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Math Help - Absolute value inequalities.

  1. #1
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    Absolute value inequalities.

    please check my work if i done correctly. I'm little confused.
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    Last edited by mr fantastic; February 2nd 2011 at 03:37 AM. Reason: Title.
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  2. #2
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    Remember that \displaystyle |X| = \sqrt{X^2}, so I would rewrite this equation as

    \displaystyle \sqrt{(2x-6)^2} = \sqrt{(4-5x)^2}

    \displaystyle (2x- 6)^2 = (4 - 5x)^2.

    Solve for \displaystyle x.
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  3. #3
    Senior Member BAdhi's Avatar
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    |2x -6|=|4 -5x| is not -(2x-6)=-(4-5x), it is (2x-6)=\pm(4-5x)
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  4. #4
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    Quote Originally Posted by jam2011 View Post
    please check my work if i done correctly. I'm little confused.
    2x-6 and 4-5x are both linear functions.

    One has a positive slope and the other has a negative slope, so the lines intersect at a single point.

    The modulus of these are V-shaped as we invert the part of the graph below the x-axis,
    hence there are now two points of intersection if the original lines do not intersect on the x-axis.

    Hence, we can look for the points of intersection of 4 lines, while eliminating points of intersection below the y-axis
    (refer to the dashed lines on the attachment).

    2x-6=4-5x\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}

    2\left(\frac{10}{7}\right)-6<0

    Therefore the lines do not intersect above the x-axis
    and are therefore not a solution of

    |2x-6|=|4-5x|

    although the value of x is a solution to the equation.


    2x-6=-(4-5x)\Rightarrow\ 2x-6=5x-4\Rightarrow\ -2=3x\Rightarrow\ x=-\frac{2}{3}

    2\left(-\frac{2}{3}\right)-6<0

    so these lines also do not intersect above the x-axis.


    -(2x-6)=4-5x\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\Rightarrow\ x=-\frac{2}{3}

    -\left[2\left(-\frac{2}{3}\right)-6\right]>0

    therefore these lines do intersect above the x-axis.


    -(2x-6)=-(4-5x)\Rightarrow\ 6-2x=5x-4\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}

    -\left[2\left(\frac{10}{7}\right)-6\right]>0

    therefore these lines also intersect above the x-axis.

    Hence the solution is the union of

    -(2x-6)=(4-5x)

    and

    -(2x-6)=-(4-5x)
    Attached Thumbnails Attached Thumbnails Absolute value inequalities.-modulus-lines.jpg  
    Last edited by Archie Meade; February 2nd 2011 at 06:40 AM.
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  5. #5
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    since Mr. prove it give (2x- 6)^2 = (4 - 5x)^2 then
    (2x-6-4+5x)(2x-6+4-5x) by x^2-y^2 = (x+y)(x-y)
    7x-10 = 0 -3x-2 = 0
    x = 10/7 x = -2/3
    Can we consider this a short cut method.?thank you a lot for your effort.God bless
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  6. #6
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    Quote Originally Posted by jam2011 View Post
    Since Mr. Prove It gives (2x- 6)^2 = (4 - 5x)^2

    then (2x-6-4+5x)(2x-6+4-5x) by using x^2-y^2 = (x+y)(x-y)

    7x-10 = 0 so x = 10/7

    -3x-2 = 0 so x = -2/3

    Can we consider this a short cut method? thank you a lot for your effort. God bless
    That's a "cute" calculation you made!

    (2x-6)^2-(4-5x)^2=0\Rightarrow\ [(2x-6)+(4-5x)][(2x-6)-(4-5x)]=0

    Yes, it's a good shortcut and locates the values of x that are the solutions of the modulus function.

    This is because the x co-ordinates of the points of intersection of the lines
    are the same as the points of intersection of the parabolas.

    The method I showed earlier will give the x and y co-ordinates if required,
    however only the x co-ordinates are requested.
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  7. #7
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    Without reference to graphs, you can also try this way......

    |2x-6|=|5x-4|

    When x>3, both 2x-6 and 5x-4 are both positive
    and when 5x<4, both 2x-6 and 5x-4 are both negative.


    Case 1: both positive within modulus

    x\ge3,\;\;5x\ge4\Rightarrow\ x\ge3

    (2x-6)=(5x-4),\;\;\;x\ge3

    -2=3x\;\;?\;\;\;x\ge3 ....No, since the minimum value of 3x is 9.


    Case 2:

    2x-6\le0,\;\;5x-4\ge0\;\;\;\Rightarrow\ 5x\ge4,\;\;x\le3

    -(2x-6)=5x-4,\;\;\;\frac{4}{5}\le x\le3

    6-2x=5x-4\Rightarrow\ 10=7x\;\;?\;\;\;\frac{4}{5}\le x\le3

    x=\frac{10}{7}


    Case 3: Both negative within modulus

    x\le\frac{4}{5}

    -(2x-6)=-(5x-4)\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\;\;?\;\;\;x\le\frac{4}{5}

    x=-\frac{2}{3}
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  8. #8
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    wow thats great...thanks a lot for all of your information. i need to study more further. God bless you sir.
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