Hey darence.
Can you show you re-arrange each equation and show that the discriminant of each equation is the above? (i.e. discriminant = (b-1)^2 - 4ac)?
I think that it is posible to resolve by prooving that x=y=z (I dont know how to ) , and than will be (b-1)^2-4ac<0 (it is supposed by problem). So, equation(s) dont have real solutions.
Sorry for my bad english. I wish you understood my idea.
after a full day,my idea:
your equations:
summing them up we get
rearranging the terms,
or,
now by completing the squares we try to make the expression in the brackets a square.In general case:
, where and
using the same principles we get,
simplifying we get,
or,
dividing both sides by ,
now, (given) also therefore, and thus the right hand side of the equation is a negative number, while the left hand side is the sum of squares and should always be positive.
This is not possible for any real number solution, and hence we conclude no real solution exist.