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Question Number 184472 by mr W last updated on 07/Jan/23

solve in R^3 { ((x+(1/y)=3)),((y+(1/z)=4)),((z+(1/x)=5)) :}

$${solve}\:{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{{x}+\frac{\mathrm{1}}{{y}}=\mathrm{3}}\\{{y}+\frac{\mathrm{1}}{{z}}=\mathrm{4}}\\{{z}+\frac{\mathrm{1}}{{x}}=\mathrm{5}}\end{cases} \\ $$

Answered by liuxinnan last updated on 07/Jan/23

actually R^3 =R z=5−(1/x) (1/z)=(x/(5x−1)) y+(x/(5x−1))=4 y=4−(x/(5x−1))=((19x−4)/(5x−1)) x+((19x−4)/(5x−1))=3 5x^2 −x+19x−4=15x−3 5x^2 +3x−1=0 x=((−3±(√(29)))/(10)) y=(1/(3−x))=(1/(3+((3∓(√(29)))/(10))))=((10)/(33∓(√(29)))) z=5−(1/x)=5+((10)/(3∓(√(29))))=((25∓5(√(29)))/(3∓(√(29))))

$${actually}\:\mathbb{R}^{\mathrm{3}} =\mathbb{R} \\ $$$${z}=\mathrm{5}−\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{{x}}{\mathrm{5}{x}−\mathrm{1}} \\ $$$${y}+\frac{{x}}{\mathrm{5}{x}−\mathrm{1}}=\mathrm{4} \\ $$$${y}=\mathrm{4}−\frac{{x}}{\mathrm{5}{x}−\mathrm{1}}=\frac{\mathrm{19}{x}−\mathrm{4}}{\mathrm{5}{x}−\mathrm{1}} \\ $$$${x}+\frac{\mathrm{19}{x}−\mathrm{4}}{\mathrm{5}{x}−\mathrm{1}}=\mathrm{3} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{19}{x}−\mathrm{4}=\mathrm{15}{x}−\mathrm{3} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{29}}}{\mathrm{10}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{3}−{x}}=\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{3}\mp\sqrt{\mathrm{29}}}{\mathrm{10}}}=\frac{\mathrm{10}}{\mathrm{33}\mp\sqrt{\mathrm{29}}} \\ $$$${z}=\mathrm{5}−\frac{\mathrm{1}}{{x}}=\mathrm{5}+\frac{\mathrm{10}}{\mathrm{3}\mp\sqrt{\mathrm{29}}}=\frac{\mathrm{25}\mp\mathrm{5}\sqrt{\mathrm{29}}}{\mathrm{3}\mp\sqrt{\mathrm{29}}} \\ $$

Commented by mr W last updated on 07/Jan/23

please recheck! x+((19x−4)/(5x−1))((5x−1)/(19x−4))=3

$${please}\:{recheck}! \\ $$$${x}+\cancel{\frac{\mathrm{19}{x}−\mathrm{4}}{\mathrm{5}{x}−\mathrm{1}}}\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{19}{x}−\mathrm{4}}=\mathrm{3} \\ $$

Commented by liuxinnan last updated on 08/Jan/23

please recheck! x+((19x−4)/(5x−1))((5x−1)/(19x−4))=3 I am sorry that

$${please}\:{recheck}! \\ $$$${x}+\cancel{\frac{\mathrm{19}{x}−\mathrm{4}}{\mathrm{5}{x}−\mathrm{1}}}\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{19}{x}−\mathrm{4}}=\mathrm{3} \\ $$$${I}\:{am}\:{sorry}\:{that} \\ $$

Commented by mr W last updated on 08/Jan/23

no problem! this happens even in the best families, as the Germans say.

$${no}\:{problem}!\:{this}\:{happens}\:{even}\:{in}\:{the}\: \\ $$$${best}\:{families},\:{as}\:{the}\:{Germans}\:{say}. \\ $$

Answered by SEKRET last updated on 07/Jan/23

×;+ { ((x+(1/y) =3)),((y+(1/z)=4)),((z+(1/x) = 5)) :} x+y+z+(1/x)+(1/y)+(1/z)+xyz+(1/(xyz)) = 60 x+y+z+(1/x)+(1/y)+(1/z)=12 xyz=a x=(a/(yz)) y=(a/(xz)) z= (a/(xy)) 12+a+(1/a)=60 a^2 +1=48a a_(1;2) =24∓5(√(23)) { (( (1/y)∙((a/z)+1)=3 →((z/(4z−1)))(((a+z)/z))=3)),(((1/z)∙((a/x)+1)=4 →((x/(5x−1)))∙(((a+x)/x))=4)),(((1/x) ((a/y)+1)=5→ ((y/(3y−1)))∙(((a+y)/y))=5)) :} { ((a+z=12z−3→ z= ((a+3)/(11)))),((a+x=20x−4 →x= ((a+4)/(19)))),((a+y=15y−5 → y= ((a+5)/(14)))) :} a_(1,2) =24±5(√(23)) {x;y;z}→{ ((28+5(√(23)))/(19)) ; ((29+5(√(23)))/(14)) ; ((27+5(√(323)))/(11)) } { x;y;z}→{((28−5(√(23)))/(19)) ; ((29−4(√(23)))/(14)) ; ((27−5(√(23)))/(11)) }

