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Math Help - |3x-1|<|-x+4|

  1. #1
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    |3x-1|<|-x+4|

    Solve in set R
    |3x-1|<|-x+4|
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by lebanon View Post
    Solve in set R
    |3x-1|<|-x+4|
    first find intervals in which you will look

     3x-1 = 0 \Rightarrow x = 1/3
     -x +4 = \Rightarrow x = 4

    so you look in intervals  (-\infty, 1/3 ] \cup [1/3 , 4 ] \cup [4 , \infty) of the interval  (-\infty, \infty)

    can you continue ? show what have you done
    Last edited by yeKciM; October 14th 2010 at 01:48 AM.
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  3. #3
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    Another way to do this: Look at the equation |3x- 1|= |-x+ 4|. Since the two absolute values are the same, either the numbers themselves are the same, 3x- 1= -x+ 4, or the differ only in sigh, 3x-1= -(-x+ 4)= x- 4.

    The first equation, 3x- 1= -x+ 4 gives 4x= 5 or x= 5/4. The secon gives 2x= -3 or x= -3/2.

    The point is that those two points are the only points at which "<" can change to ">" and vice versa.

    The number -2 is less that -3/2 and putting x= -2 in the original equation, |3(-2)-1|= |-7|= 7> |-(-2)+4)|= |6|= 6 so |3x-1|> |-x+ 4| for all x< -3/2.

    The number 0 is between -3/2 and 5/4. Putting x= 0 in the original equation, |3(0)-1|= |-1|= 1< |-(0)+ 4|= |4|= 4 so |3x-1|< |-x+ 4| for all x between -3/3 and 5/4.

    The number 2 is greater than 5/4. Putting x= 2 in the original equation, |3(2)- 1|= |5|= 5> |-(2)+ 4|= |2|= 2 so |3x-1|> |-x+ 4| for all x greater than 5/4.
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by lebanon View Post
    Solve in set R
    |3x-1|<|-x+4|
    you can square and get 9x^2-6x+1<x^2-8x+16\implies8x^2+2x-15<0, i think you can continue.
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