : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 Feb 2017, 17:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the solution set for |3x-2|<=|2x-5|

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 25 Dec 2009
Posts: 99
Followers: 1

Kudos [?]: 161 [1] , given: 3

What is the solution set for |3x-2|<=|2x-5| [#permalink]

### Show Tags

16 Jan 2010, 06:00
1
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

77% (03:06) correct 23% (07:01) wrong based on 46 sessions

### HideShow timer Statistics

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.
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [7] , given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 06:34
7
KUDOS
Expert's post
5
This post was
BOOKMARKED
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 $$x>\frac{5}{2}$$, then in this range we have no solution;

Ranges from A and B give us the solution as: $$-3\leq{x}\leq\frac{7}{5}$$.
_________________
Manager
Joined: 25 Dec 2009
Posts: 99
Followers: 1

Kudos [?]: 161 [0], given: 3

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 08:15
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.
Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1310
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 79

Kudos [?]: 1025 [1] , given: 157

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 09:49
1
KUDOS
@ 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???
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [1] , given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 11:04
1
KUDOS
Expert's post
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.
_________________
Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1310
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 79

Kudos [?]: 1025 [0], given: 157

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 11:39
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??
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [7] , given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

16 Jan 2010, 11:54
7
KUDOS
Expert's post
1
This post was
BOOKMARKED
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 $$3x-2\leq2x-5$$.
_________________
Manager
Joined: 19 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 79 [0], given: 28

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

18 Sep 2010, 06:08
Thanks Bunuel you posted excellent approach to deal with inequalities. I think you should write somthing on inequalities as well.
Kudos to you
Intern
Joined: 01 Aug 2010
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 3

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

11 Mar 2011, 10:22
Bunuel 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: $$-3\leq{x}\leq\frac{7}{5}$$.

that's a lovely and uncomplicated way....
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 162

Kudos [?]: 1737 [0], given: 376

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

11 Mar 2011, 12:42
Bunuel wrote:
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 $$3x-2\leq2x-5$$.

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [6] , given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

11 Mar 2011, 12:51
6
KUDOS
Expert's post
fluke wrote:
Bunuel wrote:
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 $$3x-2\leq2x-5$$.

My slow brain is just not getting it!!! How both LHS and RHS get negated for the range x<2/3 and only the RHS gets negated for the range B and nothing for range C. What if there were 4 or 5 ranges; what is the deciding factor as to what gets negated.

Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|\leq{-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|\leq{some \ expression}$$. For example: $$|5|=5$$;

In range A, when $$x<\frac{2}{3}$$: $$3x-2<0$$ so $$|3x-2|=-(3x-2)$$ and $$2x-5<0$$ so $$|2x-5|=-(2x-5)$$, and we get $$-3x+2\leq-2x+5$$.

In range B, when $$\frac{2}{3}\leq{x}\leq\frac{5}{2}$$: $$3x-2>0$$ so $$|3x-2|=3x-2$$ and $$2x-5<0$$ so $$|2x-5|=-(2x-5)$$, so we get $$3x-2\leq-2x+5$$.

For more check: math-absolute-value-modulus-86462.html

Hope it's clear.
_________________
Manager
Joined: 19 Dec 2010
Posts: 145
Followers: 2

Kudos [?]: 30 [0], given: 12

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

19 Mar 2011, 12:44
Great solution bunuel, kudos!
Intern
Joined: 24 Feb 2011
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 38

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

02 Sep 2012, 07:33
Bunuel,
when you say as -3<= X and as x< $$\frac{2}{3}$$ then -3<=x<=$$\frac{2}{3}$$ does this mean that the range of x that satisfies the condition x<$$\frac{2}{3}$$ is all those points on the number line which satisfy both inequalities.

Also you say ranges from A and B give us solution as -3<= x<= $$\frac{7}{5}$$ . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x<$$\frac{2}{3}$$ , $$\frac{2}{3}$$<=x<$$\frac{5}{2}$$ and x>=$$\frac{5}{2}$$.

Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?

I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.

Best wishes and many many thanks !!!
Amogh.
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [1] , given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

03 Sep 2012, 04:04
1
KUDOS
Expert's post
Amogh wrote:
Bunuel,
when you say as -3<= X and as x< $$\frac{2}{3}$$ then -3<=x<=$$\frac{2}{3}$$ does this mean that the range of x that satisfies the condition x<$$\frac{2}{3}$$ is all those points on the number line which satisfy both inequalities.

Also you say ranges from A and B give us solution as -3<= x<= $$\frac{7}{5}$$ . Can you please explain this to me. Does this mean that the final solution to this problem would be all those ranges that have a solution to their respective conditions i.e the conditions x<$$\frac{2}{3}$$ , $$\frac{2}{3}$$<=x<$$\frac{5}{2}$$ and x>=$$\frac{5}{2}$$.

Also if for example ranges from A and B were -5<=x<2 and 4<=x<10, then would the final solution be both these ranges or do they have to have an overlap?

