# Solve:

Dear Student,
consider the first two equations ​7yz+3zx=4xy,21yz-3zx=4xy
7yz+3zx=21yz-3zx$⇒$either z=0 or 7y=3x
Case 1: z=0
Putting this in equation (1) we get xy=0$⇒$either x=0 or y=0
Case A: z=0 and x=0
Putting these in equation (3) we get y=19/2
Case B: z=0 and y=0
Putting these in equation (3) we get x=19

Case 2:7y=3x
Putting this in equation (1) we get 6zx=4xy$⇒$either x=0 or 3z=2y
Case A: x=0 and 7y=3x$⇒$y=0
Putting these in equation (3) we get z=19/3
Case B: 3z=2y and 7y=3x
Putting these in equation (3) we get x+2y+2y=19$⇒$x+4y=19$⇒$x+4(3x/7)=19$⇒$19x/7=19$⇒$x=7
So y=3 and z=2
So possible solutions are (0,0,19/3),(0,19/2,0),(19,0,0) and (7,3,2)

Regards

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