Question 1033597
Let n = number of nickels, d = number of dimes, and q = number of quarters.

We have to solve the system

n+d+q = 28
0.05n+0.10d+0.25q = 2.00

The second equation is equivalent to n + 2d + 5q = 40, after simplification.

The system represented as an augmented matrix is 

{{{(matrix(2,4,1,1,1,28,1,2,5,40))}}}
~{{{(matrix(2,4,1,1,1,28,0,1,4,12))}}} ~ {{{(matrix(2,4,1,0,-3,16,0,1,4,12))}}}

==> n-3q = 16 and d+4q = 12.

==> n = 3q+16 and d = 12 - 4q.

==> The possible values of q are 0,1, 2, and 3.
The possible triples (n,d,q) are thus 
(16,12,0), (19,8,1), (22,4,2), and (25,0,3).

But since there is at least one of each type of coin, the only possible combinations are (19,8,1) and (22,4,2).