Example

Example: Solving a Coin Mixture Problem with Dimes and Nickels

Apply the problem-solving strategy to find the number of each type of coin based on their total value and a relationship involving multiplication and subtraction.

Problem: Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?

  1. Read: The coins are nickels (worth $0.05) and dimes (worth $0.10). The total value is $7.00.
  2. Identify: Find the number of nickels and dimes.
  3. Name: Let nn = the number of nickels. The number of dimes is 3n73n - 7.
TypeNumberValue ($)Total Value ($)
Nickelsnn0.050.050.05n0.05n
Dimes3n73n - 70.100.100.10(3n7)0.10(3n - 7)
  1. Translate: Add the total values to equal $7.00: 0.05n+0.10(3n7)=7.000.05n + 0.10(3n - 7) = 7.00
  2. Solve:
  • Distribute 0.100.10: 0.05n+0.30n0.70=7.000.05n + 0.30n - 0.70 = 7.00
  • Combine like terms: 0.35n0.70=7.000.35n - 0.70 = 7.00
  • Add 0.700.70 to both sides: 0.35n=7.700.35n = 7.70
  • Divide both sides by 0.350.35: n=22n = 22 Find the number of dimes: 3(22)7=667=593(22) - 7 = 66 - 7 = 59
  1. Check: Does 22(0.05)+59(0.10)=7.0022(0.05) + 59(0.10) = 7.00? 1.10+5.90=7.001.10 + 5.90 = 7.00 \checkmark
  2. Answer: Elane has 2222 nickels and 5959 dimes.

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Updated 2026-04-22

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