Can you prove this math riddle with algebra

1. Think of a number. x
2. Add five to your number
3. Subtract two from the result
4. Double your result
5. Add four to your result
6. Divide your result by two
7. Subtract your original number
8. Add six to your result
9. Multiply your result by three Is your final answer thirty-three?

1. x
2. x + 5
3. x + 5 - 2
4. 2(x + 5 - 2)
5. 2(x + 5 - 2) + 4
6. [2(x + 5 - 2) + 4]/2
7. [2(x + 5 - 2) + 4]/2 - x
8. [2(x + 5 - 2) + 4]/2 - x + 6
9. 3 * [[2(x + 5 - 2) + 4]/2 - x + 6]

3 * [[2(x + 5 - 2) + 4]/2 - x + 6]
3 * [(2x + 6 + 4)/2 - x + 6]
3 * [x + 5 - x + 6]
3 * [x - x + 5 + 6]
3 * [11]
33 <-- magic.

3(((2(x+5-2)+4)/2)-x+6) = 33
3(((2(x+3)+4)/2)-x+6) = 33
3((x+3+2)-x+6) = 33
3(x+5-x+6) = 33
3(11) = 33
33 = 33

It's an identity, meaning that no matter what value of x one picks, the result will always be 33.

1... x
2 ... x + 5
3... x+5 - 2 = x+3
4.. 2(x+3) = 2x + 6
5... 2x+6 + 4 = 2x + 10 =2(x+5)
6.... 2(x+5)/2 = x + 5
7... x+ 5 - x = 5
8... 5 + 6 = 11
9. 11 * 3 = 33

((((((((x+5)-2)x2)+4)/2)-(x))+6)3)

3(([[2((x+5)-2)+4]/2]-x)+6)

yes its 33

No, it is "I'm too smart to be answering silly riddles in Yahoo-three."

you just did it