$$\:×;+\:\begin{cases}{\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\:=\mathrm{3}}\\{\boldsymbol{\mathrm{y}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}=\mathrm{4}}\\{\boldsymbol{\mathrm{z}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:=\:\mathrm{5}}\end{cases} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}+\boldsymbol{\mathrm{xyz}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{xyz}}}\:=\:\mathrm{60} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}=\mathrm{12} \\ $$$$\:\:\boldsymbol{\mathrm{xyz}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{yz}}}\:\:\:\:\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{xz}}}\:\:\:\:\boldsymbol{\mathrm{z}}=\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{xy}}} \\ $$$$\:\:\:\:\:\mathrm{12}+\boldsymbol{\mathrm{a}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}=\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{1}=\mathrm{48}\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{a}}_{\mathrm{1};\mathrm{2}} =\mathrm{24}\mp\mathrm{5}\sqrt{\mathrm{23}} \\ $$$$\:\:\:\begin{cases}{\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\centerdot\left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{z}}}+\mathrm{1}\right)=\mathrm{3}\:\:\rightarrow\left(\frac{\boldsymbol{\mathrm{z}}}{\mathrm{4}\boldsymbol{\mathrm{z}}−\mathrm{1}}\right)\left(\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{z}}}{\boldsymbol{\mathrm{z}}}\right)=\mathrm{3}}\\{\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}\centerdot\left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{x}}}+\mathrm{1}\right)=\mathrm{4}\:\rightarrow\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{5}\boldsymbol{\mathrm{x}}−\mathrm{1}}\right)\centerdot\left(\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}}\right)=\mathrm{4}}\\{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:\left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{y}}}+\mathrm{1}\right)=\mathrm{5}\rightarrow\:\left(\frac{\boldsymbol{\mathrm{y}}}{\mathrm{3}\boldsymbol{\mathrm{y}}−\mathrm{1}}\right)\centerdot\left(\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{y}}}\right)=\mathrm{5}}\end{cases} \\ $$$$\:\:\:\:\begin{cases}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{z}}=\mathrm{12}\boldsymbol{\mathrm{z}}−\mathrm{3}\rightarrow\:\:\boldsymbol{\mathrm{z}}=\:\frac{\boldsymbol{\mathrm{a}}+\mathrm{3}}{\mathrm{11}}}\\{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}=\mathrm{20}\boldsymbol{\mathrm{x}}−\mathrm{4}\:\rightarrow\boldsymbol{\mathrm{x}}=\:\frac{\boldsymbol{\mathrm{a}}+\mathrm{4}}{\mathrm{19}}}\\{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{y}}=\mathrm{15}\boldsymbol{\mathrm{y}}−\mathrm{5}\:\rightarrow\:\boldsymbol{\mathrm{y}}=\:\frac{\boldsymbol{\mathrm{a}}+\mathrm{5}}{\mathrm{14}}}\end{cases} \\ $$$$\:\:\:\boldsymbol{\mathrm{a}}_{\mathrm{1},\mathrm{2}} =\mathrm{24}\pm\mathrm{5}\sqrt{\mathrm{23}} \\ $$$$\:\:\left\{\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\right\}\rightarrow\left\{\:\frac{\mathrm{28}+\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{19}}\:\:\:;\:\:\frac{\mathrm{29}+\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{14}}\:\:;\:\frac{\mathrm{27}+\mathrm{5}\sqrt{\mathrm{323}}}{\mathrm{11}}\:\right\} \\ $$$$\left\{\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\right\}\rightarrow\left\{\frac{\mathrm{28}−\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{19}}\:\:;\:\frac{\mathrm{29}−\mathrm{4}\sqrt{\mathrm{23}}}{\mathrm{14}}\:;\:\frac{\mathrm{27}−\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{11}}\:\right\} \\ $$