I know this is a supremely dumb and elementary question, but unfortunately inequalities and modulus happen to be my weakest areas. I cannot express in words my gratitude to you for having put up the GMAT Math Book. That thing is my Math Bible. You've made math so easy to understand.

Best wishes and many many thanks !!!
Amogh.

We consider three ranges:
A. $$x<\frac{2}{3}$$;
B. $$\frac{2}{3}\leq{x}\leq\frac{5}{2}$$;
C. $$x>\frac{5}{2}$$.

In range A we get: $$-3\leq{x}<\frac{2}{3}$$;
In range B we get: $$\frac{2}{3}\leq{x}\leq\frac{7}{5}$$;
In range C there is no solution.

So, the given inequality holds true for $$-3\leq{x}<\frac{2}{3}$$ and $$\frac{2}{3}\leq{x}\leq\frac{7}{5}$$. Now, we can combine these two ranges and write: $$-3\leq{x}\leq\frac{7}{5}$$.

Next, the solution set is $$-3\leq{x}\leq\frac{7}{5}$$, means that any x from $$-3\leq{x}\leq\frac{7}{5}$$ will satisfy $$|3x-2|\leq|2x-5|$$.

If for some question we had -5<=x<2 from one range and 4<=x<10 from another, then the solution will be both ranges (no need for overlap).

Hope it's clear.
_________________
Intern
Joined: 11 Jul 2012
Posts: 46
Followers: 0

Kudos [?]: 5 [0], given: 0

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]

### Show Tags

01 Nov 2012, 11:54
Hello Bunuel. Does this approach always work?:
|3x-2| <= |2x-5| ===> (3x-2)^2<= (2x-5)^2 ===> 5x^2+ 8x -21 <= 0 solution of the ineq (-3, 7/5)
Brother Karamazov
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [0], given: 10737

Re: What is the solution set for |3x-2|<=|2x-5| [#permalink]

### Show Tags

01 Nov 2012, 12:32
Ousmane wrote:
Hello Bunuel. Does this approach always work?:
|3x-2| <= |2x-5| ===> (3x-2)^2<= (2x-5)^2 ===> 5x^2+ 8x -21 <= 0 solution of the ineq (-3, 7/5)
Brother Karamazov

Do you mean squaring? If both parts of an inequality are non-negative (as in our case), then you can safely raise both parts to an even power (for example square).

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.
_________________
VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1096
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Followers: 38

Kudos [?]: 532 [1] , given: 70

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

02 Nov 2012, 04:58
1
KUDOS
Bunuel 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: $$-3\leq{x}\leq\frac{7}{5}$$.

Hi Bunnuel
excellent approach
My doubt in range c is whether the fraction is 5/2 or 5/7 if it is the latter can you pls explain how.

Regards
Archit
VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1096
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Followers: 38

Kudos [?]: 532 [0], given: 70

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

02 Nov 2012, 05:22
Bunuel wrote:
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 $$3x-2\leq2x-5$$.

While solving for A i understand that since (3x-2)<0 so you have multiplied the (3x-5) with negetive. But (2x-5) is >0 so i didnt understand why did you multiply it with negetive. Is it that you plugged a value 1/2 which is less than 2/3 and than analysed that we have to multiply negetive value.

Pls explain
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [0], given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

02 Nov 2012, 07:47
Archit143 wrote:
Bunuel 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: $$-3\leq{x}\leq\frac{7}{5}$$.

Hi Bunnuel
excellent approach
My doubt in range c is whether the fraction is 5/2 or 5/7 if it is the latter can you pls explain how.

Regards
Archit

Yes, it should be 5/2. Typo edited. Thank you. +1.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37083
Followers: 7243

Kudos [?]: 96336 [0], given: 10737

Re: Inequalities - Challenging and tricky One [#permalink]

### Show Tags

02 Nov 2012, 07:50
Archit143 wrote:
Bunuel wrote:
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 $$3x-2\leq2x-5$$.

While solving for A i understand that since (3x-2)<0 so you have multiplied the (3x-5) with negetive. But (2x-5) is >0 so i didnt understand why did you multiply it with negetive. Is it that you plugged a value 1/2 which is less than 2/3 and than analysed that we have to multiply negetive value.

Pls explain

For A: if $$x<\frac{2}{3}$$ then both 3x-2 and 2x-5 are negative. Just substitute x=0<2/3 to check.

Hope it helps.
_________________
Re: Inequalities - Challenging and tricky One   [#permalink] 02 Nov 2012, 07:50

Go to page    1   2    Next  [ 34 posts ]

Similar topics Replies Last post
Similar
Topics:
14 If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a 6 18 Jan 2016, 22:32
7 If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 8 31 Aug 2015, 09:42
4 Lines y=√3·x−2 and y=2√3·x−5 intersect at what height above the x axis 5 12 May 2015, 10:51
14 If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value 5 06 Nov 2014, 07:56
3 If 2x + 5y =8 and 3x = 2y, what is the value of 2x + y? 3 28 Mar 2010, 08:32
Display posts from previous: Sort by

# What is the solution set for |3x-2|<=|2x-5|

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.