(1) 5yz+3xz+xy=2xyz
(2) 3yz+8xy=2xz, xy=z(2x-3y)/8
(3) 4yz+xz-7xy=0, 7xy=4yz+xz, xy=z(4y+x)/7 and z=7xy/(4y+x)
(4) Equating xy in (2) and (3):
(2x-3y)/8=(4y+x)/7, 14x-21y=32y+8x, 6x=53y, x=53y/6
(5) Substituting for x=53y/6 in (3):
z=371y2/(6(4y+53y/6))=371y/77,
z=53y/11
Since x=53y/6, z=53y/11, we can substitute these into (1), because it contains xyz which has degree 3 (equations (2) and (3) are degree 2):
5y(53y/11)+3(53y/6)(53y/11)+53y2/6=2y(53y/6)(53y/11)
Divide through by 53y2/11:
5+53/2+11/6=53y/3, 30+159+11=106y, 200=106y, y=100/53, z=100/11, x=50/3