解下列三元一次方程组.(1)x-4y+z=-32x+y-z=18x-y-z=7;(2)4x+3y+2z=76x-4y-z=62x-y+z=1

问题描述:

解下列三元一次方程组.
(1)

x-4y+z=-3
2x+y-z=18
x-y-z=7
(2)
4x+3y+2z=7
6x-4y-z=6
2x-y+z=1

(1)

x-4y+z=-3①
2x+y-z=18②
x-y-z=7③

①+②,得3x-3y=15,
即x-y=5④,
②-③,得x+2y=11⑤,
⑤-④,得3y=6,
∴y=2,
把y=2代入④,得x=7.
再把x=7,y=2代入③,得z=-2.
∴原方程组的解为
x=7
y=2
z=-2

(2)
4x+3y+2z=7①
6x-4y-z=6②
2x-y+z=1③

①-③×2,得5y=5,
∴y=1,
②+③,得8x-5y=7,
∴x=
3
2

再把x=
3
2
,y=1代入③,得z=-1,
x=
3
2
y=1
z=-1