Question 50099
Ok, there's nothing wrong with your solution to problem a), but the solution to problem b) is another story. Let's have a look:

b)
1) 3x-y-2z = 11
2) x-2y+3z = 12
3) x+y-2z = 5
Add 1) & 3)
4) 4x-4z = 16 You are ok here.
4a) x-z = 4   You are ok here.

Now add 2) and 2 X 3)
2) x-2y+3z = 12
3a) 2x+2y-4z = 10 adding theses two, you get (or should have obtained)
5) 3x-z = 22 Here's one of your boo boo's (3z+(-4z) = -z...not +z as you had)
 Now here you could have subtracted 4a) from 5) to get x.
4a) x-z = 4
5) 3x-z = 22
6) 2x = 18
6a) x = 9 ...and you got this right!

Now to find z. You can substitute x = 9 into 4a.
4b) 9-z = 4 Solve for z.
4c) z = 5 Here's another boo boo, you had z = -5

To find y, substitute x = 9 and z = 5 into any of your original equations and solve for y. Let's use equation 1)
3x-y-2z = 11
3(9)-y-2(5) = 11
27-y-10 = 11
17-y = 11
y = 6 ...and here's you third mistake. You had y = 14.

The solution:
x = 9
y = 6
z = 5

Check:

1) 3x-y-2z = 11
3(9)-6-2(5) = 11
27-6-10 = 11
27-16 = 11 
11 = 11 OK here.

2) x-2y+3z = 12
9-2(6)+3(5) = 12
9-12+15 = 12
24-12 = 12
12 = 12 OK here too.

3) x+y-2z = 5
9+6-2(5) = 5
9+6-10 = 5
15-10 = 5
5 = 5 ...and alright here.