2x + 3y + z = 0
x − 2y − z = −3
x + y + 2z = 3

Solution:

octave:1> A = [2,3,1,0;1,-2,-1,-3;1,1,2,3]
A =

2 3 1 0
1 -2 -1 -3
1 1 2 3

octave:2> rref (A)
ans =

1 0 0 -1
0 1 0 0
0 0 1 2

That is
x = -1
y = 0
z = 2



I can't come up with anything funny, and there are still some problems left, this is the new one however:

What is a?
(2a-5)/6 = 5/12