Example

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

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

Problem: Priam has a collection of nickels and quarters, with a total value of $7.30. The number of nickels is 66 less than three times the number of quarters. How many nickels and how many quarters does he have?

  1. Read the problem. A table will help organize the information.
  2. Identify what to find: the number of nickels and the number of quarters.
  3. Name the unknowns. Let n=n = the number of nickels and q=q = the number of quarters. Organize the data into a table:
TypeNumberValue ($)Total Value ($)
Nickelsnn0.050.050.05n0.05n
Quartersqq0.250.250.25q0.25q
Total7.307.30
  1. Translate into a system of equations.

    • The "Total Value" column gives the first equation: 0.05n+0.25q=7.300.05n + 0.25q = 7.30
    • The relationship between the quantities (nickels is 66 less than three times quarters) gives the second equation: n=3q6n = 3q - 6

    The system of equations is: {0.05n+0.25q=7.30n=3q6\left\{\begin{array}{l} 0.05n + 0.25q = 7.30 \\ n = 3q - 6 \end{array}\right.

  2. Solve the system using the substitution method. Substitute 3q63q - 6 for nn in the first equation: 0.05(3q6)+0.25q=7.300.05(3q - 6) + 0.25q = 7.30

    Distribute the 0.050.05: 0.15q0.30+0.25q=7.300.15q - 0.30 + 0.25q = 7.30

    Combine like terms: 0.40q0.30=7.300.40q - 0.30 = 7.30

    Add 0.300.30 to both sides: 0.40q=7.600.40q = 7.60

    Divide by 0.400.40: q=19q = 19

    Substitute q=19q = 19 into the second equation to find nn: n=3(19)6=576=51n = 3(19) - 6 = 57 - 6 = 51

  3. Check the result. 5151 nickels at $0.05 each is $2.55. 1919 quarters at $0.25 each is $4.75. 2.55+4.75=7.302.55 + 4.75 = 7.30 \checkmark

  4. Answer the question. Priam has 5151 nickels and 1919 quarters.

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Updated 2026-05-25

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