一、解下列方程组 1、｛x+y=3,y+z=5,x+z=4｝

(1)x+y=3,y+z=5,x+z=4
三式相加：x+y+y+z+x+z=2(x+y+z)=12
x+y+z=6
x+y=3,z=3
y+z=5,y=2,x+1
(2)设x=2k,y=3k,z=5k
2x+y+3z=88
4l+3k+15k=88
22k=88,k=4
x=8,y=12,z=20