SOLUTION: Word problem; Joe has a collection of nickles and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coin

Question 198744: Word problem; Joe has a collection of nickles and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. how many dimes does he have?
10x2 + 6.05x + 9.85 = 0
using the formula I am not sure what numbers go to a,b, or c?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Whatever gave you the idea that you needed a quadratic equation for this problem? Read the problem carefully.

Let d represent the number of dimes. Let n represent the number of nickels. It will be convenient to convert the amounts from dollars and cents to just cents, that is: $6.05 is 605 cents and $9.85 is 985 cents.

Dimes are worth 10 cents each, so the value of the dimes that he has right now is 10d cents. His nickels, at 5 cents each, are worth 5n cents. Add them together:

Likewise, if you double the amount of dimes, you have 2d dimes which are worth 20d cents. 10 fewer nickels is n - 10 which are worth 5(n - 10) cents. Add them together:

which needs to be re-written as:

Now you have two equations in two variables. Multiply the first one by -1 so that the coefficients on one of them become additive inverses:

Then add this new equation to the other one we developed:

So he has 43 dimes.

Check:

43 dimes is $4.30, leaving $6.05 - $4.30 = $1.75 in nickels. $1.75 divided by .05 = 35, so 35 nickels. Twice the dimes, 86, is worth $8.60, 10 less nickels is 25, 25 nickels is worth $1.25. $8.60 plus $1.75 = $9.85. Answer checks.

John

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