Question

解三元一次方程组：(1)x+2y+3z=11x-y+4z=10x+3y+2z=2(2)x:y=3:5y:z=5:6x+y

Answer 1

(1)

x+2y+3z=11(1)

x-y+4z=10(2)

x+3y+2z=2(3)

(1)-(2)得,3y-z=1,(4)

(3)-(2)得,4y-2z=-8,即2y-z=-4,(5)

(4)-(5),得y=5,

代人得z=14,

x=-39

所以x=-39

y=5,

z=14

(2)

x:y=3:5(1)

y:z=5:6(2)

x+y+2z=40(3)

由（1）得,x=(3/5)y,

由（2）得,z=(6/5)y,

代人到（3）,得,

（3/5）y+y+2*(6/5)y=40.

解得,y=10,

代人,得x=6,z=12

所以x=6,

y=10,

z=12,

(3)

x+y-z=2（1）

y+z-x=4(2)

z+x-y=6(3）

(1)+(2)+(3),得,

x+y+z=12,(4)

(4)-(2),得x=4,

(4)-(3),得y=3,

(4)-(1),得z=5

所以：

x=4,

y=3,

z=5