Find all ordered triples (x,y,z) of real numbers which satisfy:
(x+y)(x+y+z)=120
(y+z)(x+y+z)=96
(x+z)(x+y+z)=72
You are correct there are only two triples. I realize that was misleading, sorry.
When I types up the original the caps got changed. I will edit it.
Then $\displaystyle x=X/A,~y=Y/A,~z=Z/A$ where $\displaystyle A^2=X+Y+Z $.
Take a look at this.