What is the solution to the following system? x+y+z=6 x-y+z=8 x+y-z=0

I don't really know if this is correct but I think it is x+y+z=14

Subtract the second equation from the first equation :

The xs and zs would cancel. On the left side, y–(-y)=2y. On the right side, 6–8=-2

We're left with 2y=-2
Divide both sides by 2:
Y=-1

Subtract the third equation from the second equation. The xs would cancel again. -y–y=-2y and z–(-z)=2z. So on the left side we have -2y+2z and on the right side, 8-0=8.

We get -2y+2z=8.
Plug in -1 for y:
-2(-1)+2z=8
2+2z=8
Subtract 2 from both sides:
2z=6
Divide both sides by 2:
Z=3

Plug in -1 for y and 3 for z into the first equation:
X+(-1)+3=6

Solve for x:
X+2=6
Subtract 2 from both sides:
X=4

Hope this helps! :)

I’m having a hard time understanding your question