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Math Help - algebra problem

  1. #1
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    algebra problem

    If \frac{a}{c}=\frac{b}{d}=\frac{3}{4}
    and \sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15
    find ac+bd-ad-bc

    please help
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  2. #2
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    Re: algebra problem

    108.
    a = 9 + 3k, b = 3k, c = 12 + 4k, d = 4k where k is any integer > 0
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  3. #3
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    Re: algebra problem

    Hello, earthboy!

    \text{Given: }\:\begin{Bmatrix}[1] & \dfrac{a}{c}\,=\,\dfrac{b}{d}\,=\,\dfrac{3}{4} \\ \\ [2] &\sqrt{a^2+c^2}-\sqrt{b^2+d^2}\:=\:15\end{Bmatrix}

    \text{Find: }\:[3]\;N \:=\: ac+bd-ad-bc

    \text{From [1]: }\:\begin{Bmatrix}\frac{a}{c} \,=\,\frac{3}{4} &\Rightarrow& c \,=\,\frac{4}{3}a \\ \\[-4mm] \frac{b}{d} \,=\,\frac{3}{4} & \Rightarrow & d \,=\,\frac{4}{3}b \end{Bmatrix}\;\;[4]


    \text{Substitute [4] into [3]:}

    N \;=\;a\left(\tfrac{4}{3}a\right) + b\left(\tfrac{4}{3}b\right) - a\left(\tfrac{4}{3}b\right) - b\left(\tfrac{4}{3}a\right) \;=\;\tfrac{4}{3}a^2 + \tfrac{4}{3}b^2 - \tfrac{4}{3}ab - \tfrac{4}{3}ab

    n . =\;\tfrac{4}{3}a^2 - \tfrac{8}{3}ab + \tfrac{4}{3}b^2 \;=\;\tfrac{4}{3}(a^2 - 2ab + b^2) \;=\;\tfrac{4}{3}(a-b)^2\;\;[5]


    \text{Substitute [4] into [2]:}

    . . \sqrt{a^2 + \left(\tfrac{4}{3}a\right)^2} - \sqrt{b^2 + \left(\tfrac{4}{3}b\right)^2} \;=\;15

    . . \sqrt{a^2 + \tfrac{16}{9}a^2} - \sqrt{b^2 + \tfrac{16}{9}b^2} \;=\;15 \quad\Rightarrow\quad \sqrt{\tfrac{25}{9}a^2} - \sqrt{\tfrac{25}{9}b^2} \;=\;15

    . . \tfrac{5}{3}a - \tfrac{5}{3}b \;=\;15 \quad\Rightarrow\quad a - b \;=\;9


    \text{Substitute into [5]: }\:N \;=\;\tfrac{4}{3}(a-b)^2 \;=\;\tfrac{4}{3}(9^2) \;=\;108
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