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)

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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) About Me  (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:
%28x%2By%29%2F%283a-b%29+=+%28y%2Bz%29%2F%283b-c%29+=+%28z%2Bx%29%2F%283c-a%29
= %28x%2By%2By%2Bz%2Bz%2Bx%29+%2F+%283a-b%2B3b-c%2B3c-a%29 ----using (1)
= %28x%2By%2Bz%29%2F%28a%2Bb%2Bc%29

= %28x%2By-y-z%2Bz%2Bx%29+%2F+%283a-b-3b%2Bc%2B3c-a%29 ---- by (2)
= 2x+%2F+%282a+-+4b+%2B+4c%29
= x+%2F+%28a-2b%2B2c%29
= ax+%2F+%28a%5E2+-+2ab+%2B+2ca%29 ---- result A

= %28y%2Bz-z-x%2Bx%2By%29+%2F+%283b-c-3c%2Ba%2B3a-b%29 ---- by (3)
= 2y+%2F+%282b+-+4c+%2B+4a%29
= y+%2F+%28b-2c%2B2a%29
= by+%2F+%28b%5E2+-+2bc+%2B+2ab%29 ---- result B

= %28z%2Bx-x-y%2By%2Bz%29+%2F+%283c-a-3a%2Bb%2B3b-c%29 ---- by(4)
= 2z+%2F+%282c+-+4a+%2B+4b%29
= z+%2F+%28c-2a-2b%29
= cz+%2F+%28c%5E2+-+2ca+%2B+2bc%29 ---- result C

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

= %28ax+%2B+by+%2Bcz%29+%2F+%28a%5E2+%2B+b%5E2+%2B+c%5E2%29

m.ananth@hotmail.ca