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  1. #1
    Member SengNee's Avatar
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    Help me...

    1) $\displaystyle |2x+1|<4x-2$

    2) $\displaystyle |2x-1|>x+2$

    3) $\displaystyle |\frac {x-2}{x+1}|<3$

    4) $\displaystyle |2x-1|>\frac {1}{x}$


    (Show me as many methods as possible. Thanks)
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  2. #2
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    When you have inequalities involving moduluses you need to separate it into the greater than and less than 0 cases as follows

    1) |2x+1|<4x-2
    first do the case $\displaystyle 2x+1 \geq 0$

    We want the intersection of $\displaystyle 2x+1 \geq 0$ and 2x+1<4x-2
    $\displaystyle 2x \geq -1$
    $\displaystyle x \geq -1/2$

    2x+1<4x-2
    3<2x
    x>3/2

    So the intersection is x > 3/2

    Next we look at 2x+1 < 0

    So we want the intersection of 2x+1<0 and -(2x+1)<4x-2
    solving this gives x<-1/2 and x>1/6. This has no intersection so the final answer is simply x > 3/2.
    If the intersection did exist, we would take the union of it and the first solution to get the final result.
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  3. #3
    Member SengNee's Avatar
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    1) $\displaystyle |2x+1|<4x-2$

    $\displaystyle 2x+1<4x-2$
    $\displaystyle 3<2x$
    $\displaystyle 3/2<x$ ...... i


    $\displaystyle 2x+1>2-4x$
    $\displaystyle 6x>1$
    $\displaystyle x>1/6$ ...... ii

    To fulfill both i and ii, therefore $\displaystyle 3/2<x$.


    Can I do like that?
    Last edited by SengNee; Jan 8th 2008 at 08:23 PM. Reason: Correction...
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  4. #4
    Forum Admin topsquark's Avatar
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    For
    $\displaystyle |2x+1|<4x-2$

    It would be
    $\displaystyle 2x+1<4x-2$

    and
    $\displaystyle 2x+1>2-4x$

    Note the change from < to > in the second expression.

    -Dan
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  5. #5
    Member SengNee's Avatar
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    2) $\displaystyle |2x-1|>x+2$


    $\displaystyle 2x-1>x+2$
    $\displaystyle x>3$ ... i

    $\displaystyle 2x-1<-x-2$
    $\displaystyle 3x<-1$
    $\displaystyle x<-1/3$ ... ii

    How about this?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    2) $\displaystyle |2x-1|>x+2$


    $\displaystyle 2x-1>x+2$
    $\displaystyle x>3$ ... i

    $\displaystyle 2x-1<-x-2$
    $\displaystyle 3x<-1$
    $\displaystyle x<-1/3$ ... ii

    How about this?
    Looks good to me. Can you plug in some numbers from your solution set to see if it looks good to you?

    -Dan
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  7. #7
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    linear inequalility solution

    Quote Originally Posted by SengNee View Post
    1) $\displaystyle |2x+1|<4x-2$

    2) $\displaystyle |2x-1|>x+2$

    3) $\displaystyle |\frac {x-2}{x+1}|<3$

    4) $\displaystyle |2x-1|>\frac {1}{x}$


    (Show me as many methods as possible. Thanks)
    1) 2x +1< 4x -2
    Collect like terms
    2x-4x < -2-1
    -2x < -3
    Divide both sides by -2
    X > 1
    2) 2x 1 > x +2
    2x- x > 2+1
    x> 3

    from clement
    cokhale@yahoo.com
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  8. #8
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    1.

    Help me...-easyway.jpg

    set the postive mod to equal the straightline graph to get point of intersection

    cant go wrong
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  9. #9
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    Smile inequality solution

    [QUOTE=cokhale;112884]1) 2x +1< 4x -2
    Collect like terms
    2x-4x < -2-1
    -2x < -3
    Divide both sides by -2
    X > 1
    2) 2x 1 > x +2
    2x- x > 2+1
    x> 3


    3) X -2\ x +1 < 3
    Solution
    -----------
    Multiple both sides by (x +1)
    X -2\ x +1 (x +1) < 3(x +1) = (x -2)(x +1)< 3(x +1)(x +1) = (x -2)(x +1) < 3( x +2x +1) = (x -2)(x +1)<3x +6x +3 = x -x-2< 3x +6x +3
    = 2x +7x +5<0
    2x +7x +5 =0
    2x + 5x +2x +5
    X(2x +5) +1(2x +5)
    (2x +5)(x +1)<0
    2x +5 =0
    2x = -5
    Divide both sides by 2
    X = -5\2 = -2
    Or
    X +1 =0
    X= -1
    _+____-_______+__
    -2.5 0 1 2

    Since,< is a negative solution.
    -2 < x < 2.
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