When the digits of two-digits numbers 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.

Can some one please help me here?

The tens digit is T, the units digit is U.

The number is TU, its value is 10T +U

Reverse it, subtract the origial number

10U + T - 10T -U =9

T+ U=11

Those are the guiding equations. Simplify the first, and solve.

T=5 u=6 5 + 6=11

65_56=9
Answer =56

To solve this problem, we'll use the given equations and solve them step by step.

First, let's translate the given information into equations:

1. When the digits of two-digits numbers are reversed, the new number is 9 more than the original number.
This can be written as: (10U + T) - (10T + U) = 9

2. The sum of the digits of the original number is 11.
This can be written as: T + U = 11

Now, let's simplify and solve the equations:

From equation 1, we have: 10U + T - 10T - U = 9
Simplifying further, we get: 9U - 9T = 9

From equation 2, we have: T + U = 11

We can rearrange equation 2 to get: T = 11 - U

Substitute this value of T into equation 1:

9U - 9(11 - U) = 9
Expanding and simplifying, we get: 9U - 99 + 9U = 9
Combining like terms, we get: 18U - 99 = 9
Adding 99 to both sides, we get: 18U = 108
Dividing both sides by 18, we get: U = 6

Now substitute this value of U back into T = 11 - U:

T = 11 - 6
T = 5

Therefore, the original number is 56 (T = 5, U = 6).