Maths question about ratio

Let (2x - 3y)/(y + 3z) = (y – z)/(x – z) = (x + 3z)/(2y – 3x) = k then
2x - 3y = k(y + 3z) ... (1)
y - z = k(x - z) .......... (2)
x + 3z = k(2y - 3x) ... (3)(1) + (2)*3 + (3) :
2x - 3y + 3y - 3z + x + 3z = k(y + 3z + 3x - 3z + 2y - 3x)
3x = k(3y)
k = x / y
Hence , by (2) ,
y - z = (x / y) (x - z)
y² - yz = x² - xz
x² - y² - z(x - y) = 0
(x - y) (x + y - z) = 0
x = y or x + y = z

2013-05-09 15:57:02 補充：
If k = 0 ,
2x - 3y = y - z = x + 3z = 0
⇒ x = y = z = 0
then 0/0 = k , so k ≠ 0.

2013-05-09 15:58:37 補充：
If k = 0 ,
2x - 3y = y - z = x + 3z = 0
⇒ x = y = z = 0
then 0/0 = k , so k ≠ 0.

2013-05-09 16:08:34 補充：
If y = 0 , then x = 0 since k ≠ 0 , but k = 0/(y + 3z) = 0 (contradiction) ,
so y ≠ 0.

2013-05-09 16:08:55 補充：
If y = 0 , then x = 0 since k ≠ 0 , but k = 0/(y + 3z) = 0 (contradiction) ,
so y ≠ 0.