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Math Help - Nonlinear Inequalities

  1. #1
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    Nonlinear Inequalities

    Express the solution using interval notation and graph.

    #1)
    4x/2x+3>2

    #2)
    x+2/x+3<x-1/x-2

    Thanks
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  2. #2
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    Quote Originally Posted by Ballplaya4237 View Post
    Express the solution using interval notation and graph.

    #1)
    4x/2x+3>2

    #2)
    x+2/x+3<x-1/x-2

    Thanks
    I'll show the first, the second is too ambiguous and needs to be re-formatted:

    \frac{4x}{2x + 3} > 2.

    There's a number of aproaches - the simplest would be to solve \frac{4x}{2x + 3} = 2, look at the graph to see the relevant intervals and then write down the answer. But that might be beyond your present scope.

    A slower way:

    \frac{4x}{2x + 3} - 2 > 0

    \Rightarrow \frac{4x}{2x + 3} - \frac{2(2x + 3)}{2x + 3} > 0

    \Rightarrow \frac{4x - 2(2x + 3)}{2x + 3} > 0

    \Rightarrow \frac{-6}{2x + 3} > 0

    \Rightarrow \frac{6}{2x + 3} < 0

    You therefore require 2x + 3 < 0 (think about it) => ......
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  3. #3
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    Alright, thanks. Interval notation would be (-Infinity, -3/2), correct?

    How would I go about doing the second problem?
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  4. #4
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    Quote Originally Posted by Ballplaya4237 View Post
    Alright, thanks. Interval notation would be (-Infinity, -3/2), correct?

    How would I go about doing the second problem?
    First you would comply with the implied request that was made:

    Quote Originally Posted by mr fantastic View Post
    [snip]
    the second is too ambiguous and needs to be re-formatted
    [snip]
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  5. #5
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    \frac{x+2}{x+3}<\frac{x-1}{x-2}

    Better?
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  6. #6
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    Quote Originally Posted by Ballplaya4237 View Post
    \frac{x+2}{x+3}<\frac{x-1}{x-2}

    Better?
    Much!


    \frac{x+2}{x+3} < \frac{x-1}{x-2}


    \Rightarrow \frac{x+2}{x+3} - \frac{x-1}{x-2} < 0


    Get common denominator:


    \Rightarrow \frac{(x+2)(x - 2)}{(x+3)(x - 2)} - \frac{(x-1)(x + 3)}{(x-2)(x + 3)} < 0


    \Rightarrow \frac{(x+2)(x - 2) - (x-1)(x + 3)}{(x+3)(x - 2)} < 0


    Simplify:


    \Rightarrow \frac{-1 - 2x}{(x+3)(x - 2)} < 0


    \Rightarrow \frac{2x + 1}{(x+3)(x - 2)} > 0


    So you want either:

    2x + 1 > 0 AND (x + 3)(x - 2) > 0

    OR

    2x + 1 < 0 AND (x + 3)(x - 2) < 0

    The first is only satisfied when x > 2.

    The second is only satisfied when x < -3.

    The details are left for you .....
    Last edited by mr fantastic; January 27th 2008 at 07:18 PM. Reason: Fixed a typo that led me up the garden path
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  7. #7
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    You lost me when you got the common denominator. Not sure where you got it from nor why one side is being multiplied by (x-2) and the other side by (x+2).
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  8. #8
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    Quote Originally Posted by Ballplaya4237 View Post
    You lost me when you got the common denominator. Not sure where you got it from nor why one side is being multiplied by (x-2) and the other side by (x+2).
    My mistake - the common denominator is (x + 3)(x - 2).

    Standby while I fix it up ......

    OK, all fixed. Re-read. Are you still lost? Do you understand what's happening with the common denominator business ......

    It's the same idea as getting \frac{1}{3} + \frac{1}{7} over a common denominator .....
    Last edited by mr fantastic; January 27th 2008 at 07:20 PM.
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  9. #9
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    I understand the GCF just fine but I'm completly failing to see how you simplified the numerator (x+2)(x-2)-(x-1)(x+3)
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  10. #10
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    Quote Originally Posted by Ballplaya4237 View Post
    I understand the GCF just fine but I'm completly failing to see how you simplified the numerator (x+2)(x-2)-(x-1)(x+3)
    (x+2)(x-2)-(x-1)(x+3) = x^2 -2x + 2x - 4 - (x^2 + 3x - x - 3)

    = x^2 - 4 - (x^2 + 2x - 3) = x^2 - 4 - x^2 - 2x + 3 = -1 - 2x.

    Which I hope is what I had in the numerator before I multiplied both sides of the inequality by -1 (hence reversing the inequality sign and turning -1 - 2x into -(-1 - 2x) = 1 + 2x).
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  11. #11
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    Awesome, thanks for the help.
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