Question 248703
The sum of the digits of a positive two-digit integer is 45 less than the integer. What is the tens digit of the integer?

Let I be the number. 
Let x be the tens digit and y the ones.

Then I = (10*x + y)
Also x + y = I - 45

So;

x + y = (10x + y) - 45
9x = 45
x=5.