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Math Help - Find the numerical value of x/y

  1. #1
    Senior Member DivideBy0's Avatar
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    Find the numerical value of x/y

    If I try this question in different ways I get a different answer:

    For positive x,y,z,

    \frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}

    Find the numerical value of \frac{x}{y}



    (Using the fact that \frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}),

    Solution 1:

    \frac{y+(x+y)}{x-z+(z)}=\frac{x}{y}

    \implies \frac{2y+x}{x}=\frac{x}{y}

    \implies 2y+x=y

    \frac{x}{y}=-1

    Solution 2:

    \frac{x}{y}=\frac{y+(x+y)+(x)}{x-z+(z)+(y)}=\frac{2(x+y)}{x+y}=2
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  2. #2
    Senior Member DivideBy0's Avatar
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    Nope, it works for all k, supposedly.

    \frac{a+kc}{b+kd}=\frac{a(1+k(\frac{c}{a}))}{b(1+k  (\frac{d}{b}))}=\frac{a(1+k(\frac{c}{a}))}{b(1+k(\  frac{c}{a}))}=\frac{a}{b}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Where did you get this identity from? It is not true for all k:
    \frac{1}{2} = \frac{3}{6} = \frac{1 + 3k}{2 + 6k}
    is only valid for k = 0.

    -Dan
    actually, it is true otherwise. did you try other values for k?
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  4. #4
    Senior Member DivideBy0's Avatar
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    Ooops, look how I canceled the x's in the first solution. Happens way to much.
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  5. #5
    MHF Contributor
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    And

    if (2y +x)/x = x/y

    Then (2y +x) is not equal to y.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    For positive x,y,z,

    \frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}

    Find the numerical value of \frac{x}{y}
    Take the equations like so:
    \frac{y}{x-z} = \frac{x+y}{z}
    and
    \frac{y}{x-z} = \frac{x}{y}
    (The third equation is implied by these two.)

    The first one says:
    yz = (x + y)(x - z)

    \frac{z}{y} = \left ( \frac{x}{y} + 1 \right ) \left ( \frac{x}{y} - \frac{z}{y} \right )

    and the second says
    y^2 = x(x - z)

    1 = \left ( \frac{x}{y} \right ) \left ( \frac{x}{y} - \frac{z}{y} \right )

    Now define a = \frac{x}{y} and b = \frac{z}{y} and you get the system:
    b = (a + 1)(a - b)
    and
    1 = a(a - b)

    You are after the value of a.

    -Dan
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  7. #7
    Senior Member DivideBy0's Avatar
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    thanks, lol

    Interesting approach @ topsquark

    geeze, you guys are replying so faast today
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