You can
put this solution on YOUR website!why can't you have 2 quarters and 295 dimes?
let c= # of coins, d= # of dimes, q= # of quarters
since $30 is evenly divisible by $.10, there must be an even number of quarters;
that means at least 2 but no more than 118 so 2<=q<=118
2 quarters means 295 dimes; 118 quarters means 5 dimes ... so 5<=d<=295
the fewest coins is 123 (118q + 5d) and the most is 297 (2q + 295d) so 123<=c<=297