2yz−4z+2x−2=0 2xz−2z+2y−4=0 2xy−4x−2y+2z+4=0 how to find the all values of the (x,y,z) ?


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Question Number 184258 by universe last updated on 04/Jan/23

$2 yz - 4 z + 2 x - 2 = 0$ $2 xz - 2 z + 2 y - 4 = 0$ $2 xy - 4 x - 2 y + 2 z + 4 = 0$ $how to find the all values of the (x, y, z) ?$
	Answered by SEKRET last updated on 04/Jan/23

	$(0, 1, - 1) (0, 3, 1) (1, 2, 0) (2, 1, 1) (2, 3, - 1)$
	Answered by SEKRET last updated on 04/Jan/23
	${ (( 2yz − 4z + 2x=2)),(( 2xz − 2z +2y =4)),(( 2xy−4x−2y+2z=−4)) :} z= ((2−2x)/(2y−4)) = ((1−x)/(y−2)) z= ((2(2−y))/(2(x−1))) = ((2−y)/(x−1)) z= 2x+y−2−xy ((1−x)/(y−2)) = ((2−y)/(x−1)) −(1−x)^2 =−(2−y)^2 1−x =2−y 1−x= y−2 y= 1−x y=3−x z= ((1−x)/(1−x−2))=((1−x)/(−1−x))=((x−1)/(x+1)) z= ((2−1+x)/(x−1))= ((1+x)/(x−1)) ((x−1)/(x+1))=((x+1)/(x−1)) (x−1)^2 =(x+1)^2 x−1= −x −1 2x=0 x=0 x=0 y=1 z= −1 y=3−x z=((1−x)/(3−x−2))=((1−x)/(1−x))=1 x=3−y= 2 x=2 y=1 z=1 z=2x+y−2−xy=2x−xy +y−2 z= x∙(2−y) −1(2−y) = (x−1)(2−y) z=z=z (x−1)(2−y)=((1−x)/(y−2)) (2−y)^2 =1 y= 1 y=3 x=1 y=2 z=1 x=0 y=3 z=1 y=3 x=2 z=−1 (x−1)(2−y) = ((2−y)/(x−1)) (x−1)^2 =1 x=0 x=2 x=0 y=3 z=1 x=2 y=1 z=1 x=2 y=3 z= −1 (0,1,−1) (0,3,1) (1,2,0) (2,1,1) (2,3^� −1)$
	${\begin{cases} 2 yz - 4 z + 2 x = 2 \\ 2 xz - 2 z + 2 y = 4 \\ 2 xy - 4 x - 2 y + 2 z = - 4 \end{cases}$ $z = \frac{2 - 2 x}{2 y - 4} = \frac{1 - x}{y - 2}$ $z = \frac{2 (2 - y)}{2 (x - 1)} = \frac{2 - y}{x - 1}$ $z = 2 x + y - 2 - xy$ $\frac{1 - x}{y - 2} = \frac{2 - y}{x - 1} - {(1 - x)}^{2} = - {(2 - y)}^{2}$ $1 - x = 2 - y 1 - x = y - 2$ $y = 1 - x y = 3 - x$ $z = \frac{1 - x}{1 - x - 2} = \frac{1 - x}{- 1 - x} = \frac{x - 1}{x + 1}$ $z = \frac{2 - 1 + x}{x - 1} = \frac{1 + x}{x - 1}$ $\frac{x - 1}{x + 1} = \frac{x + 1}{x - 1} {(x - 1)}^{2} = {(x + 1)}^{2}$ $x - 1 = - x - 1 2 x = 0 x = 0$ $x = 0 y = 1 z = - 1$ $y = 3 - x z = \frac{1 - x}{3 - x - 2} = \frac{1 - x}{1 - x} = 1$ $x = 3 - y = 2 x = 2 y = 1 z = 1$ $z = 2 x + y - 2 - xy = 2 x - xy + y - 2$ $z = x \cdot (2 - y) - 1 (2 - y) = (x - 1) (2 - y)$ $z = z = z$ $(x - 1) (2 - y) = \frac{1 - x}{y - 2} {(2 - y)}^{2} = 1$ $y = 1 y = 3$ $x = 1 y = 2 z = 1$ $x = 0 y = 3 z = 1$ $y = 3 x = 2 z = - 1$ $(x - 1) (2 - y) = \frac{2 - y}{x - 1}$ ${(x - 1)}^{2} = 1 x = 0 x = 2$ $x = 0 y = 3 z = 1$ $x = 2 y = 1 z = 1$ $x = 2 y = 3 z = - 1$ $(0, 1, - 1) (0, 3, 1) (1, 2, 0) (2, 1, 1) (2, \bar{3} - 1)$
	Commented by universe last updated on 04/Jan/23

	$t h a n k s s i r$
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