[racket] Delete Second

delete-second2 takes the same procedure from delete-second1, I just need to know how to use the if expression because you have to find out if the list has more than one item in order for it to delete anything.
________________________________________
From: danny.yoo at gmail.com [danny.yoo at gmail.com] on behalf of Danny Yoo [dyoo at hashcollision.org]
Sent: Monday, September 17, 2012 12:44 PM
To: Ashley Fowler
Cc: users at racket-lang.org
Subject: Re: [racket] Delete Second

On Mon, Sep 17, 2012 at 10:40 AM, Ashley Fowler
<afowler2 at broncos.uncfsu.edu> wrote:
> Basically delete-second2 has to take a input of a list, if the list has more than 1 item then it will delete the second item and return the new list, if it doesnt have more than 1 item it will just return the original list.


Let me be a bit more explicit.


Is delete-second1 any different from delete-second2?

   * If it is the exact same problem, you get to just copy your answer
from delete-second1 over to delete-second2.  :)

   *  If it is slightly different, then you can _reuse_ some of the
hard work you did to solve delete-second1.

   * If there's no similarity whatsoever, then you have to design the
function from scratch.


Which one of these cases matches?