x+y+z=100则x=100-y-z前两个12x+16y+10z=140812(100-y-z)+16y+10z=14081200-12y-12z+16y+10z=14084y-2z=20810x+12y+15z=120010(100-y-z)+12y+15z=12001000-10y-10z+12y+15z=12002y+5z=200这样就是二元一次方程组了4y-2z=208①2y+5z=200②②×2得4y+10z=400③③-①得12z=192z=16则y=60x=100-60-16=24