<s>[INST] The sum of digits in a 2-digit number 
The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number. 

My attempt:
Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at unit's place. According to question:
$$x+y=8$$
I could not figure out the other condition. Please help. Thanks in advance.
 [/INST] 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*}
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