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Math Help - System of Linear Equations?

  1. #1
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    System of Linear Equations?



    This is from my number theory class, so I'm assuming there is some sort of number theory solution here?

    Thanks.
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  2. #2
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    Hello, davismj!

    No, only algebra is needed . . .


    5) x,y,z are positive numbers such that:

    . . \begin{array}{cccccc}x + y + xy &=& 8 & [1] \\ y + z + yz &=& 15 & [2] \\ x + z + xz &=& 35 & [3] \end{array}

    Find the numbers.

    Solve [2] for z\!:\;\;z \:=\:\frac{15-y}{1 + y} \;\;[4]

    Solve [3] for z\!:\;\;z \;=\;\frac{35-x}{1+x}\;\;[5]

    Equate [4] and [5]: . \frac{15-y}{1+y} \:=\:\frac{35-x}{1+z} \quad\Rightarrow\quad y \:=\:\frac{4x-5}{9}\;\;[6]

    Substitute into [1]: . x + \frac{4x-5}{9} + x\left(\frac{4x-5}{9}\right) \:=\:8 \quad\Rightarrow\quad 4x^2 + 8x - 77 \:=\:0

    Hence:. . (2x-7)(2x+11) \:=\:0 \quad\Rightarrow\quad x \;=\;\frac{7}{2},\;-\frac{11}{2}

    Substitute into [6]: . y \;=\;1,\;-3

    Substitute into [5]: . z \;=\;7,\;-9


    There are two solutions: . (x,y,z) \;\;=\;\;\left(\frac{7}{2},\;1,\;7\right)\:\text{ and }\:\left(-\frac{11}{2},\;-3,\:\text{-}9\right)

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  3. #3
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    Wow. You have just made me feel terrible about myself for not getting that. I think I'm going to drop out of school now. Thanks, though.
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  4. #4
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    Quote Originally Posted by Soroban View Post

    Equate [4] and [5]: . \frac{15-y}{1+y} \:=\:\frac{35-x}{1+z} \quad\Rightarrow\quad y \:=\:\frac{4x-5}{9}\;\;[6]
    Hello, Soroban! Did you mean
    Equate [4] and [5]: . \frac{15-y}{1+y} \:=\:\frac{35-x}{1+x} \quad\Rightarrow\quad y \:=\:\frac{7x-10}{18}\;\;[6]
    Last edited by Intsecxtanx; March 2nd 2010 at 07:15 AM.
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  5. #5
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    Quote Originally Posted by Intsecxtanx View Post
    Hello, Soroban! Did you mean
    Equate [4] and [5]: . \frac{15-y}{1+y} \:=\:\frac{35-x}{1+x} \quad\Rightarrow\quad y \:=\:\frac{7x-10}{18}\;\;[6]
    Of course he did, patty pancakes.
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  6. #6
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    Quote Originally Posted by davismj View Post
    Of course he did, patty pancakes.
    Well, that renders the solution incorrect and means the values for x,y,z could be irrational. However, the method to go about solving is still very helpful.
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  7. #7
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    Quote Originally Posted by Intsecxtanx View Post
    Well, that renders the solution incorrect and means the values for x,y,z could be irrational. However, the method to go about solving is still very helpful.
    Sorry, I didn't read your numbers. I thought you were looking at his mistake in the denominator.

    No, his solution is correct, besides the typo.
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Another way to do this is to write

    <br /> <br /> <br />
\begin{array}{cccccc}x + y + xy = (1+x)(1+y)-1 &=& 8 &  \\ y + z<br />
+ yz  = (1+y)(1+z)-1 =&=& 15 &  \\ x + z + xz = (1+x)(1+z)-1 &=& 35 & <br />
\end{array}<br /> <br />

    and to substitute u,v,w for 1+x, 1+y, 1+z. The resulting system is easier to solve!
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