Answered by mr W last updated on 07/Jan/23

general case: { ((x+(1/y)=a ...(i))),((y+(1/z)=b ...(ii))),((z+(1/x)=c ...(iii))) :} x+(1/y)+y+(1/z)+z+(1/x)=a+b+c (x+(1/y))(y+(1/z))(z+(1/x))=abc xyz+(1/(xyz))+x+(1/y)+y+(1/z)+z+(1/x)=abc xyz+(1/(xyz))+a+b+c=abc with λ=((abc−(a+b+c))/2) (xyz)^2 −2λ(xyz)+1=0 ⇒xyz=λ±(√(λ^2 −1)) z=c−(1/x)=((cx−1)/x) y=b−(1/z)=b−(x/(cx−1))=(((bc−1)x−b)/(cx−1)) xyz=x×(((bc−1)x−b)/(cx−1))×((cx−1)/x)=(bc−1)x−b ⇒x=((xyz+b)/(bc−1)) ⇒x=((λ±(√(λ^2 −1))+b)/(bc−1)) similarly ⇒y=((λ±(√(λ^2 −1))+c)/(ca−1)) ⇒z=((λ±(√(λ^2 −1))+a)/(ab−1)) such that a solution exists, λ≥1 i.e. abc≥a+b+c+2 in current case: a=3, b=4, c=5 λ=((3×4×5−(3+4+5))/2)=24 xyz=λ±(√(λ^2 −1))=24±5(√(23)) x=((24±5(√(23))+4)/(4×5−1))=((28±5(√(23)))/(19)) y=((24±5(√(23))+5)/(5×3−1))=((29±5(√(23)))/(14)) z=((24±5(√(23))+3)/(3×4−1))=((27±5(√(23)))/(11))

$${general}\:{case}: \\ $$$$\begin{cases}{{x}+\frac{\mathrm{1}}{{y}}={a}\:\:\:...\left({i}\right)}\\{{y}+\frac{\mathrm{1}}{{z}}={b}\:\:\:...\left({ii}\right)}\\{{z}+\frac{\mathrm{1}}{{x}}={c}\:\:\:...\left({iii}\right)}\end{cases} \\ $$$${x}+\frac{\mathrm{1}}{{y}}+{y}+\frac{\mathrm{1}}{{z}}+{z}+\frac{\mathrm{1}}{{x}}={a}+{b}+{c} \\ $$$$\left({x}+\frac{\mathrm{1}}{{y}}\right)\left({y}+\frac{\mathrm{1}}{{z}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)={abc} \\ $$$${xyz}+\frac{\mathrm{1}}{{xyz}}+{x}+\frac{\mathrm{1}}{{y}}+{y}+\frac{\mathrm{1}}{{z}}+{z}+\frac{\mathrm{1}}{{x}}={abc} \\ $$$${xyz}+\frac{\mathrm{1}}{{xyz}}+{a}+{b}+{c}={abc} \\ $$$${with}\:\lambda=\frac{{abc}−\left({a}+{b}+{c}\right)}{\mathrm{2}} \\ $$$$\left({xyz}\right)^{\mathrm{2}} −\mathrm{2}\lambda\left({xyz}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{xyz}=\lambda\pm\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}} \\ $$$$ \\ $$$${z}={c}−\frac{\mathrm{1}}{{x}}=\frac{{cx}−\mathrm{1}}{{x}} \\ $$$${y}={b}−\frac{\mathrm{1}}{{z}}={b}−\frac{{x}}{{cx}−\mathrm{1}}=\frac{\left({bc}−\mathrm{1}\right){x}−{b}}{{cx}−\mathrm{1}} \\ $$$${xyz}={x}×\frac{\left({bc}−\mathrm{1}\right){x}−{b}}{{cx}−\mathrm{1}}×\frac{{cx}−\mathrm{1}}{{x}}=\left({bc}−\mathrm{1}\right){x}−{b} \\ $$$$\Rightarrow{x}=\frac{{xyz}+{b}}{{bc}−\mathrm{1}} \\ $$$$\Rightarrow{x}=\frac{\lambda\pm\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}+{b}}{{bc}−\mathrm{1}} \\ $$$${similarly} \\ $$$$\Rightarrow{y}=\frac{\lambda\pm\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}+{c}}{{ca}−\mathrm{1}} \\ $$$$\Rightarrow{z}=\frac{\lambda\pm\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}+{a}}{{ab}−\mathrm{1}} \\ $$$${such}\:{that}\:{a}\:{solution}\:{exists},\:\lambda\geqslant\mathrm{1} \\ $$$${i}.{e}.\:{abc}\geqslant{a}+{b}+{c}+\mathrm{2} \\ $$$$ \\ $$$${in}\:{current}\:{case}: \\ $$$${a}=\mathrm{3},\:{b}=\mathrm{4},\:{c}=\mathrm{5} \\ $$$$\lambda=\frac{\mathrm{3}×\mathrm{4}×\mathrm{5}−\left(\mathrm{3}+\mathrm{4}+\mathrm{5}\right)}{\mathrm{2}}=\mathrm{24} \\ $$$${xyz}=\lambda\pm\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}=\mathrm{24}\pm\mathrm{5}\sqrt{\mathrm{23}} \\ $$$${x}=\frac{\mathrm{24}\pm\mathrm{5}\sqrt{\mathrm{23}}+\mathrm{4}}{\mathrm{4}×\mathrm{5}−\mathrm{1}}=\frac{\mathrm{28}\pm\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{19}} \\ $$$${y}=\frac{\mathrm{24}\pm\mathrm{5}\sqrt{\mathrm{23}}+\mathrm{5}}{\mathrm{5}×\mathrm{3}−\mathrm{1}}=\frac{\mathrm{29}\pm\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{14}} \\ $$$${z}=\frac{\mathrm{24}\pm\mathrm{5}\sqrt{\mathrm{23}}+\mathrm{3}}{\mathrm{3}×\mathrm{4}−\mathrm{1}}=\frac{\mathrm{27}\pm\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{11}} \\ $$

