A problem in algebric geometry

1. Jun 28, 2009

mathman99

1. The problem statement, all variables and given/known data

Hello,
I am trying to solve a problem in algebraic geometry where three unknown variables x,y and z needs to be determined to further progress in this problem.
The equations involving the variables x,y and z are 10y+28z-20=A;
18z-30x+60=A;-84x-18y+204=A.
Where A=3(x^2+y^2+z^2)=3(5x+5y+11z-20)
also 3x+y+z=8.

Could you suggest an effective method to find the possible values of x,y and z.

If you have any ideas, anything will be appreciated. Thanks

2. Relevant equations

3x+y+z=8.

3. The attempt at a solution

2. Jun 28, 2009

Dick

Treat A as another variable. Use the linear equations to eliminate as many variables as you can from the set {x,y,z,A}. Substitute the results into the quadratic.

3. Jun 29, 2009

mathman99

Thanks for the hint Dick. Could you give me some elaboration on this method. Its difficult to eliminate any variable from the set {x,y,z,A}. I always get into equation of one variable in terms of other two.
Or it will be helpful to give me some e-links where I can find similar problems (Topics) discussed.

4. Jun 29, 2009

Dick

I don't see what the problem is. Start with 3x+y+z=8, so z=8-3x-y. Put that into all the other linear equations. Boom. z eliminated and one less equation, at least. Your linear equations aren't independent, so you'll find some of them turning into the same thing. There's also matrix type methods.
http://en.wikipedia.org/wiki/System_of_linear_equations

5. Jun 30, 2009

Дьявол

As I tried to solve by using
10y+28z-20=18z-30x+60
18z-30x+60=-84x-18y+204
10y+28z-20=-84x-18y+204

All of the equation turn into 3x+y+z=8 since

30x+10y+10z=80 /: 10
54x+18y+18z=144 /: 18
84x+28y+28z=224 /: 28

or the same equation 3x+y+z=8

Now the only remaining equations are
3(x^2+y^2+z^2)=3(5x+5y+11z-20)

and 3x+y+z=8