Question 313447
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 13. What is the original number?

The firs digit in each number will be multiplied by 10 because it is in the tens place.

The second number, 10y+x is 9 more than the first number, 10x+y.
So----------------->10y+x=9+10x+y
Put it in Standard form-->-9x+9y=9----> simplify, divide by 9
--------------------------> -x+y=1
Second equation--------_-->  x+y=13 now add down
-------------------------->    2y=14
--------------------------->    y=7

Now substitue the 7 into an original equation
10(7)+x=9+10x+7
70+x=16+10x
54=9x
6=x