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Math Help - algebra

  1. #1
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    algebra

    \frac {4}{a-7} = \frac {-2a}{a+3}
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  2. #2
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    <br />
\frac {4}{a-7} = \frac {-2a}{a+3}<br />
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  3. #3
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    \frac{4}<br />
{{a - 7}} = \frac{{ - 2a}}<br />
{{a + 3}}

    4\left( {a + 3} \right) =  - 2a\left( {a - 7} \right)

    4a + 12 =  - 2a^2  + 14a

    2a^2  - 10a + 12 = 0

    2\left( {a^2  - 5a + 6} \right) = 0

    2\left( {a - 3} \right)\left( {a - 2} \right) = 0

    a - 3 = 0 \Rightarrow a = 3

    or

    a - 2 = 0 \Rightarrow a = 2
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  4. #4
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    Quote Originally Posted by imbadatmath View Post
    <br />
\frac {4}{a-7} = \frac {-2a}{a+3}<br />
    Just like the last problem, we have equal ratios, so we cross multiply:

    4(a + 3) = -2a(a - 7)

    Evaluating, we get:

    4a + 12 = -2a^2 + 14a

    Bring the right side over to the left side, we get:

    2a^2 - 10a + 12 = 0

    That's a quadratic equation. If you need to solve that, go here and enter your coefficients:

    Quadratic Equation

    I get a = (2, 3).

    Plug that back into your original proportion and see if it works.
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