4yz+xz-7xy=0,3yz+8xy=2xz,5yz+3xz+xy=2xyz

Question

5yz+3xz+xy=2xyz.......(1)

3yz+8xy=2xz ...............(2)

4yz+xz-7xy=0 ..............(3)

solve the equation in three variable

Answer 1

(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