As maroon pointed out, the 'problem' with this puzzle is in the assumption that a-x=1, when infact if a=x then a-x will always be 0...
2(a-x) = a-x [x-2x = -x]...
2/a-x = a-x/a-x [divide both sides by a-x]...
2/0 = 0/0 [substitution given the proof that when a=x, a-x will always be 0]
