Nice.

First, instead of adding 3 after multiplying with 3, we can add 1 before multiplyting with 3 (as 3*1 = 3). By changing the order, we get "multiply by 3" twice in a row. Then we multiply 3*3 and get 9, so the steps can be reduced to add one, then multiply by nine.

Now, lets look a bit closer at multiplying by nine, and what digits we get:

9x = (10-1)x = 10x-x = 10(x-1)+(10-x).

E.g. for 9*7 we get 10*6+10-7 = 63.

That is, x-1 is the first digit, and 10-x is the second digit. The next step is to add the digits together.

x-1+10-x = -1+10 = 9.

Whoa, x, the number we originally picked (or rather, that number plus one) disappeared from the equation. The result is 9, no matter what x is (well, as long as x is between 1 and 10, which is why we are asked to pick a number between 1 and 9 - before adding 1).

Quod Est Demonstrandum (or something like that).