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Thread: Eliminating x and y

  1. #1
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    Post Eliminating x and y

    Hi there I have a problem that I am trying to figure out. It said that by eliminating $\displaystyle x$ and $\displaystyle y$ from the equations $\displaystyle \frac {1}{y}+\frac {1}{y}=1$, $\displaystyle x +y = a$, $\displaystyle \frac {y}{x}=m$ where $\displaystyle a =/0$. obtain a relation between $\displaystyle m $adn $\displaystyle a$. Given that a is real, determine the ranges of values of $\displaystyle a$ for which $\displaystyle m $is real. Can some one help me with this problem. Thank you in advance.
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  2. #2
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    I'm guessing the first term should be 1/x rather than another 1/y?

    Solve the second for x and substitute this expression into the other two equations. You'll be nearly done.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    I'm guessing the first term should be 1/x rather than another 1/y?

    Solve the second for x and substitute this expression into the other two equations. You'll be nearly done.
    I tried that and I am just not getting the problem solved at all.for the second I got $\displaystyle x=a-y$ but when I substitute that value for the other two eqaution every thing just seem confusing from there.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by scrible View Post
    I tried that and I am just not getting the problem solved at all.for the second I got $\displaystyle x=a-y$ but when I substitute that value for the other two eqaution every thing just seem confusing from there.
    Perhaps this is a bit sloppy (but then I tend to make problems a bit rougher than they need to be.) Start with the two equaitons:
    $\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
    and
    $\displaystyle x + y = a$

    Then we know that
    $\displaystyle \frac{x + y}{xy} = 1$

    Put the second equation in:
    $\displaystyle \frac{a}{x(a - x)} = 1$

    or
    $\displaystyle x^2 - ax + a = 0$

    Now start from x + y = a and y = mx.
    $\displaystyle x + mx = a$

    giving
    $\displaystyle x = \frac{a}{m + 1}$

    Put this into the $\displaystyle x^2 - ax + a = 0$ equation to cancel the x and you are done.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Perhaps this is a bit sloppy (but then I tend to make problems a bit rougher than they need to be.) Start with the two equaitons:
    $\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
    and
    $\displaystyle x + y = a$

    Then we know that
    $\displaystyle \frac{x + y}{xy} = 1$

    Put the second equation in:
    $\displaystyle \frac{a}{x(a - x)} = 1$

    or
    $\displaystyle x^2 - ax + a = 0$

    Now start from x + y = a and y = mx.
    $\displaystyle x + mx = a$

    giving
    $\displaystyle x = \frac{a}{m + 1}$

    Put this into the $\displaystyle x^2 - ax + a = 0$ equation to cancel the x and you are done.

    -Dan
    Thank you very much
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