Question

3x+y+z=4,3y+x+z=0,3z+x+y=6三元一次方程解加减哦

Answer 1

3x+y+z=4 (1)

3y+x+z=0 (2)

3z+x+y=6 (3)

(1)+(2)+(3)得

5x+5y+5z=10

x+y+z=2 (4)

(1)-(4)得

2x=2

∴x=1

(2)-(4)得

2y=-2

∴y=-1

(3)-(4)得

2z=4

∴z=2

∴方程组的解是

x=1

y=-1

z=2