Question 104519
x + y + 2z = 11
x + y + 3z = 14
x +2y + 3z = 5
:
Look at it; you can see that subtracting the 1st eq from the 2nd, you eliminate x & y   making it easy to solve for y:
x + y + 3z = 14
x + y + 2z = 11
------------------ Subtracting eliminates x & y
0 + 0 + 1z = 3
z = 3
:
Substitute 3 for z in the 2nd and 3rd equations:
2nd equation:
x + y + 3(3) = 14
x + y + 9 = 14
x + y = 14 - 9
x + y = 5 
:
3rd equation
x + 2y + 3(3) = 5
x + 2y + 9 = 5
x + 2y = 5 - 9
x + 2y = -4
:
Using these two unknown equations:
x + 2y = -4
x +  y = 5
------------subtracting eliminates x
0 + 1y = -9
y = -9
:
Find x using the equation x + y = 5
x - 9 = 5
x = 5 + 9
x = 14
:
We have x = 14; y = -9; z = 3
:
Check solution in the 1st equation: x + y + 2z = 11
14 - 9 + 2(3) = 
14 - 9 + 6 = 11
:
Did this give you a clue, at least?