SOLUTION: There are 237 coins in a bag that total $52.90. There are nickels and dimes in the bad. How many of each are in the bag? This is how I set up the problem, but I keep getting eit

Question 276683: There are 237 coins in a bag that total $52.90. There are nickels and dimes in the bad. How many of each are in the bag?
This is how I set up the problem, but I keep getting either negative numbers or really large numbers for my answer :-/
n + d = 237
0.05n + 0.10d = 52.90
n = 237 - d
0.05(237 - d) + 0.10d = 52.90
11.85 - 0.05d + 0.10d = 52.90
11.85 + 0.05d = 52.90
0.05d = 41.05
d = 821????
n = -584????
I tried using the substitution method, but it doesn't seem to work out when I do it.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

That's because your problem has no solution. If you have 237 coins that are some mixture of either nickels or dimes, the least amount of money you could have would be $11.85 (237 nickels and 0 dimes) and the most you could have would be $23.70 (0 nickels and 237 dimes). Hence, there is no way that the total amount of money could be as much as $52.90.

John

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