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Thread: equation problem

  1. #1
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    equation problem

    From the set of equations shown below, find the positive ratio of t to r rounded to two decimal places.
    p = 3q
    p r = q s
    p r ^2 + q s^ 2 = q t^ 2
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  2. #2
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    Hello, yzc717!

    From the set of equations below,
    find the positive ratio $\displaystyle \frac{t}{r}$ to two decimal places.

    . . $\displaystyle \begin{array}{ccc}p \;=\; 3q & {\color{blue}[1]} \\ \\[-4mm]
    pr \;=\; qs & {\color{blue}[2]}\\ \\[-4mm]
    \frac{1}{2}pr^2 + \frac{1}{2}qs^2 \;=\; \frac{1}{2}qt^ 2 & {\color{blue}[3]}\end{array}$

    [1] becomes: .$\displaystyle \frac{p}{q} \:=\:3\quad{\color{blue}[4]}$


    [2] becomes: .$\displaystyle \frac{p}{q} \:=\:\frac{s}{r}$

    Substitute [4]: .$\displaystyle 3 \:=\:\frac{s}{r} \quad\Rightarrow\quad s \:=\:3r\quad{\color{blue}[5]} $


    Multiply [3] by $\displaystyle \frac{2}{q}\!:\quad \frac{p}{q}\!\cdot\!r^2 + s^2 \:=\:t^2$

    Substitute [4] and [5]: . $\displaystyle 3\!\cdot\!r^2 + (3r)^2 \;=\;t^2 \quad\Rightarrow\quad 3r^2 + 9r^2 \;=\;t^2$

    Hence: .$\displaystyle 12r^2 \:=\:t^2 \quad\Rightarrow\quad \frac{t^2}{r^2} \:=\:12 \quad\Rightarrow\quad \frac{t}{r} \:=\:\sqrt{12} \:=\:2\sqrt{3}$

    Therefore: . $\displaystyle \frac{t}{r} \;\approx\;3.46$

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