Question

x+2y+3z=6
2x+y+z=4
x+y+2z=4​

Answer 1

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first you have to subtract eqn. 3 from 1 to get y+z=2

then put this value in eqn. 2 to get x=1

now put the value of x in eqn. 3 to get 2z+y=3 now solve this eqn. and y+z=2 to get z=1 then put this value of z in eqn. z+y=2 to get y=1

x=y=z=1

Answer 2

Answer:

x+2y+3z=6

2x+y+z=4

x+y+2z=4

(x+2y+3y)+(2x+y+z)+(x+y+2z)=6+4+4=14

4x+4y+6z=14

2x+2y+3y=7

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

x=1

2y+3z=6-1=5

2x+y+z - 2x= y+z= 4 -2=2

x+y+2z -x = y+2z= 4 -1 =3

(y+2z) -(y+z)= z=3-2=1

x=y=z=1