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Question Number 215973 by MATHEMATICSAM last updated on 23/Jan/25
(a + b) ∝ c and (b + c) ∝ a. Prove that (c + a) ∝ b.
$$\left({a}\:+\:{b}\right)\:\propto\:{c}\:\mathrm{and}\:\left({b}\:+\:{c}\right)\:\propto\:{a}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left({c}\:+\:{a}\right)\:\propto\:{b}. \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jan/25
(a + b) ∝ c and (b + c) ∝ a. Prove that (c + a) ∝ b. ((a+b)/c)=k_1 ,((b+c)/a)=k_2 ((a+b)/c)+1=k_1 +1,((b+c)/a)+1=k_2 +1 ((a+b+c)/c)=k_1 +1,((a+b+c)/a)=k_2 +1 a+b+c=ck_1 +c=ak_2 +a c=((ak_2 +a)/(k_1 +1)) c+a=((ak_2 +a)/(k_1 +1))+a=((ak_2 +a+ak_1 +a)/(k_1 +1)) c+a=a(((k_1 +k_2 +2)/(k_1 +1))) ⇒c+a∝a ?? ....
$$\left({a}\:+\:{b}\right)\:\propto\:{c}\:\mathrm{and}\:\left({b}\:+\:{c}\right)\:\propto\:{a}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left({c}\:+\:{a}\right)\:\propto\:{b}. \\ $$$$\frac{{a}+{b}}{{c}}={k}_{\mathrm{1}} ,\frac{{b}+{c}}{{a}}={k}_{\mathrm{2}} \\ $$$$\frac{{a}+{b}}{{c}}+\mathrm{1}={k}_{\mathrm{1}} +\mathrm{1},\frac{{b}+{c}}{{a}}+\mathrm{1}={k}_{\mathrm{2}} +\mathrm{1} \\ $$$$\frac{{a}+{b}+{c}}{{c}}={k}_{\mathrm{1}} +\mathrm{1},\frac{{a}+{b}+{c}}{{a}}={k}_{\mathrm{2}} +\mathrm{1} \\ $$$${a}+{b}+{c}={ck}_{\mathrm{1}} +{c}={ak}_{\mathrm{2}} +{a} \\ $$$${c}=\frac{{ak}_{\mathrm{2}} +{a}}{{k}_{\mathrm{1}} +\mathrm{1}} \\ $$$${c}+{a}=\frac{{ak}_{\mathrm{2}} +{a}}{{k}_{\mathrm{1}} +\mathrm{1}}+{a}=\frac{{ak}_{\mathrm{2}} +{a}+{ak}_{\mathrm{1}} +{a}}{{k}_{\mathrm{1}} +\mathrm{1}} \\ $$$$\:\:\:\:\:{c}+{a}={a}\left(\frac{{k}_{\mathrm{1}} +{k}_{\mathrm{2}} +\mathrm{2}}{{k}_{\mathrm{1}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{c}+{a}\propto{a}\:\:?? \\ $$$$…. \\ $$
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)
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{try}\:\mathrm{by}\:\mathrm{cancelling}\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\mathrm{by} \\ $$$$\mathrm{dividing}\:\mathrm{and}\:\mathrm{making}\:\mathrm{a}\:=\:\mathrm{mc}\:\left(\mathrm{m}\:=\:\mathrm{const}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 24/Jan/25
a = mc⇒a∝c why should we assume a∝c ?
$$\mathrm{a}\:=\:\mathrm{mc}\Rightarrow\mathrm{a}\propto\mathrm{c} \\ $$$$\mathrm{why}\:\mathrm{should}\:\mathrm{we}\:\mathrm{assume}\:\mathrm{a}\propto\mathrm{c}\:? \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jan/25
a+b=pc.....(i) b+c=qa....(ii) let c+a=rb ; r is constant.....(iii) (ii)−(i): c−a=qa−pc....(iv) (iii)+(iv) : 2c=rb+qa−pc...(v) (iii)−(iv): 2a=rb−qa+pc....(vi) (v)+(vi): 2c+2a=2rb⇒c+a=rb⇒c+a∝b
$$\: \\ $$$${a}+{b}={pc}…..\left({i}\right) \\ $$$${b}+{c}={qa}….\left({ii}\right) \\ $$$${let}\:{c}+{a}={rb}\:;\:{r}\:{is}\:{constant}…..\left({iii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${c}−{a}={qa}−{pc}….\left({iv}\right) \\ $$$$\left({iii}\right)+\left({iv}\right)\:: \\ $$$$\mathrm{2}{c}={rb}+{qa}−{pc}…\left({v}\right) \\ $$$$\left({iii}\right)−\left({iv}\right): \\ $$$$\mathrm{2}{a}={rb}−{qa}+{pc}….\left({vi}\right) \\ $$$$\left({v}\right)+\left({vi}\right): \\ $$$$\mathrm{2}{c}+\mathrm{2}{a}=\mathrm{2}{rb}\Rightarrow{c}+{a}={rb}\Rightarrow{c}+{a}\propto{b} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Jan/25
Not sure that the answer is correct
$${Not}\:{sure}\:{that}\:{the}\:{answer}\:{is}\:{correct} \\ $$

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