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


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Question Number 184472 by mr W last updated on 07/Jan/23
$solve in R^3 { ((x+(1/y)=3)),((y+(1/z)=4)),((z+(1/x)=5)) :}$
$s o l v e i n R^{3}$ ${\begin{cases} x + \frac{1}{y} = 3 \\ y + \frac{1}{z} = 4 \\ z + \frac{1}{x} = 5 \end{cases}$
	Answered by liuxinnan last updated on 07/Jan/23

	$a c t u a l l y R^{3} = R$ $z = 5 - \frac{1}{x}$ $\frac{1}{z} = \frac{x}{5 x - 1}$ $y + \frac{x}{5 x - 1} = 4$ $y = 4 - \frac{x}{5 x - 1} = \frac{19 x - 4}{5 x - 1}$ $x + \frac{19 x - 4}{5 x - 1} = 3$ $5 x^{2} - x + 19 x - 4 = 15 x - 3$ $5 x^{2} + 3 x - 1 = 0$ $x = \frac{- 3 \pm \sqrt{29}}{10}$ $y = \frac{1}{3 - x} = \frac{1}{3 + \frac{3 \mp \sqrt{29}}{10}} = \frac{10}{33 \mp \sqrt{29}}$ $z = 5 - \frac{1}{x} = 5 + \frac{10}{3 \mp \sqrt{29}} = \frac{25 \mp 5 \sqrt{29}}{3 \mp \sqrt{29}}$
	Commented by mr W last updated on 07/Jan/23

	$p l e a s e r e c h e c k!$ $x + \frac{19 x - 4}{5 x - 1} \frac{5 x - 1}{19 x - 4} = 3$
	Commented by liuxinnan last updated on 08/Jan/23

	$p l e a s e r e c h e c k!$ $x + \frac{19 x - 4}{5 x - 1} \frac{5 x - 1}{19 x - 4} = 3$ $I a m s o r r y t h a t$
	Commented by mr W last updated on 08/Jan/23

