On reversing the digits of a two digit number, number obtained is 9 less than three times theoriginal number. If difference of these two numbers is 45, find the original number.

Let's assume the original two-digit number is represented by the digits "ab", where "a" represents the tens digit and "b" represents the units digit.

According to the given information, when we reverse the digits of the original number, we obtain the number "ba".

The number obtained by reversing the digits is 9 less than three times the original number. Mathematically, we can represent this as:

10b + a = 3(10a + b) - 9

Simplifying this equation, we get:

10b + a = 30a + 3b - 9

Rearranging the terms, we have:

7b - 29a = 9

Now, we are also given that the difference between the two numbers is 45. Mathematically, this can be represented as:

(10a + b) - (10b + a) = 45

Simplifying this equation, we get:

9a - 9b = 45

Dividing both sides of this equation by 9, we have:

a - b = 5

Now, we have a system of two equations:

7b - 29a = 9 ...(1) a - b = 5 ...(2)

We can solve this system of equations to find the values of "a" and "b", which will give us the original two-digit number.

By solving these equations, we find that "a" is 4 and "b" is -1. However, since "b" represents the units digit, it cannot be negative. Therefore, we discard this solution.

We can try another approach by substituting the value of "a" from equation (2) into equation (1):

7b - 29(5 + b) = 9

Simplifying this equation, we get:

7b - 145 - 29b = 9

Combining like terms, we have:

-22b - 136 = 0

Solving for "b", we find:

b = -136 / -22

b = 6

Now, substituting the value of "b" back into equation (2), we can find the value of "a":

a - 6 = 5

a = 11

Therefore, the original two-digit number is 11.