Menu Close

Solve-the-system-a-b-1-c-1-2-1-c-b-1-a-1-3-1-a-c-1-b-1-4-1-

Tinku Tara 0 Comments
Pin




Question Number 206294 by sniper237 last updated on 11/Apr/24
Solve the system (a+b)^(−1) +c^(−1) =2^(−1) (c+b)^(−1) +a^(−1) =3^(−1) (a+c)^(−1) +b^(−1) =4^(−1)
$${Solve}\:{the}\:{system} \\ $$$$\left({a}+{b}\right)^{−\mathrm{1}} +{c}^{−\mathrm{1}} =\mathrm{2}^{−\mathrm{1}} \\ $$$$\left({c}+{b}\right)^{−\mathrm{1}} +{a}^{−\mathrm{1}} =\mathrm{3}^{−\mathrm{1}} \\ $$$$\left({a}+{c}\right)^{−\mathrm{1}} +{b}^{−\mathrm{1}} =\mathrm{4}^{−\mathrm{1}} \\ $$
Commented by MATHEMATICSAM last updated on 11/Apr/24
Q 202400 This quetion is like this one.
$$\mathrm{Q}\:\mathrm{202400}\:\mathrm{This}\:\mathrm{quetion}\:\mathrm{is}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}. \\ $$
Commented by sniper237 last updated on 11/Apr/24
Thanks
$${Thanks} \\ $$
Answered by mr W last updated on 11/Apr/24
(1/(a+b))+(1/c)=(1/2) ⇒2(a+b+c)=ca+bc ...(i) (1/(b+c))+(1/a)=(1/3) ⇒3(a+b+c)=ab+ca ...(ii) (1/(c+a))+(1/b)=(1/4) ⇒4(a+b+c)=bc+ab ...(iii) (i)+(ii)+(iii): 4.5(a+b+c)=ab+bc+ca say k=a+b+c ⇒ab=2.5(a+b+c)=2.5k ⇒bc=1.5(a+b+c)=1.5k ⇒ca=0.5(a+b+c)=0.5k (abc)^2 =2.5×1.5×0.5k^3 =((15k^3 )/8) ⇒abc=(√((15k^3 )/8))=((k(√(30k)))/4) (“−” rejected) ⇒c=((k(√(30k)))/(4×2.5k))=((√(30k))/(10)) ⇒a=((k(√(30k)))/(4×1.5k))=((√(30k))/6) ⇒b=((k(√(30k)))/(4×0.5k))=((√(30k))/2) a+b+c=((1/(10))+(1/6)+(1/2))(√(30k))=k ⇒k=((23^2 )/(30)) ⇒(√(30k))=23 ⇒a=((23)/6), b=((23)/2), c=((23)/(10)) ✓
$$\frac{\mathrm{1}}{{a}+{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{2}\left({a}+{b}+{c}\right)={ca}+{bc}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{3}\left({a}+{b}+{c}\right)={ab}+{ca}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{4}\left({a}+{b}+{c}\right)={bc}+{ab}\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\mathrm{4}.\mathrm{5}\left({a}+{b}+{c}\right)={ab}+{bc}+{ca} \\ $$$${say}\:{k}={a}+{b}+{c} \\ $$$$\Rightarrow{ab}=\mathrm{2}.\mathrm{5}\left({a}+{b}+{c}\right)=\mathrm{2}.\mathrm{5}{k} \\ $$$$\Rightarrow{bc}=\mathrm{1}.\mathrm{5}\left({a}+{b}+{c}\right)=\mathrm{1}.\mathrm{5}{k} \\ $$$$\Rightarrow{ca}=\mathrm{0}.\mathrm{5}\left({a}+{b}+{c}\right)=\mathrm{0}.\mathrm{5}{k} \\ $$$$\left({abc}\right)^{\mathrm{2}} =\mathrm{2}.\mathrm{5}×\mathrm{1}.\mathrm{5}×\mathrm{0}.\mathrm{5}{k}^{\mathrm{3}} =\frac{\mathrm{15}{k}^{\mathrm{3}} }{\mathrm{8}} \\ $$$$\Rightarrow{abc}=\sqrt{\frac{\mathrm{15}{k}^{\mathrm{3}} }{\mathrm{8}}}=\frac{{k}\sqrt{\mathrm{30}{k}}}{\mathrm{4}}\:\:\:\left(“−''\:{rejected}\right) \\ $$$$\Rightarrow{c}=\frac{{k}\sqrt{\mathrm{30}{k}}}{\mathrm{4}×\mathrm{2}.\mathrm{5}{k}}=\frac{\sqrt{\mathrm{30}{k}}}{\mathrm{10}} \\ $$$$\Rightarrow{a}=\frac{{k}\sqrt{\mathrm{30}{k}}}{\mathrm{4}×\mathrm{1}.\mathrm{5}{k}}=\frac{\sqrt{\mathrm{30}{k}}}{\mathrm{6}} \\ $$$$\Rightarrow{b}=\frac{{k}\sqrt{\mathrm{30}{k}}}{\mathrm{4}×\mathrm{0}.\mathrm{5}{k}}=\frac{\sqrt{\mathrm{30}{k}}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{30}{k}}={k} \\ $$$$\Rightarrow{k}=\frac{\mathrm{23}^{\mathrm{2}} }{\mathrm{30}}\:\Rightarrow\sqrt{\mathrm{30}{k}}=\mathrm{23} \\ $$$$\Rightarrow{a}=\frac{\mathrm{23}}{\mathrm{6}},\:{b}=\frac{\mathrm{23}}{\mathrm{2}},\:{c}=\frac{\mathrm{23}}{\mathrm{10}}\:\:\checkmark \\ $$
Answered by MATHEMATICSAM last updated on 12/Apr/24