Commented by manxsol last updated on 07/Jan/23

to make it more didactic it would be good like this. Great journal Sir.W xyz+(1/(xyz))=abc−(a+b+c) (xyz).((1/(xyz)))=1 (xyz)^2 −[abc−(a+b+c]xyz+1 xyz=((abc−(a+b+c))/2) ±((√([abc−(a+b+c)]^2 −4))/2) with λ=((abc−(a+b+c))/2) (xyz)^2 −2λ(xyz)+1

$${to}\:{make}\:{it}\:{more}\:{didactic} \\ $$$$\:{it}\:{would}\:{be}\:{good}\:{like}\:{this}. \\ $$$${Great}\:{journal}\:{Sir}.{W} \\ $$$${xyz}+\frac{\mathrm{1}}{{xyz}}={abc}−\left({a}+{b}+{c}\right) \\ $$$$\left({xyz}\right).\left(\frac{\mathrm{1}}{{xyz}}\right)=\mathrm{1} \\ $$$$ \\ $$$$\left({xyz}\right)^{\mathrm{2}} −\left[{abc}−\left({a}+{b}+{c}\right]{xyz}+\mathrm{1}\right. \\ $$$${xyz}=\frac{{abc}−\left({a}+{b}+{c}\right)}{\mathrm{2}}\:\pm\frac{\sqrt{\left[{abc}−\left({a}+{b}+{c}\right)\right]^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${with}\:\:\lambda=\frac{{abc}−\left({a}+{b}+{c}\right)}{\mathrm{2}} \\ $$$$\left({xyz}\right)^{\mathrm{2}} −\mathrm{2}\lambda\left({xyz}\right)+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Answered by ajfour last updated on 08/Jan/23

let 4−y=t=(1/z) Now x((1/x))=1 ⇒ (3−(1/(4−t)))(5−(1/t))=1 ⇒ 14t(4−t)+1=5t+3(4−t) ⇒ 14t^2 −54t+11=0 t=((27)/(14))±(√((((27)/(14)))^2 −((154)/((14)^2 )))) =((27±5(√(23)))/(14)) y=4−t=((29±5(√(23)))/(14))

$${let}\:\:\mathrm{4}−{y}={t}=\frac{\mathrm{1}}{{z}} \\ $$$${Now}\:\:\:\:\:{x}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\:\:\:\Rightarrow \\ $$$$\:\:\:\:\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{4}−{t}}\right)\left(\mathrm{5}−\frac{\mathrm{1}}{{t}}\right)=\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{14}{t}\left(\mathrm{4}−{t}\right)+\mathrm{1}=\mathrm{5}{t}+\mathrm{3}\left(\mathrm{4}−{t}\right) \\ $$$$\Rightarrow\:\:\mathrm{14}{t}^{\mathrm{2}} −\mathrm{54}{t}+\mathrm{11}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{27}}{\mathrm{14}}\pm\sqrt{\left(\frac{\mathrm{27}}{\mathrm{14}}\right)^{\mathrm{2}} −\frac{\mathrm{154}}{\left(\mathrm{14}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{\mathrm{27}\pm\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{14}} \\ $$$${y}=\mathrm{4}−{t}=\frac{\mathrm{29}\pm\mathrm{5}\sqrt{\mathrm{23}}}{\mathrm{14}} \\ $$

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