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Find the solution set (values of X) for the following

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VP
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Find the solution set (values of X) for the following [#permalink]

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New post 12 Jun 2006, 14:38
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Find the solution set (values of X) for the following inequality

|3X - 2| <= |2X - 5|.

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Manager
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Re: PS: not really easy [#permalink]

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New post 12 Jun 2006, 15:03
MA wrote:
Find the solution set (values of X) for the following inequality

|3X - 2| <= |2X - 5|.


Tricky one...did a lot of plugging in numbers.

-3<= x <= 7/5

Lower Limit
Plugged in numbers to find the lower limits. Also, realized that when x is neg, both sides of the equation (within the absolute value) are negative. Therefore, we can solve:

2-3x <= 5-2x
-3<= x


Upper Limit
Plugged in x = 1 , Worked
Plugged in x = 2, Did not work

So I know the upper limits is between 1 and 2. Looking at the problem, I also know that when x = 1, the right side of the equation is negative and the left side is positive (within the absolute value).

So to find the upper limit I reversed the right side of the equation:
3X - 2 <= 5 - 2X
x <= 7/5

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VP
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 [#permalink]

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New post 12 Jun 2006, 15:15
This wasn't solved "quickly" so might not hold for as an actual GMAT answer

Given |3x-2| <= |2x-5|

Squaring both sides,

9x^2 -12x+4 <= 4x^2-20x+25

5x^2+8x-21 <= 0

x^2 +8x/5 -21/5 <=0

x^2 + (-7x/5 + 15x/5) + ((-7/5)x(15/5) <= 0

(x - 7/5)(x+15/5) <= 0

x <= 7/5 or x <= -3 Solutions.

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VP
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New post 12 Jun 2006, 15:18
MA,
Where did you get this Q from? I need to practice more for such questions...

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VP
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 [#permalink]

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New post 13 Jun 2006, 13:07
anymore???????

haas_mba07 wrote:
MA,
Where did you get this Q from? I need to practice more for such questions...


the question was provided by one of my friend.

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 [#permalink]

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New post 13 Jun 2006, 15:18
I prefer to represent it on a XY plan. Have a look at the attachment.

I consider Y(red) = |3x-2| and Y(Blue) = |2x-5|. Thus, the problem consists of finding the delimited points: when Y(Blue) = Y(Red).

Hence, the solution is as it is said above : -3 <= x <= 7/5.
Attachments

Graph_of_X-solutions.gif
Graph_of_X-solutions.gif [ 6.95 KiB | Viewed 1025 times ]

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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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New post 13 Jun 2006, 21:03
Squaring both sides leads to
(5x-7) (x-3) <=0

That means

-3 <= x <=7/5
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Re: PS: not really easy [#permalink]

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New post 13 Jun 2006, 21:46
MA wrote:
Find the solution set (values of X) for the following inequality

|3X - 2| <= |2X - 5| (*).


Yes, squaring is one method but it's not advisable in case we can't factorize the obtained expression.

My method is breaking the | | by considering ranges of x.
We have two critical values of x which are 2/3 and 5/2
1) x> 5/2
--> (*) <=> 3x-2<=2x-5 --> x<=-3 , but x>=5/2 ---> eliminate this case
2) 2/3<=x<=5/2:
(*) <--> 3x-2<= 5-2x ---> x<=7/5. Check the precondition --> 2/3<=x<=7/5
3) x< 2/3
(*) <--> 2-3x<= 5-2x --> x>=-3. Check the condition--> -3<=x<=2/3

---> final result: -3<= x<= 7/5

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 [#permalink]

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New post 07 Jul 2006, 08:20
This can definitely be solved under 2 min.
Critical points are 2/3 and 5/2.

Interval 1: (-@, 2/3)
-3x+2<=-2x+5
x>=-3

Interval 2: <2/3, 5/2>
3x-2<=-2x+5
x<=7/5

Interval 3: (5/2, +@)
3x-2<=2x-5
x<=-3 (OUT OF INTERVAL)

Taken together:
x is from the interval <-3, 7/5>

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Senior Manager
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 [#permalink]

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New post 07 Jul 2006, 10:46
-3 <= x <= 7/5

Solved, 4 eqns, if |a| <= |b|, then
a < = -b
a < = b
a >= b
a > = -b

and plugged in values to find the solution.

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  [#permalink] 07 Jul 2006, 10:46
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