Carcass wrote:
If a, b, and c are three different numbers and \(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\) , then what is the value of \(ax + by + cz\) ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Let x/(b-c) = y/(c-a) = z/(a-b) = k
=> x = kb - kc; y = kc - ka; z = ka - kb
=> ax + by + cz
= a(kb - kc) + b(kc - ka) + c(ka - kb)
= k[ab - ac + bc - ab + ac - bc]
= 0
Answer A