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Grade 8 · lesson 4 of 4

Use Two Constraints to Recover Two Counts

A total count and a weighted total can become two equations.

About 20 minutes, plus any extended practice

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Supplied materials

  • Paper or a browser scratch area; a calculator is optional unless rounding is requested.
  • Supplied teaching case: A fictional stand sells $2 tickets and $3 tickets. It sells 5 tickets for $13. How many $3 tickets were sold? Candidate answers: (A) 2 tickets (B) 3 tickets (C) 4 tickets (D) 6 tickets.
  • Independent check inputs: A fictional stand sells $2 tickets and $3 tickets. It sells 7 tickets for $18. How many $3 tickets were sold? Candidate answers: (A) 3 tickets (B) 4 tickets (C) 5 tickets (D) 8 tickets.

Learn the idea

A system requires the same ordered pair to satisfy every equation. Substitution and elimination find that shared pair, while parallel or coincident line structure explains whether a system has none or many solutions. A total count and a weighted total can become two equations. Comparing a low-price baseline with the actual total reveals how many items must use the higher price.

Here is the method on the supplied teaching case. If all 5 tickets were $2, revenue would be $10. Each $3 ticket adds $1. The extra $3 therefore means 3 of the $3 tickets.

The counted groups must be disjoint and cover the whole supplied total. Keep count and dollar units separate; the model assumes exactly the two stated prices and no fees or refunds.

  1. Write or sketch the supplied teaching case, labeling its quantities and restrictions.
  2. Apply the method shown in the worked solution, keeping each intermediate equality or justified inference.
  3. Solve the independent check with its changed inputs before revealing feedback.
  4. Complete the three practice cases; explain which mathematical rule rules out one distractor in each case.
Worked example

See the method in action

Teaching case: A fictional stand sells $2 tickets and $3 tickets. It sells 5 tickets for $13. How many $3 tickets were sold? Candidate answers: (A) 2 tickets (B) 3 tickets (C) 4 tickets (D) 6 tickets. Solution: If all 5 tickets were $2, revenue would be $10. Each $3 ticket adds $1. The extra $3 therefore means 3 of the $3 tickets. The final answer is 3 tickets. The counted groups must be disjoint and cover the whole supplied total. Keep count and dollar units separate; the model assumes exactly the two stated prices and no fees or refunds.

Your turn

Check your reasoning

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Helpful tips
  • A total count and a weighted total can become two equations.
  • The counted groups must be disjoint and cover the whole supplied total.
A fictional stand sells $2 tickets and $3 tickets. It sells 7 tickets for $18. How many $3 tickets were sold?

Feedback appears here after you check. This is practice, not a graded school assessment.

See the answer and explanation

4 tickets

If all 7 tickets were $2, revenue would be $14. Each $3 ticket adds $1. The extra $4 therefore means 4 of the $3 tickets.

Take it further

Changed-input task: A fictional stand sells $2 tickets and $3 tickets. It sells 19 tickets for $48. How many $3 tickets were sold? Candidate answers: (A) 10 tickets (B) 11 tickets (C) 20 tickets (D) 9 tickets. Show the mathematical reason, not only the selected answer. Expected result: 10 tickets. If all 19 tickets were $2, revenue would be $38. Each $3 ticket adds $1. The extra $10 therefore means 10 of the $3 tickets.

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More practice: try new cases

Choose an answer for each problem. Explain your method on paper, then use the feedback to decide what to revisit.

More practice: try new cases
1. A fictional stand sells $2 tickets and $3 tickets. It sells 9 tickets for $23. How many $3 tickets were sold?
Need a hint?
  • A total count and a weighted total can become two equations.
  • The counted groups must be disjoint and cover the whole supplied total.
2. A fictional stand sells $2 tickets and $3 tickets. It sells 11 tickets for $28. How many $3 tickets were sold?
Need a hint?
  • A total count and a weighted total can become two equations.
  • The counted groups must be disjoint and cover the whole supplied total.
3. A fictional stand sells $2 tickets and $3 tickets. It sells 13 tickets for $33. How many $3 tickets were sold?
Need a hint?
  • A total count and a weighted total can become two equations.
  • The counted groups must be disjoint and cover the whole supplied total.

Your choices are temporary. Feedback appears after you check all the answers.

Nothing here is saved to an account or sent to a teacher. Keep your own notes or download them before leaving. The score describes these questions; it does not certify subject mastery.

Check the subject sources

These LifeStarter passages, scenarios and exercises are authored for this Academy with AI assistance and review. External resources keep their publishers’ terms. A grade label suggests a starting level; it does not establish alignment with your school’s complete requirements.