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Question Number 77604 by jagoll last updated on 08/Jan/20
given (((a−b)(b−c)(c−a))/((a+b)(b+c)(c+a)))=−(1/(30)) what the value of (b/(a+b))+(c/(b+c))+(a/(c+a)) ?
$${given} \\ $$$$\frac{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}=−\frac{\mathrm{1}}{\mathrm{30}} \\ $$$${what}\:{the}\:{value}\:{of}\: \\ $$$$\frac{{b}}{{a}+{b}}+\frac{{c}}{{b}+{c}}+\frac{{a}}{{c}+{a}}\:? \\ $$
Answered by key of knowledge last updated on 08/Jan/20
not have constat answer. if:((a−b)/(a+b))=k (and (b/(a+b))=((1−k)/2)) ((b−c)/(b+c))=j ((c/(b+c))=((1−j)/2)) ⇒((c−a)/(c+a))=−((j+k)/(1+jk)) (((a−b)(b−c)(c−a))/((a+b)(b+c)(c+a)))=−(1/(30))⇒k=((j−30j^2 ∓(√(j(900j^3 −60j^2 +j+120)))/(60j)) (b/(a+b))+(c/(b+c))+(a/(c+a)) =?=(1/(30))(((91)/2)+(1/(kj)))
$$\mathrm{not}\:\mathrm{have}\:\mathrm{constat}\:\mathrm{answer}.\: \\ $$$$\mathrm{if}:\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\mathrm{k}\:\:\:\:\left(\mathrm{and}\:\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{1}−\mathrm{k}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{b}−\mathrm{c}}{\mathrm{b}+\mathrm{c}}=\mathrm{j}\:\:\:\left(\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{1}−\mathrm{j}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\mathrm{c}−\mathrm{a}}{\mathrm{c}+\mathrm{a}}=−\frac{\mathrm{j}+\mathrm{k}}{\mathrm{1}+\mathrm{jk}} \\ $$$$\frac{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}=−\frac{\mathrm{1}}{\mathrm{30}}\Rightarrow\mathrm{k}=\frac{\mathrm{j}−\mathrm{30j}^{\mathrm{2}} \mp\sqrt{\mathrm{j}\left(\mathrm{900j}^{\mathrm{3}} −\mathrm{60j}^{\mathrm{2}} +\mathrm{j}+\mathrm{120}\right.}}{\mathrm{60j}} \\ $$$$\frac{{b}}{{a}+{b}}+\frac{{c}}{{b}+{c}}+\frac{{a}}{{c}+{a}}\:=?=\frac{\mathrm{1}}{\mathrm{30}}\left(\frac{\mathrm{91}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{kj}}\right) \\ $$$$ \\ $$
Answered by john santu last updated on 09/Jan/20
Commented by mr W last updated on 09/Jan/20
there is no unique solution! what you did is only one possible value. using the same path as you we can also get for example 1−((2b)/(a+b))=−1 ⇒(b/(a+b))=1 1−((2c)/(b+c))=−(1/5) ⇒(c/(b+c))=(3/5) 1−((2a)/(c+a))=−(1/6) ⇒(c/(b+c))=(7/(12)) ⇒(b/(a+b))+(c/(b+c))+(a/(c+a))=1+(3/5)+(7/(12))=((131)/(60)) if a,b,c ∈R, there are infinite solutions. this is clear, since we have three variables (=unknowns), but only one condition (=equation), you can not get an unique value for the function.
$${there}\:{is}\:{no}\:{unique}\:{solution}! \\ $$$${what}\:{you}\:{did}\:{is}\:{only}\:{one}\:{possible}\:{value}. \\ $$$${using}\:{the}\:{same}\:{path}\:{as}\:{you}\:{we}\:{can} \\ $$$${also}\:{get}\:{for}\:{example} \\ $$$$\mathrm{1}−\frac{\mathrm{2}{b}}{{a}+{b}}=−\mathrm{1}\:\Rightarrow\frac{{b}}{{a}+{b}}=\mathrm{1} \\ $$$$\mathrm{1}−\frac{\mathrm{2}{c}}{{b}+{c}}=−\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\frac{{c}}{{b}+{c}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{1}−\frac{\mathrm{2}{a}}{{c}+{a}}=−\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow\frac{{c}}{{b}+{c}}=\frac{\mathrm{7}}{\mathrm{12}} \\ $$$$\Rightarrow\frac{{b}}{{a}+{b}}+\frac{{c}}{{b}+{c}}+\frac{{a}}{{c}+{a}}=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{7}}{\mathrm{12}}=\frac{\mathrm{131}}{\mathrm{60}} \\ $$$$ \\ $$$${if}\:{a},{b},{c}\:\in{R},\:{there}\:{are}\:{infinite}\:{solutions}. \\ $$$${this}\:{is}\:{clear},\:{since}\:{we}\:{have}\:{three} \\ $$$${variables}\:\left(={unknowns}\right),\:{but}\:{only}\:{one} \\ $$$${condition}\:\left(={equation}\right),\:{you}\:{can}\:{not} \\ $$$${get}\:{an}\:{unique}\:{value}\:{for}\:{the}\:{function}. \\ $$
Commented by john santu last updated on 09/Jan/20
yes, i agree with your opinion sir the answer i posted is not a single answer . considering the problem is not displayed, then i take one possible answer
$${yes},\:{i}\:{agree}\:{with}\:{your}\:{opinion}\:{sir} \\ $$$${the}\:{answer}\:{i}\:{posted}\:{is}\:{not}\:{a}\:{single}\: \\ $$$${answer}\:.\:{considering}\:{the}\:{problem}\:{is} \\ $$$${not}\:{displayed},\:{then}\:{i}\:{take}\:{one}\: \\ $$$${possible}\:{answer} \\ $$

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