Unifying Accept States in NFA I recently read the first question and answer in this homework solution https://cseweb.ucsd.edu/classes/sp08/cse105/hw2s.pdf However, I am not sure if for example, you have two accepting states originally, one in which you leave the accepting state if you read 1 and the other in which you do not leave the accepting state no matter the input then how would one unite the two states. Your help is appreciated.
You can unite them exactly as said in the solution, "simply
add a new final state to the original automaton, add epsilon transitions from every old final state to the new final state, and change every old final state into a regular state."

Suppose $N$ is the original NFA at a final state $F$. Suppose it can go to one of states $E_1, \cdots, E_m$ without any input if there is an epsilon transition from $F$. Suppose on input symbols $s$, $N$ changes to one of states $S_1, \cdots, S_n$.

Let $N'$ with its unique final state $U$ be the new NFA as constructed above from $N$. When $N'$ is at its regular state $F$, what will happen?


It can stay at $F$ without further input.
It can go to any one of $E_1, \cdots, E_m$ as before because of the possible epsilon transition. 
On input $s$, it will go to one of $S_1, \cdot, S_n$. 
It can go to $U$ without any further input because of the epsilon transition from $F'$ to $U$.


Note that in case 2, 3 and 4 where $N$ has moved to state $s'\in\{E_1, \cdots, E_m, S_1, \cdots, S_n\}$, $N$ can move to another state without further input if there is epsilon transition from $s'$. Furthermore, $N$ can move to still another state without further input if there is still epsilon transition. And so on. 


In particular, if $N'$ stays at $F$ on input $s$, it can go to $U$ without any further input. 

Yes, an epsilon transition functions like teleporting magically.