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Thread: solve for x,y,z

  1. #1
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    solve for x,y,z

    If x+y+z=19,xyz=144 then find the value of x,y,z

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  2. #2
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    Re: solve for x,y,z

    2, 8, 9 is one solution ... there may be more.
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    Re: solve for x,y,z

    2 more:

    -3, -2, 24
    3, 4, 12
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    Re: solve for x,y,z

    I need mathematical explanation ...

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    Re: solve for x,y,z

    Thanks for reply

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    Re: solve for x,y,z

    Quote Originally Posted by Sajnwaj View Post
    I need mathematical explanation ...
    If you want some sort of explanation, you need to be CLEARER.
    Like:
    are x,y,z integers? Greater than 0? x=<y=<z?

    Is this a classroom assignment?
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  7. #7
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    Re: solve for x,y,z

    X,y,z are all positive integers.

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  8. #8
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    Re: solve for x,y,z

    Quote Originally Posted by Sajnwaj View Post
    If x+y+z=19,xyz=144 then find the value of x,y,z

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    Okay. So from the first equation we get z = 19 - (x + y). Thus xyz = xy(19 - (x + y)) = 144. Or 19xy - xy(x + y) = 144.

    $-x^2 y + 19 xy - x y^2 = 144$

    What's your next step?

    -Dan
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    Re: solve for x,y,z

    I don't have any great insight to offer, though I wouldn't be surprised if there is one to be had. I can offer what I think is a systematic way to generate all possible solutions, assuming that x y and z are positive integers as stated. The first step is to factor 144 completely

    144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3

    Then generate the set of possible factors of 144. I did this "by the seat of my pants," but I think the list is complete.

    \left\{ {1,2,3,4,6,8,9,12,16,18,24,36,48,72,144} \right\}

    Obviously all numbers greater than 16 aren't useful for this problem. There are not 2 other integers you can add to 18, let alone 24, to get 19. That leaves

    \left\{ {1,2,3,4,6,8,9,12,16} \right\}

    Now suppose x = 1. I then write yz = 144 and
    \begin{array}{l} y + z = 18\\ z = 18 - y \end{array}

    From which I can get a quadratic equation

    \begin{array}{l} y\left( {18 - y} \right) = 144\\ {y^2} - 18y + 144 = 0 \end{array}

    That has no real roots. This means x \ne 1 and by symmetry neither does y or z; also, none of them x y or z can be 16 because then one of the others would have to be 1.

    I haven't followed this procedure out to the bitter end, but I'm satisfied that the two solutions already given, namely \left\{ {2,8,9} \right\} and \left\{ {3,4,12} \right\} are the only solutions that exist.

    BTW, it did occur to me to use combinations with repetition to find the number of integer solutions to x + y + z = 19, but I doubt that would've saved any time.
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  10. #10
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    Re: solve for x,y,z

    If z is the smallest of the three numbers then

    z^3\leq 144

    z\leq 5

    and since z must be a divisor of 144

    z\leq 4
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