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Solve-for-a-b-and-c-1-a-1-b-c-1-2-1-b-1-c-a-1-3-1-c-1-a-b-1-4-

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Question Number 202400 by MATHEMATICSAM last updated on 26/Dec/23
Solve for a, b and c (1/a) + (1/(b + c)) = (1/2) (1/b) + (1/(c + a)) = (1/3) (1/c) + (1/(a + b)) = (1/4)
$$\mathrm{Solve}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 11/Apr/24
solution see Q206294
$${solution}\:{see}\:{Q}\mathrm{206294} \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/23
{ (((1/a) + (1/(b + c)) = (1/2))),(((1/b) + (1/(c + a)) = (1/3) )),(((1/c) + (1/(a + b)) = (1/4))) :} { (( ((a+b+c)/(a(b + c))) = (1/2)⇒((a(b + c))/(a+b+c))=2)),(( ((a+b+c)/(b(c + a))) = (1/3)⇒((b(c+a))/(a+b+c))=3 )),(( ((a+b+c)/(c(a + b))) = (1/4)⇒((c(a+b))/(a+b+c))=4)) :} { (( a(b+c)=2(a+b+c)...(i))),(( b(c+a)=3(a+b+c)...(ii) )),(( c(a+b)=4(a+b+c)...(iii))) :} { (( 6a(b+c)=12(a+b+c)...(i))),(( 4b(c+a)=12(a+b+c)...(ii) )),(( 3c(a+b)=12(a+b+c)...(iii))) :} 12(a+b+c) = 6a(b+c) = 4b(c+a) =3c(a+b) (ii)−(i): b(c+a)−a(b + c)=a+b+c (iii)−(ii): c(a+b)−b(c+a)=a+b+c ab+bc−ab−ac=ac+bc−bc−ab bc−ac=ac−ab ab+bc−2ac=0 (1/a)−(2/b)+(1/c)=0 ((6a(b + c))/(a+b+c))=((4b(c+a))/(a+b+c))=((3c(a+b))/(a+b+c))=12 6a(b + c)=4b(c+a)=3c(a+b)=12(a+b+c) b+c=((2(a+b+c))/a) c+a=((3(a+b+c))/b) a+b=((4(a+b+c))/c) 2(a+b+c)=((2(a+b+c))/a)+((3(a+b+c))/b)+((4(a+b+c))/c) 2(a+b+c)−((2(a+b+c))/a)−((3(a+b+c))/b)−((4(a+b+c))/c)=0 (a+b+c)(2−(2/a)−(3/b)−(4/c))=0 a+b+c=0 ∨ 2−(2/a)−(3/b)−(4/c)=0 (2/a)+(3/b)+(4/c)=2 2abc−2bc−3ac−4ab=0 a+b+c=((a(b + c))/2)=((b(c + a))/3)=((c(a + b))/4) s=((a(s−a))/2)=((b(s−b))/3)=((c(s−c))/4) s^3 =((a(s−a))/2)∙((b(s−b))/3)∙((c(s−c))/4) =((abc)/(24))(s−a)(s−b)(s−c) ....
$$\begin{cases}{\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:}\\{\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}}\end{cases}\: \\ $$$$\begin{cases}{\:\frac{{a}+{b}+{c}}{{a}\left({b}\:+\:{c}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{{a}\left({b}\:+\:{c}\right)}{{a}+{b}+{c}}=\mathrm{2}}\\{\:\frac{{a}+{b}+{c}}{{b}\left({c}\:+\:{a}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{{b}\left({c}+{a}\right)}{{a}+{b}+{c}}=\mathrm{3}\:\:}\\{\:\frac{{a}+{b}+{c}}{{c}\left({a}\:+\:{b}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\frac{{c}\left({a}+{b}\right)}{{a}+{b}+{c}}=\mathrm{4}}\end{cases}\:\: \\ $$$$\begin{cases}{\:{a}\left({b}+{c}\right)=\mathrm{2}\left({a}+{b}+{c}\right)…\left({i}\right)}\\{\:{b}\left({c}+{a}\right)=\mathrm{3}\left({a}+{b}+{c}\right)…\left({ii}\right)\:\:\:\:}\\{\:{c}\left({a}+{b}\right)=\mathrm{4}\left({a}+{b}+{c}\right)…\left({iii}\right)}\end{cases}\:\:\: \\ $$$$\begin{cases}{\:\mathrm{6}{a}\left({b}+{c}\right)=\mathrm{12}\left({a}+{b}+{c}\right)…\left({i}\right)}\\{\:\mathrm{4}{b}\left({c}+{a}\right)=\mathrm{12}\left({a}+{b}+{c}\right)…\left({ii}\right)\:\:\:\:}\\{\:\mathrm{3}{c}\left({a}+{b}\right)=\mathrm{12}\left({a}+{b}+{c}\right)…\left({iii}\right)}\end{cases}\:\:\: \\ $$$$\mathrm{12}\left({a}+{b}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}{a}\left({b}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{b}\left({c}+{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}{c}\left({a}+{b}\right) \\ $$$$\left({ii}\right)−\left({i}\right):\:{b}\left({c}+{a}\right)−{a}\left({b}\:+\:{c}\right)={a}+{b}+{c} \\ $$$$\left({iii}\right)−\left({ii}\right):\:{c}\left({a}+{b}\right)−{b}\left({c}+{a}\right)={a}+{b}+{c} \\ $$$${ab}+{bc}−{ab}−{ac}={ac}+{bc}−{bc}−{ab} \\ $$$${bc}−{ac}={ac}−{ab} \\ $$$$\:{ab}+{bc}−\mathrm{2}{ac}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}}−\frac{\mathrm{2}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{0} \\ $$$$\:\frac{\mathrm{6}{a}\left({b}\:+\:{c}\right)}{{a}+{b}+{c}}=\frac{\mathrm{4}{b}\left({c}+{a}\right)}{{a}+{b}+{c}}=\frac{\mathrm{3}{c}\left({a}+{b}\right)}{{a}+{b}+{c}}=\mathrm{12} \\ $$$$\mathrm{6}{a}\left({b}\:+\:{c}\right)=\mathrm{4}{b}\left({c}+{a}\right)=\mathrm{3}{c}\left({a}+{b}\right)=\mathrm{12}\left({a}+{b}+{c}\right) \\ $$$${b}+{c}=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{{a}} \\ $$$${c}+{a}=\frac{\mathrm{3}\left({a}+{b}+{c}\right)}{{b}} \\ $$$${a}+{b}=\frac{\mathrm{4}\left({a}+{b}+{c}\right)}{{c}} \\ $$$$\mathrm{2}\left({a}+{b}+{c}\right)=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{{a}}+\frac{\mathrm{3}\left({a}+{b}+{c}\right)}{{b}}+\frac{\mathrm{4}\left({a}+{b}+{c}\right)}{{c}} \\ $$$$\mathrm{2}\left({a}+{b}+{c}\right)−\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{{a}}−\frac{\mathrm{3}\left({a}+{b}+{c}\right)}{{b}}−\frac{\mathrm{4}\left({a}+{b}+{c}\right)}{{c}}=\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)\left(\mathrm{2}−\frac{\mathrm{2}}{{a}}−\frac{\mathrm{3}}{{b}}−\frac{\mathrm{4}}{{c}}\right)=\mathrm{0} \\ $$$${a}+{b}+{c}=\mathrm{0}\:\vee\:\mathrm{2}−\frac{\mathrm{2}}{{a}}−\frac{\mathrm{3}}{{b}}−\frac{\mathrm{4}}{{c}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{{a}}+\frac{\mathrm{3}}{{b}}+\frac{\mathrm{4}}{{c}}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{abc}−\mathrm{2}{bc}−\mathrm{3}{ac}−\mathrm{4}{ab}=\mathrm{0} \\ $$$$\: \\ $$$$\: \\ $$$${a}+{b}+{c}=\frac{{a}\left({b}\:+\:{c}\right)}{\mathrm{2}}=\frac{{b}\left({c}\:+\:{a}\right)}{\mathrm{3}}=\frac{{c}\left({a}\:+\:{b}\right)}{\mathrm{4}} \\ $$$${s}=\frac{{a}\left({s}−{a}\right)}{\mathrm{2}}=\frac{{b}\left({s}−{b}\right)}{\mathrm{3}}=\frac{{c}\left({s}−{c}\right)}{\mathrm{4}} \\ $$$${s}^{\mathrm{3}} =\frac{{a}\left({s}−{a}\right)}{\mathrm{2}}\centerdot\frac{{b}\left({s}−{b}\right)}{\mathrm{3}}\centerdot\frac{{c}\left({s}−{c}\right)}{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{{abc}}{\mathrm{24}}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right) \\ $$$$…. \\ $$
Answered by Frix last updated on 26/Dec/23
Simply solve the 1^(st) for a, insert and solve the 2^(nd) for b, insert and solve the 3^(rd) for c a=((23)/(10)) b=((23)/6) c=((23)/2)
$$\mathrm{Simply}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{for}\:{a},\:\mathrm{insert}\:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:{b},\:\mathrm{insert}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{for}\:{c} \\ $$$${a}=\frac{\mathrm{23}}{\mathrm{10}}\:\:\:\:\:{b}=\frac{\mathrm{23}}{\mathrm{6}}\:\:\:\:\:{c}=\frac{\mathrm{23}}{\mathrm{2}} \\ $$

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