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Solving a system of equations .

  1. Aug 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve simultaneously:
    6(x + y) = 5xy,
    21(y + z) = 10yz,
    14(z + x) = 9zx

    2. Relevant equations
    -


    3. The attempt at a solution
    Obviously one of the solution is (0,0,0) But I'm more interested in finding the other.
    Expanding these three equations , I get -
    6x+6y=5xy
    21y+21z=10yz
    14z+14x=9zx

    But what now? I've no experience solving these type of systems . Please give hints.
     
  2. jcsd
  3. Aug 8, 2013 #2

    pasmith

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    Multiply each equation by the unknown which doesn't appear in it so that the right hand side of each equation is then a multiple of [itex]xyz[/itex]. You can then eliminate [itex]xyz[/itex] in three ways to end up with three linear simultaneous equations for [itex]xy[/itex], [itex]xz[/itex] and [itex]yz[/itex].

    Having solved those, use the fact that [itex](xy)(xz)(yz) = (xyz)^2[/itex] to find [itex]x[/itex], [itex]y[/itex] and [itex]z[/itex] up to a sign.
     
  4. Aug 8, 2013 #3

    Ray Vickson

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    Even better: if p = xyz, one can express xy, xz and yz as numerical multiples of p (by solving linear equations with p in the right-hand-sides), so---if x, y and z are non-zero---one gets a unique solution for (x,y,z), with no sign ambiguities.

    Also: explore what must happen if one of the variables, say x, is zero.
     
  5. Aug 9, 2013 #4
    Thanks for the hints , but I used a different method to solve this equation. From eq. 1 I got y=(6x)/(5x-6) .
    I substituted that in eq. 2 to get x in terms of z. ( x=(126z)/(126+45z) ) then I substituted that into eq. 3 to get a quadratic equation in z with solutions z=0,7. From z=7 , I got x= 2 . Then I got y=3 from the relationship between x and y obtained earlier from eq. 1 . But I felt that the arithmetic involved was quite daunting and there must be a better way to solve it. I tried using the method that pasmith recommended but am feeling a bit lost. Can anyone please explain a simpler method to solve this system ?
     
  6. Aug 9, 2013 #5

    Ray Vickson

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    Let ##a = xy, \;b = yz, \; c = zx \:\text{ and }\: p = xyz.## Multiply the first equation by z, the second by x and the third by y to get
    [tex]6b+6c=5p\\
    21a+21c=10p \\
    14a+14b=9p[/tex]
    This is a simple 3x3 linear system which can be solved using grade-school methods, to get
    ## a = p/7,\: b = p/2, \: c = p/3.## Assuming ##x,y,z,\neq 0## we have
    [tex] xy = xyz/7 \Longrightarrow z = 7 \\
    yz = xyz/2 \Longrightarrow x = 2\\
    zx = xyz/3 \Longrightarrow y=3 [/tex]
     
  7. Aug 10, 2013 #6
    That's Awesome ! Thanks a lot !
     
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