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Question Number 202540 by MATHEMATICSAM last updated on 29/Dec/23

If (b/(a + b)) = ((3a − b − c)/(2b + c − a)) = ((3c − a)/(2a − b + 3c)) [a + b + c ≠ 0] then show that a = b = c.

Ifba+b=3a−b−c2b+c−a=3c−a2a−b+3c[a+b+c≠0]thenshowthata=b=c.

Answered by som(math1967) last updated on 29/Dec/23

(b/(a+b))=((3a−b−c)/(2b+c−a))=((3c−a)/(2a−b+3c)) ((3b)/(3a+3b))=((3a−b−c)/(2b+c−a))=((3c−a)/(2a−b+3c)) ∴ each ratio =((3b+3a−b−c+3c−a)/(3a+3b+2b+c−a+2a−b+3c)) =((2(a+b+c))/(4(a+b+c)))=(1/2) [(a+b+c)≠0] ∴(b/(a+b))=(1/2)⇒2b=a+b∴a=b again ((3c−a)/(2a−b+3c))=(1/2) ⇒6c−2a=2a−a+3c [∵ a=b] ⇒3c=3a ∴c=a ∴a=b=c

ba+b=3a−b−c2b+c−a=3c−a2a−b+3c3b3a+3b=3a−b−c2b+c−a=3c−a2a−b+3c∴eachratio=3b+3a−b−c+3c−a3a+3b+2b+c−a+2a−b+3c=2(a+b+c)4(a+b+c)=12[(a+b+c)≠0]∴ba+b=12⇒2b=a+b∴a=bagain3c−a2a−b+3c=12⇒6c−2a=2a−a+3c[∵a=b]⇒3c=3a∴c=a∴a=b=c

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