# SOLUTION: What I have is a money mixture - I can't remember how to figure it. The problem is John has a change jar which he empties occasionally. This time he has all nickels and dimes. He h

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 Question 24108: What I have is a money mixture - I can't remember how to figure it. The problem is John has a change jar which he empties occasionally. This time he has all nickels and dimes. He has nine times as many dimes as nickels, and the value of the dimes equals \$5.10 more than the value of the nickels. How many dimes and nickels does John have. I have two equations, 9d=n or d=n+\$5.10 we are supposed to use a method of substitution or similar to solve. Please help! Three people here, and no clues. - NickyAnswer by rapaljer(4667)   (Show Source): You can put this solution on YOUR website!I think you need to say Let n = number of nickels d = number of dimes. He has nine times as many dimes as nickes, so the number of dimes is 9 times the number of nickels. This means that d = 9n Value of the dimes = 10d Value of the nickels = 5n Value of the dimes equals 510 cents more than the value of the nickels 10d = 5n + 510 This is a good place to substitute, since d=9n 10d = 5n + 510 10*9n = 5n+ 510 90n= 5n + 510 85n= 510 n = 6 Nickels 9n = 54 Dimes Check. Value of 6 nickels = \$.30 Value of 54 dimes = \$5.40 Dimes are worth \$5.10 more than the nickels so it checks! R^2 at SCC