SOLUTION: Harry has a collection of coins consisting of dimes and quarters whose value is $17.60. The number of quarters exceeds the number of dimes by 8. Find the number of coins of each ki

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Question 1012061: Harry has a collection of coins consisting of dimes and quarters whose value is $17.60. The number of quarters exceeds the number of dimes by 8. Find the number of coins of each kind in the collection.
Answer by ValorousDawn(53) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say our number of dimes is represented by the variable d, and quarters by the variable q.
If the number of quarters exceeds that of dimes by 8, we have the equation q=d%2B8.
We also have numeric values. Since a quarter is worth 25 cents, the total value is 0.25q. The same applies for dimes with 0.10d. We are told it adds to 17.60, so we have the equation 0.25q%2B0.10d=17.60
We now have a system of linear equations. You can solve this with substitution.
Solved by pluggable solver: Linear System solver (using determinant)
Solve:


Any system of equations:


has solution

or



(d=56, q=48}

Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
We'll use substitution. After moving -1*q to the right, we get:
1%2Ad+=+8+-+-1%2Aq, or d+=+8%2F1+-+-1%2Aq%2F1. Substitute that
into another equation:
0.1%2A%288%2F1+-+-1%2Aq%2F1%29+%2B+0.25%5Cq+=+17.6 and simplify: So, we know that q=48. Since d+=+8%2F1+-+-1%2Aq%2F1, d=56.

Answer: system%28+d=56%2C+q=48+%29.