Example

Solving a Quarters and Dimes Mixture Problem Using a System of Equations

Apply the seven-step problem-solving strategy and a table to solve a coin mixture application involving quarters and dimes using a system of linear equations.

Problem: Matilda has a handful of quarters and dimes, with a total value of $8.55. The number of quarters is 33 more than twice the number of dimes. How many dimes and how many quarters does she have?

  1. Read the problem. A table will help organize the information.
  2. Identify what to find: the number of dimes and the number of quarters.
  3. Name the unknowns. Let qq be the number of quarters and dd be the number of dimes. Organize the data into a table:
TypeNumberValue ($)Total Value ($)
Quartersqq0.250.25q
Dimesdd0.100.10d
Total8.55
  1. Translate into a system of equations.
  • The "Total Value" column gives the first equation: 0.25q+0.10d=8.550.25q + 0.10d = 8.55
  • The relationship between the quantities (quarters is 33 more than twice dimes) gives the second equation: q=2d+3q = 2d + 3

The system of equations is: {0.25q+0.10d=8.55q=2d+3\left\{\begin{array}{l} 0.25q + 0.10d = 8.55 \\ q = 2d + 3 \end{array}\right.

  1. Solve the system using the substitution method. Substitute 2d+32d + 3 for qq in the first equation: 0.25(2d+3)+0.10d=8.550.25(2d + 3) + 0.10d = 8.55

Distribute the 0.25: 0.50d+0.75+0.10d=8.550.50d + 0.75 + 0.10d = 8.55

Combine like terms: 0.60d+0.75=8.550.60d + 0.75 = 8.55

Subtract 0.75 from both sides: 0.60d=7.800.60d = 7.80

Divide by 0.60: d=13d = 13

Substitute d=13d = 13 into the second equation to find qq: q=2(13)+3=26+3=29q = 2(13) + 3 = 26 + 3 = 29

  1. Check the result.
  • 2929 quarters at $0.25 each is $7.25.
  • 1313 dimes at $0.10 each is $1.30.
  • 7.25+1.30=8.557.25 + 1.30 = 8.55 checkmark
  1. Answer the question. Matilda has 1313 dimes and 2929 quarters.

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Updated 2026-06-24

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Ch.4 Systems of Linear Equations - Intermediate Algebra @ OpenStax

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