solve-in-R-3-x-1-y-3-y-1-z-4-z-1-x-5-

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

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

\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

g e n e r a l c a s e :

{\begin{cases} x + \frac{1}{y} = a \dots (i) \\ y + \frac{1}{z} = b \dots (i i) \\ z + \frac{1}{x} = c \dots (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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