Theater Ticket Sales: Solving a System of Equations
A] Show how you represent your variables, choose x and y.
B] Show how you write the systems of equations.
C] Find the solution - Show all work.
What This Problem Teaches
- Variable identification — recognizing the two unknown quantities and assigning meaningful variables
- Systems setup — translating word problems into two equations with two unknowns
- Substitution method — solving by isolating one variable and substituting
- Revenue calculations — understanding how unit price × quantity = total revenue
- Solution verification — checking answers against both original constraints
Visualizing the Setup
| Ticket Type | Price Each | Number Sold | Total Revenue |
|---|---|---|---|
| Adult | $9.00 | x | 9x |
| Child | $5.00 | y | 5y |
| Total | — | 400 | $3,100 |
This table helps us see the structure clearly: we need equations for both the total ticket count and total revenue.
Solution: Method 1 — The Substitution Approach
Part A — Variable Representation
We need to define variables for the two unknown quantities:
Let y = number of child tickets sold
Part B — Writing the System of Equations
We create two equations based on the given constraints:
Equation 2 (total revenue): 9x + 5y = 3100
Equation 1 comes from: adult tickets + child tickets = total tickets
Equation 2 comes from: (price per adult)(number of adults) + (price per child)(number of children) = total revenue
Part C — Finding the Solution
Step 1 — Isolate y from the first equation
From x + y = 400, we solve for y:
Step 2 — Substitute into the second equation
Replace y in the revenue equation with (400 - x):
Step 3 — Expand and simplify
Distribute the 5 and combine like terms:
4x + 2000 = 3100
4x = 1100
x = 275
Step 4 — Find y using substitution
Substitute x = 275 back into y = 400 - x:
Solution: Method 2 — The Elimination Approach
Starting with the same system, we can eliminate one variable by scaling equations:
9x + 5y = 3100 ——— (2)
Step 1 — Eliminate x by scaling equation (1)
Multiply equation (1) by -9 to make the x-coefficients opposites:
9x + 5y = 3100 ———— (2)
Step 2 — Add the equations
When we add, the x terms cancel out:
-4y = -500
y = 125
Step 3 — Find x using back-substitution
Substitute y = 125 into equation (1):
x = 275
Both methods give us the same answer: 275 adult tickets and 125 child tickets.
Verification
Let's check our answer against both original constraints:
Check total revenue: $9(275) + $5(125) = $2,475 + $625 = $3,100 ✓
Both equations are satisfied, confirming our solution is correct.
Common Pitfalls
Some students write x for children and y for adults, then forget their choice when setting up equations. Always write your variable definitions clearly at the start.
Common error:
9 + 5 = 14x = 3100 (adding prices instead of calculating revenue)Correct:
9x + 5y = 3100 (price × quantity for each type)
Students sometimes solve for x but forget to substitute back to find y. Always find both unknowns and state them clearly in context.
The Pattern Behind This
All two-category mixture problems follow the same structure:
Equation 1: x + y = total quantity
Equation 2: (value₁)x + (value₂)y = total value
This pattern applies whether you're mixing tickets, coins, investments, or chemical solutions. The key insight is recognizing that you need two different constraints — one about quantities and one about values — to solve for two unknowns.
Recognizing This Problem in the Wild
Look for these signal phrases that indicate a system of equations problem:
- "Two types" or "two kinds" of items with different values
- A total count AND a total value given
- "How many of each..." (asking for two unknown quantities)
- Different prices per unit for each category
Four "What-If?" Problems
Let x = adult tickets and y = child tickets
Equations: x + y = 400 and 10x + 6y = 3600
From equation 1: y = 400 - x
Substitute: 10x + 6(400 - x) = 3600
10x + 2400 - 6x = 36004x = 1200x = 300
y = 400 - 300 = 100
Check: 300 + 100 = 400 ✓ and $10(300) + $6(100) = $3,600 ✓
Answer: 300 adult tickets, 100 child tickets
Let a = adult, c = child, s = senior
Given: c = s, so we can substitute s = c
a + c + c = 500 → a + 2c = 5009a + 5c + 7c = 3300 → 9a + 12c = 3300
From equation 1: a = 500 - 2c
Substitute: 9(500 - 2c) + 12c = 3300
4500 - 18c + 12c = 3300-6c = -1200, so c = 200s = 200, a = 500 - 2(200) = 100
Answer: 100 adult, 200 child, 200 senior tickets
Period 1 (first 200): Adults $9, Children $5
Period 2 (next 200): Adults $7.20, Children $4 (20% off)
Let x = adult tickets in each period
Period 1: x adults, (200-x) children
Period 2: x adults, (200-x) children
9x + 5(200-x) + 7.2x + 4(200-x) = 26409x + 1000 - 5x + 7.2x + 800 - 4x = 2640
7.2x + 1800 = 26407.2x = 840x = 116.67... Wait, this should be whole!
Actually, let a = total adults, so a/2 in each period
Revenue = 9(a/2) + 5(200-a/2) + 7.2(a/2) + 4(200-a/2) = 2640
This gives us a = 240 total adult tickets
Let c = child ticket price
Then adult price = c + 3 (since adults cost $3 more)
Total revenue = 320(adult price) + 180(child price)320(c + 3) + 180c = 4260
320c + 960 + 180c = 4260500c = 3300c = 6.60
Child tickets: $6.60
Adult tickets: $6.60 + $3.00 = $9.60
Check: 320($9.60) + 180($6.60) = $3072 + $1188 = $4260 ✓
Answer: Adult tickets $9.60, child tickets $6.60
Frequently Asked Questions
2026-08-21