SOLUTION: Hi, I've tried this problem many many times but still cannot figure it out. Please help me. Thanks. Joe has 45 coins which consisted of nickels and dimes, and quarters. These co

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Hi, I've tried this problem many many times but still cannot figure it out. Please help me. Thanks. Joe has 45 coins which consisted of nickels and dimes, and quarters. These co      Log On

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Question 436632: Hi, I've tried this problem many many times but still cannot figure it out. Please help me. Thanks.
Joe has 45 coins which consisted of nickels and dimes, and quarters. These coins were worth $6.30. The number of dimes exceeded the number of nickels by 5. How many nickels, dimes and quarters did he have?
This is as far as I got, I don't know how to get one variable by itself...
5n + 10d + 25q = 630
n + d = 45
d = n + 5
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
n+d+q=45
n+(n+5)+q=45
2n+5+q=45
q=40-2n
5n+10(n+5)+25(40-2n)=630
5n+10n+50+1000-50n=630
420=35n
n=12
Joe has 12 nickels, 17 dimes, and 16 quarters..