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Using lagrange i got something like $$3x = 4z = 6y$$

And the constraint is $$z^2 = x^2 + y^2$$

Where do you get from here?

I usually get $x=y=z$, but here i got $3$ variables with different values.

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    What's the system you're trying to solve? Since you mention Lagrange, it seems more likely that you're trying to optimize a function given a constraint.2012-11-15
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    To add up to the comment above, the problem as written now given these two simultaneous equations has a single trivial solution: $x=y=z=0$.2012-11-15
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    ah, actually it is -6z = -8x = 3y2012-11-15
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    is there a way to solve this, like a method? or do you think i've made a mistake somehow?2012-11-15
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    You haven't given the actual problem you are trying to solve. But there is only only solution to the equations you gave above, as is mentioned about and below.2012-11-15
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    4x-3y+8z=5 given the constraint z^2 = x^2 + y^22012-11-15
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    What do you mean by that? Both equations have the form of constraints. Are you minimizing or maximizing something?2012-11-15
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    i need to find the extreme values of f(x) = 4z-3y+8z=52012-11-15
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    Your next-to-last comment should be part of the question-it is what you (seem to) really want. But neither one is the function you want to maximize, they are both constraints (see the equal sign).2012-11-16

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