A problem in algebric geometry

[mathman99]

Homework Statement

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

Homework Equations

3x+y+z=8.

The Attempt at a Solution

 

[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.

 

[mathman99]

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.

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.

 

[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

 

[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