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a hell of inequality

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a hell of inequality [#permalink] New post 13 Jul 2007, 06:30
SOLVE |3x-2|<=|2x-5|
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Re: a hell of inequality [#permalink] New post 13 Jul 2007, 07:03
boubi wrote:
SOLVE |3x-2|<=|2x-5|


3X-2X <= -5+2
X <= -3

To test it, I tried a couple a couple of answers, and it seems like -3 is in fact the breaking point.
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 [#permalink] New post 13 Jul 2007, 07:16
that is not the right answer... :)
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 [#permalink] New post 13 Jul 2007, 07:28
7/5 works too.

Kevin is right, ans is between -3 and 7/5.
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 [#permalink] New post 13 Jul 2007, 07:37
post calculation and reasoning
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 [#permalink] New post 13 Jul 2007, 07:38
i got 2/3<x<5/2 .. OA?
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 [#permalink] New post 13 Jul 2007, 07:43
boubi wrote:
post calculation and reasoning


Well, in this case we have 4 different inequalities + <= +, -<=-, +<=-, -<=+.

The first 2 give us x=-3, the others give us x=7/5. This must be our range.
What's the A?
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Re: a hell of inequality [#permalink] New post 13 Jul 2007, 10:28
boubi wrote:
SOLVE |3x-2|<=|2x-5|


Let me tell u a faster way to solve such problems where both sides of the ineq has abs values.
Simply square both sides and solve the quadratic eqn.

You will get (x+3)*(x - 7/5) <= 0
which give -3 <= x <= 7/5
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 [#permalink] New post 13 Jul 2007, 11:02
SOLVE |3x-2|<2x>= 7/5; x <= -3

Impossible or an empty set.
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Re: a hell of inequality [#permalink] New post 13 Jul 2007, 14:00
dahcrap wrote:
boubi wrote:
SOLVE |3x-2|<=|2x-5|


Let me tell u a faster way to solve such problems where both sides of the ineq has abs values.
Simply square both sides and solve the quadratic eqn.

You will get (x+3)*(x - 7/5) <= 0
which give -3 <= x <= 7/5


right on the spot, just need to square both sides
OA IS x E [-3;7/5]

dahcrap how did you come up from 5X^2+8X-16 to (X+3)*(X - 7/5)?
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Re: a hell of inequality [#permalink] New post 13 Jul 2007, 14:16
boubi wrote:
dahcrap wrote:
boubi wrote:
SOLVE |3x-2|<=|2x-5|


Let me tell u a faster way to solve such problems where both sides of the ineq has abs values.
Simply square both sides and solve the quadratic eqn.

You will get (x+3)*(x - 7/5) <= 0
which give -3 <= x <= 7/5


right on the spot, just need to square both sides
OA IS x E [-3;7/5]

dahcrap how did you come up from 5X^2+8X-16 to (X+3)*(X - 7/5)?


For this use

[-b +- sqrt(b^2 - 4ac)]/2a
where b = 8, c =16, a=5
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 [#permalink] New post 13 Jul 2007, 14:20
(3x-2)^2 <= (2x-5)^2
9x^2 -12x + 4 <= 4x^2 -20x + 25
Moving everything to LHS:
5x^2+ 8x - 21 <= 0

Factoring above inequation, you get the correct answer per dahcrap.
It is a good method for solving when two absolutes are involved
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 [#permalink] New post 13 Jul 2007, 14:20
:lol: sooo you didn't escaped the hard way !!! we ggot to find some less fastidious wayy!!
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 [#permalink] New post 13 Jul 2007, 14:24
boubi wrote:
:lol: sooo you didn't escaped the hard way !!! we ggot to find some less fastidious wayy!!


