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Math Help - proportions - heavy algebra needed

  1. #1
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    proportions - heavy algebra needed

    Found this problem looking in my old textbook:

    If proportions - heavy algebra needed-codecogseqn-10.gif, find the numerical value of the ratio proportions - heavy algebra needed-codecogseqn-11.gif.

    I started cross multiplying & I got stuck quite easily. How do I finish this?
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  2. #2
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    Re: proportions - heavy algebra needed

    Use a common denominator.
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  3. #3
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    Re: proportions - heavy algebra needed

    I agree with your advice, but would I then solve for a/b?
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  4. #4
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    Re: proportions - heavy algebra needed

    \frac{4a-9b}{4a} = \frac{a-2b}{b}

    \frac{4a-9b}{4a}\cdot \frac{b}{b} = \frac{a-2b}{b} \cdot \frac{4a}{4a}

    \frac{4ab-9b^2}{4ab} = \frac{4a^2-8ab}{4ab}

    \frac{4a^2-8ab}{4ab} - \frac{4ab-9b^2}{4ab} = 0

     \frac{4a^2-8ab-4ab+9b^2}{4ab}=0

     \frac{4a^2-12ab+9b^2}{4ab}=0

     4a^2-12ab+9b^2=0

     (2a-3b)(2a-3b)=0

    2a=3b\neq0

    a=1.5b

    \frac{a}{b} = \frac{1.5b}{b} = 1.5
    Thanks from zachd77
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  5. #5
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    Re: proportions - heavy algebra needed

    thanks I remember attempting this problem a few years ago, didn't turn out well.
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  6. #6
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    Re: proportions - heavy algebra needed

    Alternatively:
    (4a-9b)/4a= (a-2b)/b Just cross multiply and we get
    4ab 9b^2 = 4a^2 8 ab
    Or 4a^2 12 ab + 9 b^2 = 0
    ( 2a )^2 2 ( 2a) ( 3b) + ( 3b)^2 = 0
    ( 2a 3b)^2 = 0
    That gives a = 1.5 b
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