+ Reply to Thread
Page 1 of 3
1 2 3 LastLast
Results 1 to 10 of 22

Thread: 3 Inequations

  1. #1
    Grrr... arjmen is a TestMagic guru. Show your respect! arjmen is a TestMagic guru. Show your respect! arjmen's Avatar
    Join Date
    Apr 2005
    Posts
    1,268

    3 Inequations

    Find the range of x for these inequations:

    1) |x^2 - 7x + 12| > x^2 - 7x + 12

    2) 2 <= |x+3| < 11

    3) |x+8| < |x-3|

  2. #2
    Eager! maria1_03 just joined TestMagic. maria1_03's Avatar
    Join Date
    Oct 2004
    Location
    Bucharest
    Posts
    65
    Quote Originally Posted by arjmen
    Find the range of x for these inequations:

    1) |x^2 - 7x + 12| > x^2 - 7x + 12

    2) 2 <= |x+3| < 11

    3) |x+8| < |x-3|

    1) x is between (3;4)

    2) x is between (-14 ; -5) U [-1; 8)

    3) x < - 2.5

  3. #3
    An Urch Guru Pundit Swami Sage chix475ntu radiates success. chix475ntu's Avatar
    Join Date
    May 2005
    Location
    Jersey City
    Posts
    952
    1) 3<x<4
    2) -14<x<=-5 U -1<=x<8
    3)x<-2.5

  4. #4
    Trying to make mom and pop proud 4567 just joined TestMagic.
    Join Date
    May 2005
    Posts
    28
    Could someone please write out how to solve the first one for old-timers such as myself

  5. #5
    Within my grasp! Synergy79 just joined TestMagic.
    Join Date
    Feb 2005
    Posts
    126
    I agree with the first two. Could somebody please explain how the third eq was resolved?

  6. #6
    Within my grasp! dsaqwert 's dreams are becoming reality.
    Join Date
    Mar 2005
    Posts
    386
    Quote Originally Posted by Synergy79
    I agree with the first two. Could somebody please explain how the third eq was resolved?
    |x+8| < |x-3|
    x<=-8
    -x-8<-x+3
    -8<3
    x<=-8

    -8<x<3
    x+8<-x+3
    -8<x<-2.5

    x>3
    x+8<x-3
    8<-3

    overall x<-2.5

  7. #7
    Grrr... arjmen is a TestMagic guru. Show your respect! arjmen is a TestMagic guru. Show your respect! arjmen's Avatar
    Join Date
    Apr 2005
    Posts
    1,268
    Solutions:

    1) |x^2 - 7x + 12| > x^2 - 7x + 12

    If |y| > y, clearly y<0

    Similarly,
    x^2 - 7x + 12 < 0
    => (x-4)(x-3) < 0
    Therefore, 3 < x < 4

    -------------------------------------------------------------------------

    2) 2 <= |x+3| < 11

    Textbook approach:
    Break into 2 inequations
    i) |x+3| < 11
    implies -11 < (x+3) < 11
    implies -14 < x < 8

    ii) |x+3| >= 2
    implies x >= -1 OR x <= -5

    Merging the 2 solution sets gives,
    -14 < x <= -5 OR -1 <= x < 8

    Number line method:
    http://mathforum.org/library/drmath/view/53126.html

    I find that sketching a quick graph, helps me see the answer immediately

    -------------------------------------------------------------------------

    3) |x+8| < |x-3|
    Clearly, x<0

    We have 2 inequations:
    i) -(x+8) < -(x-3) --------> where (x+8) <= 0
    implies -8 < 3 ---> Insightful!

    ii) (x+8) < -(x-3) ---------> where (x+8) > 0
    implies x < -5/2

    Therefore, x < -5/2
    Last edited by arjmen; 07-22-2005 at 09:10 AM. Reason: formatting

  8. #8
    Eager! daipayan_b just joined TestMagic.
    Join Date
    Mar 2005
    Posts
    71
    In Solution 3,

    Why is -(x-3) the RHS in both inequalities? why not (x-3).

    |x+8| < |x+3|

    => (x+8) < (x+3) ; -(x+8) < (x+3)

    OR

    (x+8) < -(x+3) ; -(x+8) < -(x+3)

    Please explain...

  9. #9
    An Urch Guru Pundit Swami Sage chix475ntu radiates success. chix475ntu's Avatar
    Join Date
    May 2005
    Location
    Jersey City
    Posts
    952
    |x+8| < |x+3|
    From the above, there can be 4 equations possible.
    -(x+8)=-(x+3) ==> (x+8)=(x+3)
    -(x+8)=(x+3) ==> (x+8)=-(x+3)
    (x+8)=-(x+3)
    (x+8)=(x+3)

    From the above you can see that there are actually 2 equations only. How you want to put - and + on both side of equations depends from each person I think

  10. #10
    TOEFL : GRE :: Dimas 's dreams are becoming reality. Dimas's Avatar
    Join Date
    Apr 2005
    Location
    Some where
    Posts
    735

    Post

    Q1:Q2: nothing added to arjmen

    Q3: I recommend solving it geometrically. Much easier for linear absolute values.
    Attached Files

+ Reply to Thread
Page 1 of 3
1 2 3 LastLast

Similar Threads

  1. Inequations - Data Sufficiency
    By gmataspirenew in forum GMAT Data Sufficiency
    Replies: 4
    Last Post: 11-08-2008, 08:19 PM
  2. Inequations
    By CrackXam in forum GMAT Problem Solving
    Replies: 6
    Last Post: 08-02-2007, 07:26 PM
  3. inequations
    By happyking in forum GMAT Data Sufficiency
    Replies: 5
    Last Post: 11-21-2003, 06:51 PM

Bookmarks

What you can do

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts

SEO by vBSEO 3.5.0 RC2