Solving linear systems with 3 variables

x+4y=6
2x+3z=12
4y+z=10

x+4y=6 (1)
2x+3z=12 (2)
4y+z=10 (3)
(1) gives 2x+8y=12, or 2x=12-8y (4)
(2) gives 2x=12-3z (5)
Equate RHS of (4) and (5), 12-8y=12-3z, giving 8y=3z.
(3) gives 8y=20-2z (6).
With 8y=3z in (6) gives 3z=20-2z=, z=4; 8y=12, y=3/2.
(1) gives x+4(3/2)=6, x=0.
Check: (1), 0 + 4(3/2) = 6; (2), 0 + 12 =12, 4(3/2) + 4 = 10.
Solution:x=0, y=3/2 and z=4.