Hello, can someone check my answer the the question below?

When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. WHat is the original number?

My answer
Let t = the tens digit of the original number
Let u = the units digit of the original number

10t+u
10u+t

10u+t = 10t+u+9
u+t=11

10u+t = 10t+u+9
10u - u+t - 10t = 9
9u - 9t = 9

u+t = 11
u-t = 1
2u = 12
u = 6

u+t = 11
6+t = 11
t = 5

The original number is 56

Check:
10u+t = 10t+u+9
60+5 = 50+6+9
65 = 65
u+t = 11
6+5 = 11
11 = 11

Thanks in advance for the help. I am just about finished with Algebra, which means I am close to HS graduation.

good job

In the verification, I would have checked if the answer satisfies the wording of the problem.
- original number 56
- number reversed 65
is this 9 more than the original ? YES

is the sum of the digits of 56 equal to 11? YES

Suppose your equations had been incorrect, but the solution to those incorrect equations were found by correct steps
Substituting the answers would satisfy the incorrect equations, but would not necessarily have been the correct answer.