Question

if , 5a-3b/a = 4a+b-2c/a+4b-2c = a+2b-3c/4a-4c then prove that 6a = 4b = 3c

Answer 1

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Answer 2

Answer:

(5a-3b)/a=(4a+b-2c)/(a+4b-2c)=(a+2b-3c)/(4a-4c)

=> {(5a-3b)*2}/a={(4a+b-2c)*2}/(a+4b-2c)={(a+2b-3c)*2}/(4a-4c)

=> (10a-6b)/a=(8a+2b-4c)/(a+4b-2c)=(2a+4b-6c)/(4a-4c)

=> [(10a-6b)/a]-1=[(8a+2b-4c)/(a+4b-2c)]-1=[(2a+4b-6c)/(4a-4c)-1

=>  (9a-6b)/a=(7a-2b-2c)/(a+4b-2c)=(-2a+4b-2c)/(4a-4c)

=>  (9a-6b)/a={(7a-2b-2c)*(-1)}/{(a+4b-2c)*(-1)}=(-2a+4b-2c)/(4a-4c)

=> (9a-6b)/a=(-7a+2b+2c)/(-a-4b+2c)=(-2a+4b-2c)/(4a-4c)=[(9a-6b)+(-7a+2b+2c)+(-2a+4b-2c)] / [a+(-a-4b+2c)+(4a-4c)]

=> (9a-6b)/a=(-7a+2b+2c)/(-a-4b+2c)=(-2a+4b-2c)/(4a-4c)=0

=> 9a-6b=0,  => 6a=4b and also from another ratio (-2a+4b-2c)/(4a-4c)=0 we get 2a= c i.e. 6a=3b , Hence 6a=4b=3c (proved)