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How would I work this out algebracally

  1. Jan 4, 2014 #1
    These 2 equations are part of a calculus problem:
    6z-2yz-z^2=0
    6y-y^2-2yz=0

    I have already done the problem, and I can see from some inspection that the solutions to these equations are z=0,y=0 z=6,y=0 z=0,y=6 and z=y=2

    My question is how can this be worked out in a formal algebraic way? Thanks for any help
     
  2. jcsd
  3. Jan 4, 2014 #2

    lurflurf

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    Homework Helper

    Such problems can be difficult in general. This one can be factored.

    6z-2yz-z^2=0
    6y-y^2-2yz=0

    z(6-2y-z)=0
    y(6-y-2z)=0

    We see there are four solutions and they are easy to work out. By zero product property

    z=0
    y=0

    z=0
    (6-y-2z)=0

    (6-2y-z)=0
    y=0

    (6-2y-z)=0
    (6-y-2z)=0
     
  4. Jan 4, 2014 #3
    Thanks,
    (6-y-2z)=0
    -2z=-6+y
    2z=6-y
    z=2.1 , y=1.8

    but that won't satisfy 6-2y-z=0...

    So it's a case of guesswork? These sort of equations come up a lot in the calculus I'm doing at the moment, in my textbook they never show how they solve these equations , they just say "y is 2", or whatever, without giving the algebra behind it.

    While I'm on this subject I had a similar problem recently that involved 2 quadratic curves, one in terms of x and the other in terms of y,

    Quick example: y=3x^2-20,x=y^2+x-10

    If anyone knows some clues or tricks for solving these I'd love to hear them, the only way I can solve them atm with my limited time and knowledge is Wolfram Alpha

    Cheers
     
  5. Jan 4, 2014 #4

    lurflurf

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    ^I don't follow your work.
    There is no guessing.
    You need to solve two equations at a times as in my above post.
    Maybe drawing the graph will help.
     
  6. Jan 4, 2014 #5
    Sorry the equations I gave were wrong, I'll post them another time :)
     
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