(1/(a + b)) + (1/c) = (1/2) ⇒ ((c + a + b)/(ac + bc)) = (1/2) ⇒ ((ac + bc)/(a + b + c)) = 2 .... (1) From the other two equations same way we can get ((ac + ab)/(a + b + c)) = 3 .... (2) ((ab + bc)/(a + b + c)) = 4 .... (3) (1) + (2) + (3) ((2(ab + bc + ca) )/(a + b + c)) = 9 ⇒ ((ab + bc + ca)/(a + b + c)) = 4.5 .... (4) (4) − (1) ((ab )/(a + b + c)) = 2.5 .... (5) (4) − (2) ((bc)/(a + b + c)) = 1.5 .... (6) (4) − (3) ((ca)/(a + b + c)) = 0.5 .... (7) (5) ÷ (6) ((ab)/(bc)) = ((2.5)/(1.5)) ⇒ (a/c) = (5/3) ⇒ c = ((3a)/5) (6) ÷ (7) ((bc)/(ca)) = ((1.5)/(0.5)) ⇒ (b/a) = 3 ⇒ b = 3a ((ab)/(a + b + c)) = 2.5 = (5/2) ⇒ ((a × 3a)/(a + 3a + ((3a)/5))) = (5/2) ⇒ ((3a^2 )/((23a)/5)) = (5/2) ⇒ ((15a)/(23)) = (5/2) ⇒ a = (5/2) × ((23)/(15)) = ((23)/6) b = 3a = 3 × ((23)/6) = ((23)/2) c = ((3a)/5) = (3/5) × ((23)/6) = ((23)/(10))
$$\frac{\mathrm{1}}{{a}\:+\:{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{{c}\:+\:{a}\:+\:{b}}{{ac}\:+\:{bc}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{{ac}\:+\:{bc}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{2}\:….\:\left(\mathrm{1}\right) \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{equations}\:\mathrm{same}\: \\ $$$$\mathrm{way}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get} \\ $$$$\frac{{ac}\:+\:{ab}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{3}\:….\:\left(\mathrm{2}\right) \\ $$$$\frac{{ab}\:+\:{bc}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{4}\:….\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)\:+\:\left(\mathrm{2}\right)\:+\:\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{2}\left({ab}\:+\:{bc}\:+\:{ca}\right)\:}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{9} \\ $$$$\Rightarrow\:\frac{{ab}\:+\:{bc}\:+\:{ca}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{4}.\mathrm{5}\:….\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{4}\right)\:−\:\left(\mathrm{1}\right) \\ $$$$\frac{{ab}\:}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{2}.\mathrm{5}\:….\:\left(\mathrm{5}\right)\: \\ $$$$\left(\mathrm{4}\right)\:−\:\left(\mathrm{2}\right) \\ $$$$\frac{{bc}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{1}.\mathrm{5}\:….\:\left(\mathrm{6}\right) \\ $$$$\left(\mathrm{4}\right)\:−\:\left(\mathrm{3}\right) \\ $$$$\frac{{ca}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{0}.\mathrm{5}\:….\:\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{5}\right)\:\boldsymbol{\div}\:\left(\mathrm{6}\right) \\ $$$$\frac{{ab}}{{bc}}\:=\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{1}.\mathrm{5}}\:\Rightarrow\:\frac{{a}}{{c}}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\:\Rightarrow\:{c}\:=\:\frac{\mathrm{3}{a}}{\mathrm{5}} \\ $$$$\left(\mathrm{6}\right)\:\boldsymbol{\div}\:\left(\mathrm{7}\right) \\ $$$$\frac{{bc}}{{ca}}\:=\:\frac{\mathrm{1}.\mathrm{5}}{\mathrm{0}.\mathrm{5}}\:\Rightarrow\:\frac{{b}}{{a}}\:=\:\mathrm{3}\:\Rightarrow\:{b}\:=\:\mathrm{3}{a} \\ $$$$\frac{{ab}}{{a}\:+\:{b}\:+\:{c}}\:=\:\mathrm{2}.\mathrm{5}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{{a}\:×\:\mathrm{3}{a}}{{a}\:+\:\mathrm{3}{a}\:+\:\frac{\mathrm{3}{a}}{\mathrm{5}}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{a}^{\mathrm{2}} }{\frac{\mathrm{23}{a}}{\mathrm{5}}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{15}{a}}{\mathrm{23}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\:×\:\frac{\mathrm{23}}{\mathrm{15}}\:=\:\frac{\mathrm{23}}{\mathrm{6}} \\ $$$${b}\:=\:\mathrm{3}{a}\:=\:\mathrm{3}\:×\:\frac{\mathrm{23}}{\mathrm{6}}\:=\:\frac{\mathrm{23}}{\mathrm{2}} \\ $$$${c}\:=\:\frac{\mathrm{3}{a}}{\mathrm{5}}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:×\:\frac{\mathrm{23}}{\mathrm{6}}\:=\:\frac{\mathrm{23}}{\mathrm{10}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *