This doesn't scroll vertically on the Blackberry browser. You have to sweep right to find the special scroll bar, click on that (hoping you don't miss), and then sweep back left to read the new text. A few iterations and I gave up. Maybe I'll read it later, maybe not, but the page is, for me, a usability disaster.
Designers - please consider using widely tested techniques and systems so you are at least aware of the people you exclude.
It's in the intersection of linear algebra and number theory, and knownm to be difficult. In effect it's a Diophantine equation, and most problems in that field have no generic techniques to solve them. I think Integer Programming is known to be NP-Hard, but I can't check that from here.
So no, there is no general technique apart from ingenuity, clever guesswork, and trial and error.
If you get there the puzzle is basically solved but for a bit of number crunching - you know n1, n2 and n3 are positive integers, n1 <= 10, so for  either n1=n2=n3 (trivial 0-post-lunch-price solution) or n1 - n2 = 3, n1 - n3 = 8. From there you substitute back into  and get 3 * (x-y) = 6y, so x = 3y.
You now have simple relations between n1, n2 and n3; as well as between x and y. The remaining step is to tie x or y to one of the n's by substituting into initial equations. For example from :
n1 * 2y + 10y = 35
2y = 35 / (n1 + 5)
y must be integral in pennies; that part is tricker to derive and the simplest solution is to just notice that since n1 >= 0, and n3 >= 0, and n1 = n3 + 8, then 8 <= n1 <= 10, and try 8, 9 and 10 for n1. With n1 = 9, you get y = 1.25. That's the only 'guess and check' part of the problem.
Quite often swathes of the search space can be eliminated or winnowed by using modulo arguments. You can also sometimes fold these problems into smaller spaces by considering equivalence arguments. There are many ad hoc techniques that no one has yet classified or unified, they are scattered across many mathematical disciplines.
It seems reasonable to require "real world" conditions. Non-zero prices, whole cent prices, integral chickens per sale, non-negative chickens per sale, etc. With those, the solution is unique, and in part, that's what makes this problem interesting, hard, surprising, and with its own elegance.