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

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. - Nicky
Answer by rapaljer(4671) (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

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