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Manager
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a hell of inequality [#permalink]
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13 Jul 2007, 07:30
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SOLVE 3x2<=2x5



Manager
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Re: a hell of inequality [#permalink]
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13 Jul 2007, 08:03
boubi wrote: SOLVE 3x2<=2x5
3X2X <= 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.



Manager
Joined: 24 Jun 2006
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that is not the right answer...



Manager
Joined: 22 May 2007
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7/5 works too.
Kevin is right, ans is between 3 and 7/5.



Manager
Joined: 24 Jun 2006
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post calculation and reasoning



Director
Joined: 16 May 2007
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i got 2/3<x<5/2 .. OA?



Manager
Joined: 22 May 2007
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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?



Senior Manager
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Re: a hell of inequality [#permalink]
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13 Jul 2007, 11:28
boubi wrote: SOLVE 3x2<=2x5
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



Director
Joined: 06 Sep 2006
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SOLVE 3x2<2x>= 7/5; x <= 3
Impossible or an empty set.



Manager
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Re: a hell of inequality [#permalink]
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13 Jul 2007, 15:00
dahcrap wrote: boubi wrote: SOLVE 3x2<=2x5 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+8X16 to (X+3)*(X  7/5)?



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Re: a hell of inequality [#permalink]
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13 Jul 2007, 15:16
boubi wrote: dahcrap wrote: boubi wrote: SOLVE 3x2<=2x5 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+8X16 to (X+3)*(X  7/5)?
For this use
[b + sqrt(b^2  4ac)]/2a
where b = 8, c =16, a=5



Manager
Joined: 25 Jul 2006
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(3x2)^2 <= (2x5)^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



Manager
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sooo you didn't escaped the hard way !!! we ggot to find some less fastidious wayy!!



Senior Manager
Joined: 03 Jun 2007
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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



SVP
Joined: 01 May 2006
Posts: 1796

3x2<=2x5
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 3x2 = 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
3x2<=2x5
<=> (3*x2) <= (2*x5)
<=> 3*x  2 >= 2*x5
<=> x >= 3 >>>> thus, 3 =< x =< 2/3 is an interval of solutions
o If 2/3 < x < 5/2, then
3x2<=2x5
<=> (3*x2) <= (2*x5)
<=> 5*x <= 7
<=> x <= 7/5 >>>> thus, 2/3 < x <= 7/5 is an interval of solutions
o If x >= 5/2, then
3x2<=2x5
<=> (3*x2) <= (2*x5)
<=> x <= 3 >>>> Impossible as x >= 5/2.... No interval of solutions here.
Finally, we have, by unioning the 2 intervals:
3 <= x <= 7/5



Manager
Joined: 25 Jul 2006
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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.
3x2 <= 2x5
1. First you find the pivotal/critical points:
i.e.
a. you set 3x2 = 0, so x = 2/3
b. you set 2x5 = 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 2x5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign)
or x <= 7/5
c. For x > 5/2
3x2 <= 2x5
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, 22:09, edited 1 time in total.



SVP
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Posts: 1796

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. 3x2 <= 2x5 1. First you find the pivotal/critical points: i.e. a. you set 3x2 = 0, so x = 2/3 b. you set 2x5 = 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 2x5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign) or x <7> 5/2 3x2 <= 2x5 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



Manager
Joined: 25 Jul 2006
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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...



Manager
Joined: 24 Jun 2006
Posts: 129

Fig master of absolutes!! that is the least we can expect from an MBA HEC



Manager
Joined: 25 Jul 2006
Posts: 97

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: ?
x2  x3 = x5
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}







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