SOLUTION: (x+y)/(3a-b)=(y+z)/(3b-c)=(z+x)/(3c-a) prove that (x+y+z)/(a+b+c)=(ax+by+cz)/(a^2+b^2+c^2)

Question 481709: (x+y)/(3a-b)=(y+z)/(3b-c)=(z+x)/(3c-a)
prove that
(x+y+z)/(a+b+c)=(ax+by+cz)/(a^2+b^2+c^2)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
You have to apply the rule of identities.

a1/b1 = a2/b2 = a3/b3
= (a1+a2+a3)/(b1+b2+b3) ---- 1
= (a1-a2+a3)/(b1-b2+b3) ---- 2
= (a2-a3+a1)/(b2-b3+b1) ---- 3
= (a3-a1+a2)/(b3-b1+b2) ---- 4
..............

apply to the given equations:

= ----using (1)
=

= ---- by (2)
=
=
= ---- result A

= ---- by (3)
=
=
= ---- result B

= ---- by(4)
=
=
= ---- result C

= -------- by adding results A,B,C

=

m.ananth@hotmail.ca
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