Answer:
The difference of the original two-digit number and the number with reversed digits is 98 - 89 = 9
Step-by-step explanation:
Let the digit at unit(one's) place = y
Let the digit at ten's place = x
So, the two-digit number will be: 10x +y
According to given condition:
Five times the sum of the digits of a two-digit number is 13 less than the original number. This statement can be written as:
5(x +y) = (10x +y) - 13
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.
if digits are reserved the two digit will be 10y + x
4(x + y) = (10y + x) -21
So, we have 2 equations, solving these 2 equations we can find the 2 digit number.
5(x +y) = (10x +y) - 13
5x + 5y = 10x +y -13
Rearranging:
5x -10x +5y -y = -13
-5x + 4y = -13 (eq1)
4(x + y) = (10y + x) -21
4x +4y = 10y +x =-21
4x -x +4y -10y = -21
3x -6y = -21 (eq2)
Solving:
Multiply eq 1 with 3 and eq 2 with 2
-15x + 12 y = -39
6x - 12y = -42
_______________
-9x = -81
x = -81/-9
x = 9
Putting in eq(1)
-5x + 4y = -13
-5(9) +4y = -13
-45 + 4y = -13
4y = -13 + 45
4y = 32
y = 32/4
y = 8
So, y =8 and x = 9
The 2 digit number is:
10x + y = 10(9) + 8 = 90+8 = 98
The reversed 2 digit number is:
10y + x = 10(8) + 9 = 80+9 = 89
The difference of the original two-digit number and the number with reversed digits is 98 - 89 = 9
