That was OK.  You eliminated p by subtracting the 1st from the second: 

   10d + 5n + p = 121
 -(  d +  n + p =  21)
 ---------------------
    9d + 4n     = 100 

No that's no good because you eliminated p from the first two,
so you must also eliminate p from a different pair of equations 
both of which have p in them.  What you got does not contain p.  So
you must first eliminate p from the 1st and 3rd, by subtracting the 
3rd from the 1st:   


     d +  n + p =  21
   -(    5n + p =  41)
   -------------------
     d - 4n     = -20

Now take what you got from subtracting the first two: 

    9d + 4n     = 100

and put that with it and add the two equations:

    9d + 4n     = 100
     d - 4n     = -20
    -----------------
   10d          =  80
              d =   8

Now substitute 8 for d in  

         d - 4n = -20
         8 - 4n = -20 
            -4n = -28
              n =   7

Now substitute both d = 8 and n = 7 in one of the original equations, 
say, the easiest one:

      d + n + p = 21
      8 + 7 + p = 21
         15 + p = 21 
              p = 6

So the solution is (d,n,p) = (8,7,6)

Now we check:

8   dimes = 8 coins
7 nickels = 7 coins
6 pennies = 6 coins
-------------------
           21 coins, that checks!  

8 dimes   = 80 cents
7 nickels = 35 cents
6 pennies =  6 cents
--------------------
           121 cents or $1.21, that checks!


If you take the dimes away, you have only the 7 nickels
and the 6 pennies

7 nickels = 35 cents
6 pennies =  6 cents
--------------------
            41 cents, that checks!

So the answer is correct.

Edwin