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Math Help - Ordered Triples?

  1. #1
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    Ordered Triples?

    I was given a problem to solve this evening that seems to have lost my understanding.

    What are the ordered triples of real numbers (x,y,z) that satisfy

    (x+y)(x+y+z)=120
    (y+z)(x+y+z)=96
    (x+z)(x+y+z)=72


    Now I have no idea how to complete this problem or how it would be done. Sadly, an in-depth explanation would be useful if possible.

    Our educator said that there were two sets, and I was able to find one.

    (4,6,2) and this value was achieved without well knowledge of what I was doing.
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  2. #2
    Kep
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    I think that if you add the equations in pairs you will notice some interesting things
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  3. #3
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    72=9(8)=(-9)(-8)
    96=12(8)=(-12)(-8)
    120=15(8)=(-15)(-8)

    Hence one possible set of solutions gives equations..
    however, this set gives no solution

    x+y=15\ and\ -15
    z+y=12\ and\ -12
    x+z=9\ and\ -9

    easily solveable

    2nd set

    72=12(6)=-12(-6)
    96=12(8)=-12(-8)
    120=12(10)=-12(-10)
    Last edited by Archie Meade; January 5th 2010 at 03:43 PM.
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  4. #4
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    Ordered pair definition: (a,b) \equiv \left\{ {\{ a\} ,\{ a,b\} } \right\}.

    Ordered triple definition: (a,b,c) \equiv \left( {(a,b),c} \right).
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  5. #5
    Kep
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    Plato, although correct I don't see that that answers the question. We're hardly doing axiomatic set theory on this thread...
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    Unfortunately I am unable to do anything mostly because I lack the understanding of this even to the most basic. This is due to the fact that I am unable to understand my teacher's method of .. teaching!

    I was hoping that someone in this thread could give me an entirely in depth solution because it may be easier to comprehend.
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  7. #7
    Kep
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    Quote Originally Posted by jd3005 View Post
    I was hoping that someone in this thread could give me an entirely in depth solution because it may be easier to comprehend.
    I don't have time for a full worked method but here are the solutions according to Wolfram Alpha (easy to verify)

    (x+y)(x+y+z)=120, (y+z)(x+y+z)=96, (x+z)(x+y+z)=72 - Wolfram|Alpha
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  8. #8
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    Quote Originally Posted by jd3005 View Post
    unable to understand my teacher's method of .. teaching!
    You're not alone.

    Look at the factors and notice that (x+y+z) is common to all 3 products.
    x, y and z are certain values but x+y+z is the constant resulting from adding them together.

    Hence, we can examine 120, 96 and 72 for a common factor.

    120=2(60), 96=2(48), 72=2(36)

    The values in brackets are the sum of only 2 of the values, while the common one is (x+y+z).
    We may as well look for larger values for (x+y+z) then.
    120=4(30)=6(20)=8(15)=10(12).
    Do the same for 96 and 72 and work them out using the three that have common factors.
    I'm sorry if this is too short as I have to dash.
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  9. #9
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    Thank you for fulfilling my requests.
    Last edited by jd3005; January 5th 2010 at 04:02 PM.
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  10. #10
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    Quote Originally Posted by Kep View Post
    Plato, although correct I don't see that that answers the question. We're hardly doing axiomatic set theory on this thread...
    I had absolutely no intention of providing a complete answer to the question.
    Rather, I provided basic definitions that seemed to be asked for.
    There are some here who do simply provide complete, ready to handin, solutions.
    But I consider those as academic frauds.
    Is not the real purpose of a site such as this to aid in learning mathematics?
    If not then it is no more that a free brothel.
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  11. #11
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    Of course, I'm not to fond of complete solutions, either, but this one has one point worth mentioning...

    This way isn't too painful.

    From the first, notice that 1) (x+y+z) = \frac{120}{x+y}

    We already know x + y \ne 0, why?

    Substituting into the other two, and a little algebra, gives:

    2) \frac{y+z}{x+y} = \frac{96}{120}

    3) \frac{x+z}{x+y} = \frac{72}{120}

    Solving each for 'z' gives:

    4) z = \frac{120}{x+y} - (x+y)

    5) z = \frac{1}{5}(4x-y)

    6) z = \frac{1}{5}(3y-2x)

    This lends itself to three different equations in only x and y. Just pick two. Perhaps some redundancy will drop out. I chose 4 vs 6 and 5 vs 6 to produce:

    7) A Quadratic looking things that I'll get back to.

    8) A nice linear equation that simplifies to 2y = 3x

    Substituting this into the quadratic (after solving for y), gives a lovely expresion in x alone. This expression reduces to x^{2} = 16

    What say you? Can you take it from here?

    Really, this is just a bunch of algebra. The real trick was right up front. I resisted the temptation to solve for x, y, or z and used (x+y+z) as my first focus of substitution.

    Of course, this Brute Force algebra version entirely overlooks the far more elegant concept of common factors. On the other hand, Brute Force gave a more natural result that included both solutions.
    Last edited by TKHunny; January 5th 2010 at 04:15 PM.
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  12. #12
    Kep
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    Quote Originally Posted by Plato View Post
    I had absolutely no intention of providing a complete answer to the question.
    Rather, I provided basic definitions that seemed to be asked for.
    I think that it was clear that the questioner did not need a set-theoretic definition of an ordered triple. They just wanted help solving equations.

    Of course, I agree that helpful hints are far more pedagogical than complete solutions
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  13. #13
    Super Member bigwave's Avatar
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     <br />
(x+y)(x+y+z)=120<br />
     <br />
(y+z)(x+y+z)=96<br />
     <br />
(x+z)(x+y+z)=72<br />

    since (x+y+Z) is constant and we can set it = 12 (factor of 120,96,72)
    as suggested by archie meads

    then

    10(12) = 120
    8(12) = 96
    6(12) = 72

    and

    x + y = 10
    y + z = 8
    x + z = 6

    mult (-1) thru eq 2 and add

    x + y = 10
    - y - z = -8
    x + z = 6
    ---------------------------
    2x = 8

    substituting
    so x = 4 y = 6 and z = 2

    since (x+y+z) can also = -12

    we have also
    x = -4 y = -6 and z = -2
    Last edited by bigwave; January 5th 2010 at 04:14 PM. Reason: wording
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  14. #14
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    I understood after Archie Meade lead me to the common factors. But thank you immensely for your assistance. Bigwave, your solution differs from mine, as I added an extra step although they both seam able to work.
    Last edited by jd3005; January 5th 2010 at 04:19 PM.
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  15. #15
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    Quote Originally Posted by Plato View Post
    I had absolutely no intention of providing a complete answer to the question.
    Rather, I provided basic definitions that seemed to be asked for.
    There are some here who do simply provide complete, ready to handin, solutions.
    But I consider those as academic frauds.
    Is not the real purpose of a site such as this to aid in learning mathematics?
    If not then it is no more that a free brothel.
    The information you provided was meaningless in this context and you know it, stop being facecious.
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