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Question Number 100976 by  M±th+et+s last updated on 29/Jun/20

find all possible values of x,y,z in terms of a,b,c gor a triplet (x,y,z) that satisfy x+(1/y)=a y+(1/z)=b z+(1/x)=c

$${find}\:{all}\:{possible}\:{values}\:{of}\:{x},{y},{z}\:{in}\:{terms} \\ $$$${of}\:{a},{b},{c}\:{gor}\:{a}\:{triplet}\:\left({x},{y},{z}\right)\:{that}\:{satisfy} \\ $$$$ \\ $$$${x}+\frac{\mathrm{1}}{{y}}={a} \\ $$$$ \\ $$$${y}+\frac{\mathrm{1}}{{z}}={b} \\ $$$$ \\ $$$${z}+\frac{\mathrm{1}}{{x}}={c} \\ $$

Commented by john santu last updated on 29/Jun/20

⇒x = ((ay−1)/y) ∧ y = ((bz−1)/z) ⇒ z+ (y/(ay−1)) = c ∧ y = ((bz−1)/z) ⇒ z + (((bz−1)/z)/(a(((bz−1)/z))−1)) = c z + ((bz−1)/(abz−a−z)) = c z + ((bz−1)/(z(ab−1)−a)) = c z(z(ab−1)−a)+bz−1 = c(z(ab−1)−a) (ab−1)z^2 +(b−a)z−1=(abc−c)z−ac (ab−1)z^2 +(b−a+c−abc)z+ac−1=0 z = ((abc+a−b−c ± (√((b−a+c−abc)^2 −4(ac−1)(ab−1))))/(2(ab−1)))

$$\Rightarrow{x}\:=\:\frac{{ay}−\mathrm{1}}{{y}}\:\wedge\:\mathrm{y}\:=\:\frac{{bz}−\mathrm{1}}{{z}} \\ $$$$\Rightarrow\:{z}+\:\frac{{y}}{{ay}−\mathrm{1}}\:=\:{c}\:\wedge\:\mathrm{y}\:=\:\frac{{bz}−\mathrm{1}}{{z}} \\ $$$$\Rightarrow\:{z}\:+\:\frac{\frac{{bz}−\mathrm{1}}{{z}}}{{a}\left(\frac{{bz}−\mathrm{1}}{{z}}\right)−\mathrm{1}}\:=\:{c}\: \\ $$$${z}\:+\:\frac{{bz}−\mathrm{1}}{{abz}−{a}−{z}}\:=\:{c}\: \\ $$$${z}\:+\:\frac{{bz}−\mathrm{1}}{{z}\left({ab}−\mathrm{1}\right)−{a}}\:=\:{c}\: \\ $$$${z}\left({z}\left({ab}−\mathrm{1}\right)−{a}\right)+{bz}−\mathrm{1}\:=\:{c}\left({z}\left({ab}−\mathrm{1}\right)−{a}\right) \\ $$$$\left({ab}−\mathrm{1}\right){z}^{\mathrm{2}} +\left({b}−{a}\right){z}−\mathrm{1}=\left({abc}−{c}\right){z}−{ac} \\ $$$$\left({ab}−\mathrm{1}\right){z}^{\mathrm{2}} +\left({b}−{a}+{c}−{abc}\right){z}+{ac}−\mathrm{1}=\mathrm{0} \\ $$$${z}\:=\:\frac{{abc}+{a}−{b}−{c}\:\pm\:\sqrt{\left({b}−{a}+{c}−{abc}\right)^{\mathrm{2}} −\mathrm{4}\left({ac}−\mathrm{1}\right)\left({ab}−\mathrm{1}\right)}}{\mathrm{2}\left({ab}−\mathrm{1}\right)} \\ $$

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