x+2y+3z=10,x-y+4z=10,x+3y+2z=2

问题描述:

x+2y+3z=10,x-y+4z=10,x+3y+2z=2

x=-42
y=-4
z=12

x+2y+3z=10,(1)
x-y+4z=10,(2)
x+3y+2z=2 (3)
(1)-(2)得:
3y-z=0
z=3y (4)
(3)-(2)得:
4y-2z=-8 (5)
(4)代入(5)得:
-2y=-8
y=4
代入(4)得:
z=12
代入(1)得:
x=10-8-36
x=-34
所以,方程组的解是:x=-34;y=4;z=12