No way in hell for THIS problem
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 [#permalink] New post 13 Jul 2007, 14:36
|3x-2|<=|2x-5|

First of all, we have to see for which values each absolute is equal to 0. This will provide us the interval to study.

o 3x-2 = 0
<=> x = 2/3

o 2*x - 5 = 0
<=> x = 5/2

So, we must study the inequality with :
o x < 2/3
o 2/3 < x < 5/2
o x > 5/2

o If x =< 2/3, then
|3x-2|<=|2x-5|
<=> -(3*x-2) <= -(2*x-5)
<=> 3*x - 2 >= 2*x-5
<=> x >= -3 >>>> thus, -3 =< x =< 2/3 is an interval of solutions

o If 2/3 < x < 5/2, then
|3x-2|<=|2x-5|
<=> (3*x-2) <= -(2*x-5)
<=> 5*x <= 7
<=> x <= 7/5 >>>> thus, 2/3 < x <= 7/5 is an interval of solutions

o If x >= 5/2, then
|3x-2|<=|2x-5|
<=> (3*x-2) <= (2*x-5)
<=> x <= -3 >>>> Impossible as x >= 5/2.... No interval of solutions here.

Finally, we have, by unioning the 2 intervals:
-3 <= x <= 7/5
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 [#permalink] New post 13 Jul 2007, 14:40
When more than two absolutes are present though, it becomes complicated to use the squaring approach. Another method seems to work better. I'll briefly solve your given problem using this other method.

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

1. First you find the pivotal/critical points:
i.e.
a. you set 3x-2 = 0, so x = 2/3
b. you set 2x-5 = 0, so x = 5/2

2. If you take the 2 values, 2/3 and 5/2 on the number line, you get three domains
a. left of 2/3
b. b/w 2/3 and 5/2
c. right of 5/2

Next, you modify the intial inequation for these three domains, by taking out the modulus signs.

a. So, for x < 2/3 the inequation becomes,
-3x+2 <= -2x +5
(since both terms become negative for x < 2/3, you reverse their signs)
-3 <= x

b. For x b/w 2/3 and 5/2
3x -2 <= -2x + 5
(since 2x-5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign)
or x <= 7/5

c. For x > 5/2
3x-2 <= 2x-5
x <= -3 (not valid since x is assumed to be > 5/2 above)

Combining the above three results, you get
-3 <= x <= 7/5
------------------

Credit for this method goes to user "Fig" who has posted several problems/solutions using this approach. Please search and look at all those problems if you have any questions. Manhattan GMAT also uses this approach.

Last edited by oops on 13 Jul 2007, 21:09, edited 1 time in total.
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 [#permalink] New post 13 Jul 2007, 14:43
oops wrote:
When more than two absolutes are present though, it becomes complicated to use the squaring approach. Another method seems to work better. I'll briefly solve your given problem using this other method.

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

1. First you find the pivotal/critical points:
i.e.
a. you set 3x-2 = 0, so x = 2/3
b. you set 2x-5 = 0, so x = 5/2

2. If you take the 2 values, 2/3 and 5/2 on the number line, you get three domains
a. left of 2/3
b. b/w 2/3 and 5/2
c. right of 5/2

Next, you modify the intial inequation for these three domains, by taking out the modulus signs.

a. So, for x < 2/3 the inequation becomes,
-3x+2 <= -2x +5
(since both terms become negative for x < 2/3, you reverse their signs)
-3 <= x

b. For x b/w 2/3 and 5/2
3x -2 <= -2x + 5
(since 2x-5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign)
or x <7> 5/2
3x-2 <= 2x-5
x <3> 5/2 above)

Combining the above three results, you get
-3 <= x <= 7/5
------------------

Credit for this method goes to user "Fig" who has posted several problems/solutions using this approach. Please search and look at all those problems if you have any questions. Manhattan GMAT also uses this approach.


Well ;)... Finally, I should not post :)... All is in your post :D :)
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 [#permalink] New post 13 Jul 2007, 14:48
Gosh, Fig (master of absolutes!), didn't see you also responded to this mesg - we must've been typing it up simultaneously - credit goes to you - glad to see you back...
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 [#permalink] New post 13 Jul 2007, 14:54
Fig master of absolutes!! that is the least we can expect from an MBA HEC 8-)
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 [#permalink] New post 13 Jul 2007, 15:04
For example, here is a question using more than two absolutes. It is from Manhattan GMAT (will post their soln later).

Question
Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?
|x-2| - |x-3| = |x-5|

1. {-6, -5, 0, 1, 7, 8|
2. { -4, -2, 0, 10/3, 4, 5}
3. {-4, 0, 1, 4, 5, 6}
4. {-1, 10/3, 3, 5, 6, 8}
5. {-2, -1, 1, 3, 4, 5}
  [#permalink] 13 Jul 2007, 15:04
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