you are correct.
you need 3 equations in 3 unknowns to get an exact answer.
with 2 equations in 3 unknowns, you would have to solve for 2 of the variables in terms of one of the variables and then make assumptions about the that variable that would fix the value of the other 2 variables.
i did that with this problem and got the following:
n = 6 - .75d
q = 26 - .25d
from that, i made assumptions about the number of dime and got the following:
dimes nickels quarters coins money invalid
0 6 26 32 $6.80
4 3 25 32 $6.80
8 0 24 32 $6.80
12 -3 23 32 $6.80 *****
16 -6 22 32 $6.80 *****
20 -9 21 32 $6.80 *****
24 -12 20 32 $6.80 *****
28 -15 19 32 $6.80 *****
32 -18 18 32 $6.80 *****
36 -21 17 32 $6.80 *****
40 -24 16 32 $6.80 *****
44 -27 15 32 $6.80 *****
48 -30 14 32 $6.80 *****
52 -33 13 32 $6.80 *****
56 -36 12 32 $6.80 *****
60 -39 11 32 $6.80 *****
64 -42 10 32 $6.80 *****
68 -45 9 32 $6.80 *****
when creating the table, it became apparent that i needed to increment the number of dimes in groups of 4 to get integer values for the other coins.
all possible combinations gave me 32 coins for a total of $6.80, but the ones marked as invalid did this by using negative values of coins.
for example:
with 68 dimes, i got -45 nickels and 9 quarters.
68 - 45 + 9 = 32
68*10 - 45*.05 + 9*.25 = 6.80 - 2.25 + 2.25 = 6.80
the numbers came out right but only if the number of nickels was negative which can't be so the answer was marked as invalid.
the possible answers became the first 3 in the list.
there would be no way to narrow it down further than that without some additional information being provided such as the number of dimes being 0, 4, or 8.