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Given, (x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a). Show that (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²).

Given, 
(x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a).
Show that (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²). 

Grade:10

3 Answers

Arun
25750 Points
4 years ago
Step-by-step explanation:
( x + y ) / ( 3a - b ) .........( 1 )
( y + z ) / ( 3b - c )  .........( 2 )
( z + x ) / ( 3c - a ) .........( 3 )
Add equation 1, 2 and 3
( 2x + 2y + 2z ) / (3a - b + 3b - c + 3c - a )
=  ( 2x + 2y + 2z ) / ( 2a + 2b + 2c )
= [ 2 ( x + y + z ) ] / [ 2 ( a + b + c )
= ( x+ y + z ) / ( a + b +c )
= ( x / a ) = ( y / b ) = ( z / c )
⇒ ax / a²  =  by / b² = cz / c²
= ( ax + by + cz ) / ( a² + b² + c² )
It proves ( x+ y + z ) / ( a + b +c ) = ( ax + by + cz ) / ( a² + b² + c² )
 
Sayantan Garai
117 Points
4 years ago
I cannot understand the following step:
 
" ( x+ y + z ) / ( a + b +c )
= ( x / a ) = ( y / b ) = ( z / c )"
 
Please show the full step without any jumps. 
Aditya Gupta
2081 Points
4 years ago
hello sayanta a much better way to solve is:
let (x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a) = k
by componendo and dividendo, we have
k= [(x+y) + (y+z) + (z+x)]/[(3a – b) + (3b – c) + (3c – a)]
or k= (x+y+z)/(a+b+c).......(*)
also x+y= k(3a – b)....(1)
y+z= k(3b – c)......(2)
and z+x= k(3c – a).......(3)
solving these 3 eqns we have
x= k(a+b+c) – k(3b – c)= k(a – 2b + 2c)
y= k(b – 2c + 2a)
and z= k(c – 2a + 2b)
so, ax+by+cz= k(a^2 – 2ab + 2ac + b^2 – 2bc + 2ab + c^2 – 2ac + 2bc)= k(a^2 + b^2 + c^2)
or (ax+by+cx) / (a²+b²+c²) = k.......(#)
from (*) and (#), we have
k= (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²). 
kindly approve :))

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