Five times the sum of the digits of a two-digit number is 13 less than the original number. if you reverse the digits in the two-digit number, four times the sum of its two digits is 21 less than the reversed two-digit number. (hint: you can use variables to represent the digits of a number. if a two-digit number has the digit x in tens place and y in one’s place, the number will be 10x + y. reversing the order of the digits will change their place value and the reversed number will 10y + x.) the difference of the original two-digit number and the number with reversed digits is .

Let the original 2-digit number be xy.
Because 5 times the sum of the digits is 13 less than the number, therefore
5(x + y) = 10x + y - 13 
5x + 5y = 10x + y - 13
-5x + 4y = - 13             (1)

The number with reversed digits is yx.
Because 4 times the sum of the digits is 21 less than the reversed 2-digit number, therefore
4(x + y) = 10y + x - 21
4x + 4y = 10y + x - 21
3x - 6y = -21
x - 2y = -7
x = 2y - 7                 (2)

Substitute (2) into (1).
-5(2y - 7) + 4y = -13
-10y + 35 + 4y = -13
-6y = -48
y = 8

From (3), obtain
x = 2*8 - 7 = 9

Answers:
The original 2-digit number is 98
The reversed 2-digit number is 89
The difference between the original and the reversed 2-digit numbers is
98 - 89 = 9

Agreed i also had the same answer