Re: difficult system of equation

Am 08.07.05 01:27 schrieb James Waldby:
> leox wrote:
>
>>How to find a real number solution of the system
>>4*x*y+y^2+2*z^2=-3,
>>4*x*z+x^2+2*z^2=1,
>>8*x*z+y^2+2*z^2=1
>
> and
>
>>yes, (-1,1,0) and (1,-1,0) are solutions but how to get it?
>
>
> Here is how I got the latter solution without pencil or computer -
> I misread that middle x as a y, and concluded that x*z = 0;
> that x*y < 0 from 1st equation, so z=0; that x=1 from 2nd
> equation (reading it correctly this time); and that y=-1 from
> eq. 1. Of course (1,-1,0) checks ok in the correctly-read
> equations...

I needed a pencil (well, a keyboard ;) ), and sometimes
it is already sufficient to write such a system in a
proper manner to see a first path to a solution easier:

A) 4*x*y +y^2 +2*z^2 = -3,
B) 4*x*z +x^2 +2*z^2 = 1,
C) 8*x*z +y^2 +2*z^2 = 1

-------- This already shows, things are simple:----------------

A)-C)
D) 4*x*y- 8*x*z = -4

C)-B)
E) 4*x*z -x^2 +y^2 = 0

-------------- only to get rid of z ----------------------------

1/2 D) + E)
F) 2*x*y -x^2 +y^2 = -2

....

Gottfried Helms
.

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