Math With AmarA C A D E M Y

Intermediate · 10 minute lesson

Recover two ticket counts from attendance and receipts

Use simultaneous equations to combine a head count with a revenue total.

Lesson 17 of 30 in Algebra. Take the time you need; the lesson estimate is a guide.

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01 · Read and understand

What you will learn

  • Define two unknown counts.
  • Eliminate one unknown using two equations.
  • Check counts and receipts separately.

Before you start

Solve linear equations and multiply each term of an equation by a constant.

Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.

Start with a question

A fictional event sells 26 tickets: adult tickets at $8 and student tickets at $5. Receipts total $163. How many of each ticket were sold?

Why this math matters

One total can leave many possibilities. A second independent relationship can identify a unique combination. Here the attendance count and money total measure different aspects of the same sales, so both must be satisfied simultaneously.

Make a representation of your own.Sketch the quantities or relationships in this question before working through the solution. The cover image sets the learning scene; it does not show this problem’s exact values.

Set up the model

A useful answer starts with clear assumptions:

  • Each attendee purchased exactly one ticket of one of the two types.
  • There were no discounts, refunds, complimentary tickets, or added fees.

02 · Work through the example

Follow the reasoning, one step at a time.

Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.

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The complete worked example, one idea at a time.

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Recover two ticket counts from attendance and receipts

Paused

Question: Start with the question. Paused.

Question

Start with the question

A fictional event sells 26 tickets: adult tickets at $8 and student tickets at $5. Receipts total $163. How many of each ticket were sold?

Before you calculate

Read what is known and what you need to find. Make a prediction before moving to the first calculation.

Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.

Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.

  1. Record both relationships

    a + s = 26; 8a + 5s = 163

    The first equation counts tickets; the second totals dollars. The coefficients in the money equation are prices per ticket.

  2. Align and eliminate

    5a + 5s = 130; (8a + 5s) − (5a + 5s) = 33

    Multiply the attendance equation by five, then subtract it from the money equation. The student-ticket terms cancel, leaving 3a = 33.

  3. Recover and check both counts

    a = 11; s = 26 − 11 = 15; 8(11) + 5(15) = 163

    The counts sum to twenty-six and the receipts sum to $163. Both conditions hold with whole nonnegative values.

The result

The event sold 11 adult tickets and 15 student tickets.

If all twenty-six tickets had cost $5, receipts would be $130. Each adult ticket adds $3 above that baseline; the $33 extra therefore corresponds to eleven adult tickets.

Common mistakes to catch

  • Use the same definitions of a and s in both equations.
  • A fractional ticket count can reveal inconsistent data rather than a quantity to round.

03 · Practice independently

Try it before revealing the answer.

Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.

Practice 1

At the same prices, 20 tickets bring in $124. Find both counts.

Show a hint

Twenty student-priced tickets would total $100.

Reveal answer and explanation

8 adult and 12 student tickets

The $24 excess over $100 is $3 per adult ticket, so a = 8 and s = 12.

Practice 2

If both ticket types cost $5, can 10 tickets and $50 receipts determine the split?

Show a hint

Consider whether the receipts add a new restriction.

Reveal answer and explanation

No; several splits fit.

Any nonnegative whole counts summing to ten also produce $50. The money equation is five times the count equation, so it adds no independent information.

Take the idea with you

Before solving a system, identify which distinct fact each equation contributes. Repeating the same fact does not add information.

04 · Reflect and continue

Can you explain it in your own words?

Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.

Next lesson

Up next: Find every count that fits a package-mass range

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