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Math Help - System of equations

  1. #1
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    System of equations

    Hi,

    My question is how do you solve analitically:

    sqrt(x)+y=7 and sqrt(y)+x=11 without solving a 4th degree polynomial equation?

    Thanks
    Last edited by kolg; December 26th 2009 at 02:19 PM.
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  2. #2
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    Quote Originally Posted by kolg View Post
    Hi,

    My question is how do you solve:

    sqrt(x)+y=7 and sqrt(y)+x=11 without solving a 4th degree polynomial equation?

    Thanks
    graphing ...
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  3. #3
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    Dear kolg,

    \sqrt{x}+y=11------------(1)

    \sqrt{y}+x=7-------------(2)

    From (1); y<11 and x\geq{0}

    From (2); x<7 and y\geq{0}

    Therefore, 0\leq{y}<11 and 0\leq{x}<7

    Also from (1) and (2) it could be seen that, \sqrt{x}\in{N} and \sqrt{y}\in{N}

    Therefore x=4 is the only solution for x.

    When, x=4\Rightarrow{y=9}

    Therefore x=4, y=9 is the only solution for this system of equations.
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  4. #4
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    Hi

    Thanks for the reply but how do you know that srqrt(x) and sqrt(y) are integers? Can't they be racionals or irracionals?

    Thanks
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  5. #5
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    Dear Klog,

    If \sqrt{x} is rational but not an integer,

    \sqrt{x}=\frac{p}{q} where q\neq{1}

    x=\frac{p^2}{q^2}

    Therefore, \sqrt{y}=7-\frac{p^2}{q^2}=(\sqrt{7}-\frac{p}{q})(\sqrt{7}-\frac{p}{q})

    Hence \sqrt{y} is irrational.

    Then x+\sqrt{y} is irrational. (Contradiction)

    Therefore \sqrt{x} is irrational or an integer.(according to our assumption.)

    But if \sqrt{x} is irrational y+\sqrt{x} is irrational. (Contradiction)

    So \sqrt{x} is an integer.

    Same proof goes to \sqrt{y} as well.

    Hope this helps.
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