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.
the difference of the original two-digit number and the number with reversed digits is

The answer is 9

u - the unit digit
t - the tenth digit
So, the original two digit number will be: 10t + u.
The reversed two-digit number will be: 10u + t

Five times the sum of the digits of a two-digit number is 13 less than the original number is: 5(u + t) = 10t + u - 13
four times the sum of its two digits is 21 less than the reversed two-digit number: 4(u + t) = 10u + t - 21

Thus, we have two equations:
5(u + t) = 10t + u - 13
4(u + t) = 10u + t - 21

Multiply the factor:
5u + 5t = 10t + u - 13
4u + 4t = 10u + t - 21

Rearrange:
5u + 5t - 10t - u = -13
4u + 4t - 10u - t = -21

4u - 5t = -13
-6u + 3t = -21

Multiply the first equation by 3, and the second by 2:
4u · 3 - 5t · 3 = -13 · 3
-6u · 2 + 3t · 2 = -21 · 2

12u -15t = -39
-12u + 6t = -42

Add these two equations to make one equation:
12u - 15t - 12u + 6t = -39 - 42

12u can be canceled out:
-9t = - 81

t = -81/-9
t = 9

Replace t in one of the equations:
4u - 5 · 9 = -13
4u - 45 = -13
4u = -13 + 45
4u = 32
u = 32/4
u = 8

Therefore, t = 9, u = 8
The original two digit number will be: 10t + u = 10·9 + 8 = 90 + 8 = 98
The reversed two-digit number will be: 10u + t = 10·8 + 9 = 80 + 9 = 89

The difference of the original two-digit number and the number with reversed digits is 98 - 89 = 9

1/5 i hope this