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Solve for 3 variables with 2 equations

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the system of equations for x, y, and z.
    -------------------------
    [tex]4x^4+8y^4+z^4=9[/tex]
    [tex]x^8+y^8+z^8=1 [/tex]
    --------------------------

    2. Relevant equations
    3. The attempt at a solution
    Both equations can be rearranged and factored (difference of squares). I've tried it and it doesn't seem to help solve the problem. Any hints? Thanks :)
     
    Last edited: May 2, 2012
  2. jcsd
  3. May 2, 2012 #2

    tiny-tim

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    welcome to pf!

    hi staples! welcome to pf! :smile:

    (i assume that's x8, not xy?)

    have you tried the obvious substitution? :wink:
     
  4. May 2, 2012 #3

    Mark44

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    Is the exponent on x in the 2nd equation a typo? Just checking.
     
  5. May 2, 2012 #4
    Sorry about that. That was a typo. It's corrected now.
    tiny-tim: I don't see it :frown:
     
  6. May 2, 2012 #5

    tiny-tim

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    u = x4 … ?
     
  7. May 2, 2012 #6

    Mark44

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    And similar substitutions for y4 and z4.
     
  8. May 2, 2012 #7
    So I substituted x^4 = u, y^4 = v, z^4 = w.
    I still don't see how it would help. The only thing I could see to isolate one variable in the first equation and substitute that in to the second equation. That still leaves us with 2 variables. I don't understand how we can solve for 3 variables with 2 equations. I always thought if you have n variables you need at least n equations. Please explain. Thanks
     
  9. May 2, 2012 #8

    SammyS

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    What does each of the resulting equations describe?

    [itex]4u+8v+w=9[/itex]

    [itex]u^2+v^2+w^2=1[/itex]

    How would you describe their intersection, if they intersect?
     
  10. May 2, 2012 #9

    Mark44

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    If you have n variables you need at least n equations to get a unique solution (a single point). Since you have only two equations in three variables, there won't be a unique solution. This means that the solution set will be all the points along some curve, assuming the two equations represent surfaces that intersect.
     
  11. May 2, 2012 #10
    So like Marks says, the intersection will be infinite over a certain interval(s)? I don't understand what you mean by what they represent? 3 dimensional planes/surfaces? :uhh: Thanks.
     
  12. May 3, 2012 #11

    Mark44

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    In (u, v, w) space, the two equations are surfaces that might or might not intersect. If they intersect, they will do so at an infinite number of points - all the points along the curve of intersection.
     
  13. May 3, 2012 #12
    I think I get it now. Divide the first equation by 9 and compare it to the second equation. This shows that u,v,w must be equal to their coefficients because of the square in the second equation. So u=4/9, v=8/9, w=1/9. Now we take the (positive and negative) fourth root of these values. So there are 8 solutions (WolframAlpha: http://www.wolframalpha.com/input/?i=%284x^4%2B8y^4%2Bz^4%3D9%29+%28x^8%2By^8%2Bz^8%3D1%29)

    Is this correct? Can I just "compare" the two equations like I did? Thanks.

    P.S. Mark: I think you're under the impression that I'm doing higher level math courses. I am only in high school and we've never worked with 3d coordinate systems...
     
  14. May 3, 2012 #13

    tiny-tim

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    the first equation represents a plane, the second represents a sphere :smile:

    (so what will be the shape of the points where they intersect?)
     
  15. May 3, 2012 #14

    HallsofIvy

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    Normally, you cannot solve two equations for specific values of three unknown numbers but here you can. It is crucial to this problem that [itex]8^2+ 4^2+ 1^2= 64+ 16+ 1= 81= 9^2[/itex]
     
  16. May 3, 2012 #15
    I'm guessing the answer is a circle or ellipse depending on how they intersect. But I highly doubt it that the teacher wants an answer using the geometry of the equations. After all we've never learned the equation of a sphere. I'm pretty sure he's looking for an algebraic approach...

    I'm going to look in to what you said HallsofIvy. So I guess, my "solution" is wrong? A less subtle hint would be great :) I appreciate everyone's effort to help me. Thanks.
     
    Last edited: May 3, 2012
  17. May 3, 2012 #16

    tiny-tim

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    a plane always intersects a sphere in a circle :smile:
    yes, but knowledge of the geometry points you towards an easy algebraic approach :wink:

    try substituting a b c for u v w so that a = constant, a2 + b2 + c2 = 1 :smile:
     
  18. May 3, 2012 #17
    |a|,|b|,|c| ≤ 1 ?
     
  19. May 3, 2012 #18
    Can someone please help me with this question? Thanks.
     
  20. May 3, 2012 #19

    tiny-tim

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    on the intersection, yes

    conversion from u v and w axes to a b and c axes is just a rotation of the axes

    (eg if instead of the usual x and y axes you use p = (x+y)/√2 and q = (x-y)/√2, you've simply rotated the axes by 45°, and x2 + y2 = 1 becomes p2 + q2 = 1)

    here's another geometrical snippet that may help …

    4u + 8v + w = 9 is a plane whose normal is the direction (4,8,1) …

    if we call the endpoints of the diameter of the sphere parallel to (4,8,1) the north and south poles, then the solutions (in u,v,w coordinates) are a circle of latitude on the sphere :wink:
     
  21. May 3, 2012 #20

    SammyS

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    ... unless the plane is tangent to the sphere.
     
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