Without using algebra:

If all 23 coins had been nickels, the amount of money would have been only
23×$0.05 or $1.15.  Therefore the extra $1.85-$1.15 = $0.70 must have come
from the dimes, each of which contributed an extra $0.05 per coin. So we
divide $0.70 by $0.05, getting 14.  So there were 14 dimes and 23-14=9
nickels. 

Answer: 14 dimes and 9 nickels.

-----------------------------------------------------

Using algebra:

coins  | number  | value per coin | value of all coins |
--------------------------------------------------------
dimes  |    x    |     $0.10      |      $0.10x        |
nickels|    y    |     $0.05      |      $0.05y        |
--------------------------------------------------------
       |   23    |                |      $1.85



Clear the second equation of decimals by multiplying through by 100
and dropping the "$"



Divide the second equation through by 5



Solve the first equation for y:

              y = 23-x

Substitute into the second equation

2x+(23-x) = 37
  2x+23-x = 37
        x = 14

Substitute in 

       y = 23-x
       y = 23-14
       y = 9

Answer: 14 dimes and 9 nickels.

[Easier without algebra!!]

Edwin