Question

x+2y+3z=2 3x+y+2z=3 2x+3y+z=4求这个三元一次方程组的解!？

Answer 1

x+2y+3z=2 (1)
3x+y+2z=3 (2)
2x+3y+z=4 (3)
(2)×2-(1)
5x+z=4 z=4-5x
(2)×3-(1)×2
7x-y=5 y=7x-5
y=7x-5 z=4-5x代入(3)
2x+3(7x-5)+(4-5x)=4,整理,得
18x=15
x=5/6
y=7x-5=35/6 -5=5/6
z=4-5x=4-25/6=-1/6
x=5/6 y=5/6 z=-1/6,10,
x+2y+z=12 ①
3x-y+z=6 ②
x+y+z=12 ③
①-③的
y=0
②-①的
2x-3y=-6
2x=-6
x=-3
代入③得
-3+0+z=12
z=15
所以它的解为
x=-3
y=0
z=15,2,x+2y+3z=2
3x+y+2z=3
2x+3y+z=4
三式相加得6x+6y+6z=9
所以x+y+z=9/6=3/2,2,