(a + b) ∝ c and (b + c) ∝ a. Prove that (c + a) ∝ b.


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Question Number 215973 by MATHEMATICSAM last updated on 23/Jan/25

$(a + b) \propto c and (b + c) \propto a . Prove that$ $(c + a) \propto b .$
	Answered by Rasheed.Sindhi last updated on 24/Jan/25

	$(a + b) \propto c and (b + c) \propto a . Prove that$ $(c + a) \propto b .$ $\frac{a + b}{c} = k_{1}, \frac{b + c}{a} = k_{2}$ $\frac{a + b}{c} + 1 = k_{1} + 1, \frac{b + c}{a} + 1 = k_{2} + 1$ $\frac{a + b + c}{c} = k_{1} + 1, \frac{a + b + c}{a} = k_{2} + 1$ $a + b + c = {c k}_{1} + c = {a k}_{2} + a$ $c = \frac{{a k}_{2} + a}{k_{1} + 1}$ $c + a = \frac{{a k}_{2} + a}{k_{1} + 1} + a = \frac{{a k}_{2} + a + {a k}_{1} + a}{k_{1} + 1}$ $c + a = a (\frac{k_{1} + k_{2} + 2}{k_{1} + 1})$ $\Rightarrow c + a \propto a ? ?$ $. . . .$
	Commented by MATHEMATICSAM last updated on 24/Jan/25

	$Can you try by cancelling a + b + c by$ $dividing and making a = mc (m = const)$
	Commented by Rasheed.Sindhi last updated on 24/Jan/25

	$a = mc \Rightarrow a \propto c$ $why should we assume a \propto c ?$
	Answered by Rasheed.Sindhi last updated on 24/Jan/25

	$a + b = p c . . . . . (i)$ $b + c = q a . . . . (i i)$ $l e t c + a = r b; r i s c o n s t a n t . . . . . (i i i)$ $(i i) - (i) :$ $c - a = q a - p c . . . . (i v)$ $(i i i) + (i v) :$ $2 c = r b + q a - p c . . . (v)$ $(i i i) - (i v) :$ $2 a = r b - q a + p c . . . . (v i)$ $(v) + (v i) :$ $2 c + 2 a = 2 r b \Rightarrow c + a = r b \Rightarrow c + a \propto b$
	Commented by Rasheed.Sindhi last updated on 24/Jan/25

	$N o t s u r e t h a t t h e a n s w e r i s c o r r e c t$
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