	$n o p r o b l e m! t h i s h a p p e n s e v e n i n t h e$ $b e s t f a m i l i e s, a s t h e G e r m a n s s a y .$
	Answered by SEKRET last updated on 07/Jan/23
	$×;+ { ((x+(1/y) =3)),((y+(1/z)=4)),((z+(1/x) = 5)) :} x+y+z+(1/x)+(1/y)+(1/z)+xyz+(1/(xyz)) = 60 x+y+z+(1/x)+(1/y)+(1/z)=12 xyz=a x=(a/(yz)) y=(a/(xz)) z= (a/(xy)) 12+a+(1/a)=60 a^2 +1=48a a_(1;2) =24∓5(√(23)) { (( (1/y)∙((a/z)+1)=3 →((z/(4z−1)))(((a+z)/z))=3)),(((1/z)∙((a/x)+1)=4 →((x/(5x−1)))∙(((a+x)/x))=4)),(((1/x) ((a/y)+1)=5→ ((y/(3y−1)))∙(((a+y)/y))=5)) :} { ((a+z=12z−3→ z= ((a+3)/(11)))),((a+x=20x−4 →x= ((a+4)/(19)))),((a+y=15y−5 → y= ((a+5)/(14)))) :} a_(1,2) =24±5(√(23)) {x;y;z}→{ ((28+5(√(23)))/(19)) ; ((29+5(√(23)))/(14)) ; ((27+5(√(323)))/(11)) } { x;y;z}→{((28−5(√(23)))/(19)) ; ((29−4(√(23)))/(14)) ; ((27−5(√(23)))/(11)) }$
	$\times; + {\begin{cases} x + \frac{1}{y} = 3 \\ y + \frac{1}{z} = 4 \\ z + \frac{1}{x} = 5 \end{cases}$ $x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + xyz + \frac{1}{xyz} = 60$ $x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 12$ $xyz = a x = \frac{a}{yz} y = \frac{a}{xz} z = \frac{a}{xy}$ $12 + a + \frac{1}{a} = 60$ $a^{2} + 1 = 48 a a_{1; 2} = 24 \mp 5 \sqrt{23}$ ${\begin{cases} \frac{1}{y} \cdot (\frac{a}{z} + 1) = 3 \to (\frac{z}{4 z - 1}) (\frac{a + z}{z}) = 3 \\ \frac{1}{z} \cdot (\frac{a}{x} + 1) = 4 \to (\frac{x}{5 x - 1}) \cdot (\frac{a + x}{x}) = 4 \\ \frac{1}{x} (\frac{a}{y} + 1) = 5 \to (\frac{y}{3 y - 1}) \cdot (\frac{a + y}{y}) = 5 \end{cases}$ ${\begin{cases} a + z = 12 z - 3 \to z = \frac{a + 3}{11} \\ a + x = 20 x - 4 \to x = \frac{a + 4}{19} \\ a + y = 15 y - 5 \to y = \frac{a + 5}{14} \end{cases}$ $a_{1, 2} = 24 \pm 5 \sqrt{23}$ ${x; y; z} \to {\frac{28 + 5 \sqrt{23}}{19}; \frac{29 + 5 \sqrt{23}}{14}; \frac{27 + 5 \sqrt{323}}{11}}$ ${x; y; z} \to {\frac{28 - 5 \sqrt{23}}{19}; \frac{29 - 4 \sqrt{23}}{14}; \frac{27 - 5 \sqrt{23}}{11}}$
	Answered by mr W last updated on 07/Jan/23
	$general case: { ((x+(1/y)=a ...(i))),((y+(1/z)=b ...(ii))),((z+(1/x)=c ...(iii))) :} x+(1/y)+y+(1/z)+z+(1/x)=a+b+c (x+(1/y))(y+(1/z))(z+(1/x))=abc xyz+(1/(xyz))+x+(1/y)+y+(1/z)+z+(1/x)=abc xyz+(1/(xyz))+a+b+c=abc with λ=((abc−(a+b+c))/2) (xyz)^2 −2λ(xyz)+1=0 ⇒xyz=λ±(√(λ^2 −1)) z=c−(1/x)=((cx−1)/x) y=b−(1/z)=b−(x/(cx−1))=(((bc−1)x−b)/(cx−1)) xyz=x×(((bc−1)x−b)/(cx−1))×((cx−1)/x)=(bc−1)x−b ⇒x=((xyz+b)/(bc−1)) ⇒x=((λ±(√(λ^2 −1))+b)/(bc−1)) similarly ⇒y=((λ±(√(λ^2 −1))+c)/(ca−1)) ⇒z=((λ±(√(λ^2 −1))+a)/(ab−1)) such that a solution exists, λ≥1 i.e. abc≥a+b+c+2 in current case: a=3, b=4, c=5 λ=((3×4×5−(3+4+5))/2)=24 xyz=λ±(√(λ^2 −1))=24±5(√(23)) x=((24±5(√(23))+4)/(4×5−1))=((28±5(√(23)))/(19)) y=((24±5(√(23))+5)/(5×3−1))=((29±5(√(23)))/(14)) z=((24±5(√(23))+3)/(3×4−1))=((27±5(√(23)))/(11))$
	$g e n e r a l c a s e :$ ${\begin{cases} x + \frac{1}{y} = a . . . (i) \\ y + \frac{1}{z} = b . . . (i i) \\ z + \frac{1}{x} = c . . . (i i i) \end{cases}$ $x + \frac{1}{y} + y + \frac{1}{z} + z + \frac{1}{x} = a + b + c$ $(x + \frac{1}{y}) (y + \frac{1}{z}) (z + \frac{1}{x}) = a b c$ $x y z + \frac{1}{x y z} + x + \frac{1}{y} + y + \frac{1}{z} + z + \frac{1}{x} = a b c$ $x y z + \frac{1}{x y z} + a + b + c = a b c$ $w i t h λ = \frac{a b c - (a + b + c)}{2}$ ${(x y z)}^{2} - 2 λ (x y z) + 1 = 0$ $\Rightarrow x y z = λ \pm \sqrt{λ^{2} - 1}$ $z = c - \frac{1}{x} = \frac{c x - 1}{x}$ $y = b - \frac{1}{z} = b - \frac{x}{c x - 1} = \frac{(b c - 1) x - b}{c x - 1}$ $x y z = x \times \frac{(b c - 1) x - b}{c x - 1} \times \frac{c x - 1}{x} = (b c - 1) x - b$ $\Rightarrow x = \frac{x y z + b}{b c - 1}$ $\Rightarrow x = \frac{λ \pm \sqrt{λ^{2} - 1} + b}{b c - 1}$ $s i m i l a r l y$ $\Rightarrow y = \frac{λ \pm \sqrt{λ^{2} - 1} + c}{c a - 1}$ $\Rightarrow z = \frac{λ \pm \sqrt{λ^{2} - 1} + a}{a b - 1}$ $s u c h t h a t a s o l u t i o n e x i s t s, λ ⩾ 1$ $i . e . a b c ⩾ a + b + c + 2$ $i n c u r r e n t c a s e :$ $a = 3, b = 4, c = 5$ $λ = \frac{3 \times 4 \times 5 - (3 + 4 + 5)}{2} = 24$ $x y z = λ \pm \sqrt{λ^{2} - 1} = 24 \pm 5 \sqrt{23}$ $x = \frac{24 \pm 5 \sqrt{23} + 4}{4 \times 5 - 1} = \frac{28 \pm 5 \sqrt{23}}{19}$ $y = \frac{24 \pm 5 \sqrt{23} + 5}{5 \times 3 - 1} = \frac{29 \pm 5 \sqrt{23}}{14}$ $z = \frac{24 \pm 5 \sqrt{23} + 3}{3 \times 4 - 1} = \frac{27 \pm 5 \sqrt{23}}{11}$
	Commented by manxsol last updated on 07/Jan/23

	$t o m a k e i t m o r e d i d a c t i c$ $i t w o u l d b e g o o d l i k e t h i s .$ $G r e a t j o u r n a l S i r . W$ $x y z + \frac{1}{x y z} = a b c - (a + b + c)$ $(x y z) . (\frac{1}{x y z}) = 1$ ${(x y z)}^{2} - [a b c - (a + b + c] x y z + 1$ $x y z = \frac{a b c - (a + b + c)}{2} \pm \frac{\sqrt{{[a b c - (a + b + c)]}^{2} - 4}}{2}$ $w i t h λ = \frac{a b c - (a + b + c)}{2}$ ${(x y z)}^{2} - 2 λ (x y z) + 1$
	Answered by ajfour last updated on 08/Jan/23

	$l e t 4 - y = t = \frac{1}{z}$ $N o w x (\frac{1}{x}) = 1 \Rightarrow$ $(3 - \frac{1}{4 - t}) (5 - \frac{1}{t}) = 1$ $\Rightarrow 14 t (4 - t) + 1 = 5 t + 3 (4 - t)$ $\Rightarrow 14 t^{2} - 54 t + 11 = 0$ $t = \frac{27}{14} \pm \sqrt{{(\frac{27}{14})}^{2} - \frac{154}{{(14)}^{2}}}$ $= \frac{27 \pm 5 \sqrt{23}}{14}$ $y = 4 - t = \frac{29 \pm 5 \sqrt{23}}{14}$
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