Question 259103
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The system of equations is

{{{system(n+d+q=16,
5n+10d+25q=185)}}}

Solving that system for n and d in terms of q you get:

{{{system(n=3q-5,d=21-4q)}}}

To have a positive number of nickels, we must have at least 
2 quarters.  To have a positive number of dimes, we have to 
have no more than 5 quarters

So if we have 2 quarters, 
then we have n=3(2)-5 or 1 nickel, and d=21-4(2)=13 dimes 

If we have 3 quarters, 
then we have n=3(3)-5 or 4 nickels, and d=21-4(3)=9 dimes

If we have 4 quarters, 
then we have n=3(4)-5 or 7 nickels, and d=21-4(4)=5 dimes

If we have 5 quarters, 
then we have n=3(5)-5 or 10 nickels, and d=21-4(5)=1 dimes

So there are four solutions:

2 quarters, 13 dimes, and 1 nickle
3 quarters, 9 dimes, and 4 nickles
4 quarters, 5 dimes, and 7 nickles
5 quarters, 1 dime, and 10 nickles

Edwin</pre>