The sum digit of two digit …

Let the digit at unit's place be x and the digit at ten's place be y. Then,
Number = 10y + x
Number formed by reversing the digits = 10x + y
According to the given conditions
The sum of the digits of a two digit number is 8 .
So, x + y =8 ....... (i)
 and the difference between the number and that formed by reversing the digits is 18
So, (10y + x) - (10x + y)=18
{tex}\Rightarrow{/tex} 10 y + x  - 10x - y = 18
{tex}\Rightarrow{/tex} 9y - 9x = 18
{tex}\Rightarrow{/tex} 9(y - x) =18
{tex}\Rightarrow{/tex}y - x =2 ........ (ii)
Add equations (i) and (ii), we get
(x + y) + (y - x) = 8 + 2
{tex}\Rightarrow{/tex}x + y + y - x = 10
{tex}\Rightarrow{/tex} 2y = 10
{tex}\Rightarrow{/tex} y = 5
Put y = 5 in (i) we get
x + y =8
{tex}\Rightarrow{/tex} x + 5 = 8
{tex}\Rightarrow{/tex} x = 3
So, two digit number = 10y + x =10{tex}\times{/tex}5 + 3 = 53
Hence given two digit number is 53.