solve the following system of equations?

Question

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

Answer 1

sum the first two and obtain 2x+2y=4 x+y=2 (4)
sum the first two and the opposite of three and get 3y-z=1 (5)
add (5) to the first and you'll get x+3y=4 (6)
compare (4) and (6) to obtain y=1; x=1; and z=2

Answer 2

x=3-z first equation
Substitute into second equation
3-z+2y-z=1
2y-2z=-2
y=-1+Z
Substitute expression for x and y into 3rd equation
2(3-z) -(-1+z) +z =3
6-2z +1-z +z = 3
-2z =-4
z=2
x=3-z
x=1
y=-1+z
y=1
solution is x=1,y=1,z=2

Answer 3

x = 3 - z

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

6 - 2z - y + z = 3
- y - z = - 3
y + z = 3

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

x = 1, y = 1 , z = 2