Question 275504
if you take a certain two-digit number and reverse its digits to get another two-digit number, then add these two numbers together, their sum is 132. what is the original number?

Let the digit in the units place be y
& in the tens place be x

So the number will be 10x+y

On reversing the digits
the number becomes 10y+x

The sum of the two = 132

10x+y + 10y+x= 132

11x+11y=132

x+y=12

To satisfy this equation only three pairs are possible
48, 57, 66

All the three numbers wheen reversed and added give you 132