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 q=q = the number of quarters and d=d = the number of dimes. Organize the data into a table:
TypeNumberValue ($)Total Value ($)
Quartersqq0.250.250.25q0.25q
Dimesdd0.100.100.10d0.10d
Total8.558.55
  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.

  2. 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 0.250.25: 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 0.750.75 from both sides: 0.60d=7.800.60d = 7.80

    Divide by 0.600.60: 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

  3. 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

  4. Answer the question. Matilda has 1313 dimes and 2929 quarters.

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Updated 2026-05-26

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