Theater Ticket Sales: Solving a System of Equations

Skills This Problem Builds

• Translating a story into two equations. One equation counts things (tickets) and one counts value (dollars). This pairing shows up in almost every "two types of item" problem.
• Choosing and declaring variables. Part A of the problem is not busywork. A clearly defined variable is what keeps the later algebra honest.
• Substitution and elimination as two routes to the same place. You'll see why they must agree, and how to pick the more convenient one.
• Reasoning about weighted averages. The answer is pulled toward the cheaper or dearer ticket depending on how the revenue compares with the average price.
• Checking against the story. A solution must satisfy both equations and be a sensible number of tickets.

The Setup: Organize Before You Write Equations

Two ticket types, each with a price, a count, and a revenue. Putting them in a table first turns a paragraph of words into a grid you can read off, and the two equations come straight from the table.

Solution: Method 1 — Substitution

Substitution works well here because the ticket-count equation is already almost solved for one variable. We'll follow the three parts the problem asks for.

Part A — Represent the variables

Name exactly what each letter counts, with its unit. "x = adults" is too vague. Say what is being counted.

Part B — Write the system of equations

The first equation says the two kinds of tickets together make 400. The second says the money from each kind, price times quantity, adds up to $3,100.

Both equations are measured in different units: equation (1) is in tickets, equation (2) is in dollars. Never mix them. That is why you need two equations rather than one.

Part C, Step 1 — Solve equation (1) for y

Subtract x from both sides to express the child tickets in terms of the adult tickets:

Part C, Step 2 — Substitute into equation (2)

Wherever y appears in the revenue equation, write 400 − x instead. The parentheses matter: the 5 multiplies the whole quantity.

Part C, Step 3 — Solve for x

Part C, Step 4 — Back-substitute to find y

So 275 adult tickets and 125 child tickets were sold. Notice the number 4 in 4x: it's the difference between the two prices ($9 − $5). That's not a coincidence, and Method 3 and the pattern section both build on it.

Solution: Method 2 — Elimination

Elimination is different in structure: instead of solving for one variable, you add or subtract whole equations so a variable disappears. Both equations have a 5y-compatible term waiting to be matched.

Step 1 — Line up the equations

Step 2 — Multiply equation (1) so the y-coefficients match

Multiply every term of (1) by 5. This is legal because it applies the same operation to both sides.

Step 3 — Subtract (1′) from (2)

The 5y terms cancel, leaving one equation in one unknown.

Step 4 — Find y

Put x = 275 into the simpler equation (1): 275 + y = 400, so y = 125. Same answer as substitution, as it must be.

Solution: Method 3 — The "Everyone Buys a Child Ticket" Baseline

This approach needs no variables at all. It's the reasoning hiding inside the algebra, and it's worth knowing because it lets you do the problem on the back of an envelope.

1. Pretend all 400 tickets were child tickets. Revenue would be 400 × $5 = $2,000.
2. Compare with reality. The actual revenue was $3,100, so there is a shortfall of $3,100 − $2,000 = $1,100.
3. Each swap closes part of the gap. Changing one child ticket into an adult ticket adds $9 − $5 = $4.
4. Count the swaps needed.$1,100 ÷ $4 = 275 swaps, meaning 275 adult tickets. The remaining 400 − 275 = 125 are child tickets.

Ancient problem-solvers used exactly this "assume all one kind, then correct" idea long before algebraic notation existed. The algebra in Methods 1 and 2 is the same computation: the 2000 and the 4x appear in both.

Verification: Test Both Equations

A solution to a system must satisfy every equation. Checking only one is the classic way to miss an error.

Here is what the verified answer looks like. The bar is drawn to scale, with 400 tickets across its full width:

Sanity Check: The Average-Price Argument

You can predict roughly where the answer should land before doing any algebra. The average revenue per ticket is $3,100 ÷ 400 = $7.75. That has to sit between the child price ($5) and the adult price ($9), and it does.

$7.75 is closer to $9 than to $5, so adults should be the majority, and 275 out of 400 (about 69%) fits. In fact, $7.75 is exactly 69% of the way from $5 to $9 (since 2.75 ÷ 4 = 0.6875), which is the same 275/400.

What Trips People Up

If You See These Words…

This problem belongs to the most common family in applied algebra: two unknown quantities, two kinds of information. Look for:

• A total count ("sold 400 tickets," "30 coins," "50 animals") combined with a total value ("for a total of $3,100," "worth $4.10," "100 legs").
• Two items with different unit values ("each adult ticket…, each child ticket…").
• A question that asks "how many of each?"

The same structure hides behind very different stories: nickels and dimes, two grades of coffee mixed to a target price, or chickens and rabbits counted by legs. If you can fill in a table like the one above, you can write the system.

The Pattern Behind Every "Total Count, Total Value" Problem

Let there be N items in total with total value R. Item A is worth p each and item B is worth q each (with p > q). The system is:

Substituting b = N − a and solving gives a formula you can use directly:

The numerator, R − qN, is the "shortfall over the all-cheap baseline," and the denominator, p − q, is the gain per swap. That is Method 3 in symbols. For our theater: a = (3100 − 5·400) ÷ (9 − 5) = 1100 ÷ 4 = 275.

When the Answer Isn't a Whole Number

Suppose a typo changed the revenue to $3,107 while everything else stayed the same. The method doesn't change at all:

You can't sell three-quarters of a ticket. A messy result for something that must be counted means the problem is inconsistent, not that your algebra is broken (provided you've rechecked your setup). Here, with $5 and $9 tickets, every possible revenue for 400 tickets is 2000 + 4x, so it must be a multiple of 4 more than 2,000, and 3,107 is not.

Beyond the Textbook

• Event budgeting. Box offices use exactly this reasoning to work backward from revenue and attendance to the ticket mix, which tells them how many discounted seats they gave away.
• Pharmacy and chemistry. Replace "adult" and "child" with a 10% solution and a 30% solution, and the same pair of equations (amount and amount-of-substance) governs mixtures.
• Retail and inventory. A shop that knows the number of items sold and the total sales can estimate how many were premium-priced versus standard-priced.

A Brief History

The "total count, total value" puzzle is ancient. The Chinese text Sunzi Suanjing (roughly 4th–5th century CE) poses the famous chickens-and-rabbits problem: given the number of heads and the number of feet, find how many of each. The classic solution is the same "assume all one kind, then adjust" trick used in Method 3. Methods of eliminating variables from simultaneous equations also appear in the Chinese Nine Chapters on the Mathematical Art, many centuries before Western algebra adopted them. These problems stay in the curriculum because they teach the central idea of algebra: one story can produce two constraints.

Extend Your Thinking

Four variations, roughly in order of difficulty. Try each on paper first, then reveal the solution to compare.

Frequently Asked Questions