The sum of digits in a 2-digit number

First note that $y \ne 0$ since otherwise we would have $x + y = x + 0 = 8$, and so $10x + y = 80$, but $80$ doesn't satisfy the second condition.

Therefore we must have $1 \le y \le 9$. This means that when we add $9$ to $10x + y$, the tens digit must increase by $1$ and the ones digit decreases by $1$. So then $10x + y + 9 = 10(x+1) + (y-1)$. Since the digits are equal, we have $x+1 = y-1$. Now you just have a system of two equations in two variables:

\begin{align*} x+y &= 8\\ x+1 &= y-1 \end{align*}

Let, number is of the form $10a+b$, then according to question:

$a+b=8$ and digits of $10a+b+9$ are equal.

Notice that adding $9$ to the give number will increase its tens digit by $1$ and decrease its unit digit by $1$. So,

$a+1=b-1\implies a+2=b$.

Hence, we have $a+2+a=8\implies a=3\implies b=5 \implies 10a+b=35$

The number is $35$, since $$x+y=8$$ and if $9$ is added to any number the ones digit must decrease by $1$ and the tens digit must increase by $1$ if and only if the unit digit is not $0$ hence by adding $9$ $$x+1=y-1$$ $$x-y=-2.$$ By solving both the equations we have $$x=3$$ and $$y=5$$ $$**OR**$$ The numbers whose sum of digits is equal to $8$ the numbers are $$17,26,35,44,53,62,71,80$$ and in these only $35$ is the number whose digits become equal on adding $9$