give : x,y,z∈R x+y+xy=8 y+z+yz=15 z+x+zx=35 ⇒x+y+z+xyz=?


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Question Number 169600 by bounhome last updated on 04/May/22

$g i v e : x, y, z \in R$ $x + y + x y = 8$ $y + z + y z = 15$ $z + x + z x = 35$ $\Rightarrow x + y + z + x y z = ?$
Commented by infinityaction last updated on 04/May/22

$(1 + x) (1 + y) = 9 . . . . . (1)$ $(1 + y) (1 + z) = 16 . . . . . . . (2)$ $(1 + x) (1 + z) = 36 . . . . . . . (3)$ ${(1 + x)}^{2} {(1 + y)}^{2} {(1 + z)}^{2} = 9 \times 16 \times 36$ $(1 + x) (1 + y) (1 + z) = 72 . . . . (4)$ $e q .^{n} 4 / e q .^{n} 1$ $z + 1 = 8 \Rightarrow z = 7$ $s i m i l a r l y$ $y = 1 a n d x = \frac{7}{2}$ $x + y + z + x y z = 36$
Commented by Tawa11 last updated on 04/May/22

$Great sir$
	Answered by ajfour last updated on 04/May/22
	$let y=sx , z=tx x(1+s+sx)=8 x(s+t+stx)=15 x(t+1+tx)=35 s=(((8/x)−1)/(1+x)) , t=((((35)/x)−1)/(1+x)) ((43−2x)/(1+x))+(((8−x)(35−x))/((1+x)^2 ))=15 ⇒ (43−2x)(1+x) +(8−x)(35−x)=15(1+x)^2 ⇒ −2x^2 +41x+43+280 +x^2 −43x=15x^2 +30x+15 ⇒ 16x^2 +32x−308=0 4x(4x+8)=4×7×11 ⇒ 4x=14 ⇒ x=(7/2), −((11)/2) s=(((8/x)−1)/(1+x)) , t=((((35)/x)−1)/(1+x)) with x=(7/2) s=((((16)/7)−1)/(9/2))=(2/7) ; t=2 x+y+z+xyz=x(1+s+t)+stx^3 =(7/2){1+(2/7)+2+((7/2))^2 ((4/7))} =(7/2)+1+7+((49)/2)=36.$
	$l e t y = s x, z = t x$ $x (1 + s + s x) = 8$ $x (s + t + s t x) = 15$ $x (t + 1 + t x) = 35$ $s = \frac{\frac{8}{x} - 1}{1 + x}, t = \frac{\frac{35}{x} - 1}{1 + x}$ $\frac{43 - 2 x}{1 + x} + \frac{(8 - x) (35 - x)}{{(1 + x)}^{2}} = 15$ $\Rightarrow (43 - 2 x) (1 + x)$ $+ (8 - x) (35 - x) = 15 {(1 + x)}^{2}$ $\Rightarrow - 2 x^{2} + 41 x + 43 + 280$ $+ x^{2} - 43 x = 15 x^{2} + 30 x + 15$ $\Rightarrow 16 x^{2} + 32 x - 308 = 0$ $4 x (4 x + 8) = 4 \times 7 \times 11$ $\Rightarrow 4 x = 14 \Rightarrow x = \frac{7}{2}, - \frac{11}{2}$ $s = \frac{\frac{8}{x} - 1}{1 + x}, t = \frac{\frac{35}{x} - 1}{1 + x}$ $w i t h x = \frac{7}{2}$ $s = \frac{\frac{16}{7} - 1}{\frac{9}{2}} = \frac{2}{7}; t = 2$ $x + y + z + x y z = x (1 + s + t) + {s t x}^{3}$ $= \frac{7}{2} {1 + \frac{2}{7} + 2 + {(\frac{7}{2})}^{2} (\frac{4}{7})}$ $= \frac{7}{2} + 1 + 7 + \frac{49}{2} = 36 .$
	Commented by Tawa11 last updated on 04/May/22

	$Great sir$
	Answered by mr W last updated on 04/May/22

	$x + y + x y + 1 = 8 + 1 = 9$ $(x + 1) (y + 1) = 9 . . . (i)$ $(y + 1) (z + 1) = 16 . . . (i i)$ $(z + 1) (x + 1) = 36 . . . (i i i)$ $(i) \times (i i) \times (i i i) :$ $(x + 1) (y + 1) (z + 1) = \pm 3 \times 4 \times 6 . . . (i v)$ $(i v) / (i) :$ $z + 1 = \pm \frac{3 \times 4 \times 6}{9} = \pm 8 \Rightarrow z = 7 o r - 9$ $x + 1 = \pm \frac{3 \times 4 \times 6}{16} = \pm \frac{9}{2} \Rightarrow x = \frac{7}{2} o r - \frac{11}{2}$ $y + 1 = \pm \frac{3 \times 4 \times 6}{36} = \pm 2 \Rightarrow y = 1 o r - 3$ $x + y + z + x y z = \frac{7}{2} + 1 + 7 + \frac{7}{2} \times 1 \times 7 = 36$ $o r$ $x + y + z + x y z = - \frac{11}{2} - 3 - 9 - \frac{11}{2} \times 3 \times 9 = - 166$
	Commented by Tawa11 last updated on 04/May/22

	$Great sir$
	Answered by ajfour last updated on 06/May/22

	$z = \frac{35 - x}{1 + x} = \frac{15 - y}{1 + y}$ $\Rightarrow 35 + 35 y - x = 15 + 15 x - y$ $\Rightarrow 36 y + 20 = 16 x$ $\Rightarrow 9 y + 5 = 4 x$ $(9 y + 5) + 4 y + y (9 y + 5) = 32$ $9 y^{2} + 18 y - 27 = 0$ $y^{2} + 2 y + 1 = 4$ $\Rightarrow y + 1 = \pm 2$ $y = - 3, 1$ $c o r r e s p o n d i n g x = - \frac{11}{2}, \frac{7}{2}$ $z = \frac{15 - y}{1 + y} = - 9, 7$ $x + y + z + x y z = - 17 - \frac{1}{2} - \frac{297}{2}$ $= - 166$ $o r = 11 + \frac{1}{2} + \frac{49}{2} = 36$
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