Solving linear systems w/ 3 variables

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

my solution:
(-x -y -z = 3) -1
x + y + z = -3

x + y + z = -3
+ x + y - z = 2
----------------------------
2x + 2y + 0 = -1 eqn. 4

eqn. 2 + eqn. 3
-x -y -z = 3
+ x -y -z =4
------------------------
0-2y-2z=7 eqn. 5

then add eqn. 4 to eqn. 5? I'm not sure how. I hope what I'm doing above is right and I don't know what to do next. help me please?

1st equation—z in terms of x & y:
z = x + y - 2

2nd equation—y in terms of x, substitute z:
- x - y - x - y + 2 = 3
2y = - 2x - 1
y = - x - 0.5

1st equation—z, substitute y:
= x - x - 0.5 - 2
= - 2.5

3rd equation—x, substitute y & z:
x + x + 0.5 + 2.5 = 4
2x = 1
x = 0.5

2nd equation—y, substitute x:
= - 0.5 - 0.5
= - 1

Answer: x = 0.5, y = - 1, z = - 2.5

Check 1st, 2nd & 3rd equations, respectively, substituting values:
1st equation:
= 0.5 - 1 + 2.5
= 2

2nd equation:
= - 0.5 + 1 + 2.5
= 3

3rd equation:
= 0.5 + 1 + 2.5
= 4

You can immediately add eqn 1 and eqn 2 to get -2z=5 so z=-5/2. You can then plug in z and get a system of two equation and two unknowns (which I assume you can solve since you're doing three equations).

Also, what you're doing seems right... if you add equation 4 and equation 5, you would get 2x-2z=4. I don't know where you plan to go from here though.

x + y - z = 2 eqn. 1
-x -y -z = 3 eqn. 2
============== add them
-2z = 5
z = -5/2

x + y - z = 2 eqn. 1
x -y -z = 4 eqn. 3
============== subtract them
2y = -2
y = -1

-x -y -z = 3 eqn. 2
x -y -z = 4 eqn. 3
============ subtract them
2x = 1
x = 1/2

...checking your solution:

eqn. 2 + eqn. 3
-x -y -z = 3 <========== first multiply by -1
+ x -y -z =4......................to eliminate z and y
------------------------
0-2y-2z=7 eqn. 5