25q + 10d + 5n = 250
  q +   d +  n =  18

Let's eliminate d by multiplying the 
second equation through by -10 and
adding the two equations:

 25q + 10d +  5n = 250
-10q - 10d - 10n = 180
----------------------
 15q       -  5n =  70

Divide through by 5

  3q - n = 14
 3q - 14 = n 

    

   
   

So the smallest possible number of quarters is 5 

Substitute q = 5 

n = 3q - 14 = 3(5) - 14 = 15 - 14 = 1  

Substitute q = 5 and n = 1 in   

q + d + n = 18
5 + d + 1 = 18
    d + 6 = 18
        d = 12

So 1 solution is 5 quarters, 1 nickel and 12 dimes
----------------------------------------------------

Substitute q = 6 

n = 3q - 14 = 3(6) - 14 = 18 - 14 = 4  

Substitute q = 6 and n = 4 in   

q + d + n = 18
6 + d + 4 = 18
   d + 10 = 18
        d = 8

So 1 solution is 6 quarters, 4 nickels and 8 dimes

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

Substitute q = 7 

n = 3q - 14 = 3(7) - 14 = 21 - 14 = 7  

Substitute q = 7 and n = 7 in   

q + d + n = 18
7 + d + 7 = 18
   d + 14 = 18
        d = 4

So 1 solution is 7 quarters, 7 nickels and 4 dimes

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


Substitute q = 8 

n = 3q - 14 = 3(8) - 14 = 24 - 14 = 10  

Substitute q = 8 and n = 10 in   

q + d +  n = 18
8 + d + 10 = 18
    d + 18 = 18
         d = 0

So 1 solution is 8 quarters, 10 nickels and 0 dimes

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

We can't use 9 quarters for that's $2.25, and we'd
have to make the remaining 25 cents with 9 nickels 
and dimes, which is impossible because even 9 
nickels alone is 45 cents!!

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

So there are 4 solutions:

5 quarters, 1 nickel, and 12 dimes.           
6 quarters, 4 nickels, and 8 dimes.
7 quarters, 7 nickels, and 4 dimes.
8 quarters, 10 nickels, and 0 dimes.

Edwin