x+2y+z=8 (1)2x+3y+z=11 (2)x+3y+3z=16 (3)(3)-(1),得 y+2z=82*(1)-(2),得 y+z=5两个等式相减,z=3 所以y=2,x=1方程组的解是 x1=1,x2=2,x3=3 .
