Solving a Car Rental System: Small & Large Cars

Systems of Equations 9th-10th Grade
PROBLEM
A group of college students are going to a lake house for the weekend and plan on renting small cars and large cars to make the trip. Each small car can hold 4 people and each large car can hold 6 people. The students rented 3 more large cars than small cars, which altogether can hold 58 people. Determine the number of small cars rented and the number of large cars rented.

What This Problem Teaches

  • How to translate word constraints into algebraic equations
  • Setting up and solving two-variable linear systems
  • Using substitution method when one equation easily isolates a variable
  • Recognizing capacity problems where items contribute differently to a total
  • Verifying solutions make sense in the original real-world context

Solution: Method 1 — The Substitution Approach

This problem gives us two pieces of information that we can turn into two equations. Let's define our variables first, then build our system.

Step 1 — Define variables

Let s = number of small cars rented
Let l = number of large cars rented

Step 2 — Write the constraint equation

We're told "The students rented 3 more large cars than small cars." This translates directly to:

l = s + 3

Step 3 — Write the capacity equation

Each small car holds 4 people, each large car holds 6 people, and the total capacity is 58 people:

4s + 6l = 58

Step 4 — Substitute the first equation into the second

Since l = s + 3, we can replace l in the capacity equation:

4s + 6(s + 3) = 58
4s + 6s + 18 = 58
10s + 18 = 58
10s = 40
s = 4

Step 5 — Find the number of large cars

Now that we know s = 4, we can find l:

l = s + 3 = 4 + 3 = 7

Solution: Method 2 — The Capacity Breakdown Strategy

Instead of using substitution, let's think about the problem structurally. Since there are 3 more large cars than small cars, we can think of the fleet as having equal pairs plus 3 extra large cars.

Step 1 — Isolate the extra capacity

The 3 extra large cars contribute 3 × 6 = 18 people to the total. Remove this from the total capacity:

58 - 18 = 40 people remain

Step 2 — Set up equal pairs

Now we have an equal number of small and large cars accounting for 40 people. If there are n pairs, then:

n × 4 + n × 6 = 40
10n = 40
n = 4

Step 3 — Calculate total cars

We have 4 equal pairs plus 3 extra large cars:

Small cars: 4
Large cars: 4 + 3 = 7
Answer: 4 small cars and 7 large cars were rented.

Verification

Let's check both conditions from the original problem:

Constraint check

Do we have 3 more large cars than small cars?
7 - 4 = 3 ✓

Capacity check

Do the cars hold exactly 58 people?
4 small cars × 4 people + 7 large cars × 6 people = 16 + 42 = 58 ✓

Car TypeNumber of CarsPeople per CarTotal People
Small4416
Large7642
Total11—58

Common Mistakes to Avoid

✗ Writing l = 3s instead of l = s + 3
The phrase "3 more large cars than small cars" means you add 3 to the number of small cars. If there are 4 small cars, then large cars = 4 + 3 = 7, not 4 × 3 = 12.
✗ Using the wrong capacity contributions
Don't mix up which car holds which number of people. Small cars hold 4 people, large cars hold 6 people. The equation is 4s + 6l = 58, not 6s + 4l = 58.
✗ Solving for one variable and forgetting the other
The problem asks for both the number of small cars AND the number of large cars. Don't stop after finding s = 4; you need to calculate l = s + 3 = 7.

The Pattern Behind This

This is a classic constraint-capacity system. The general structure is:

  • One equation expressing a relationship between quantities
  • One equation expressing how those quantities contribute to a total
General form:
Constraint: y = x + k (or some other linear relationship)
Capacity: ax + by = total

This pattern appears in many contexts: workers with different productivity rates, investments with different returns, containers with different volumes. The substitution method works particularly well when the constraint equation easily isolates one variable.

Why This Matters

Capacity and resource allocation problems like this appear frequently in real life:

  • Event planning: Mixing round tables (8 people) and rectangular tables (6 people) for a wedding
  • Transportation logistics: Combining different vehicle types to move cargo or passengers efficiently
  • Manufacturing: Using machines with different production rates to meet a target output
  • Staffing decisions: Balancing full-time and part-time employees to cover required work hours

Four "What-If?" Problems

1
Different Constraint
A group rents small cars (4 people each) and large cars (6 people each). They rent 2 more large cars than small cars, and the total capacity is 44 people. How many of each type of car did they rent?
Step 1 — Set up variables and equations

Let s = small cars, l = large cars
Constraint: l = s + 2
Capacity: 4s + 6l = 44

Step 2 — Substitute

4s + 6(s + 2) = 44
4s + 6s + 12 = 44
10s = 32
s = 3.2

Step 3 — Find large cars

l = s + 2 = 3.2 + 2 = 5.2

Step 4 — Check

This gives fractional cars, which isn't realistic. The numbers in this problem don't work out to whole vehicles, suggesting the total capacity or constraint needs adjustment for a real scenario.

Mathematical Answer

3.2 small cars and 5.2 large cars (though this would need whole numbers in practice)

2
Add a Third Vehicle
A group rents small cars (4 people), large cars (6 people), and vans (8 people). They rent 2 more large cars than small cars, and the same number of vans as small cars. The total capacity is 66 people. Find the number of each vehicle type.
Step 1 — Define variables

Let s = small cars, l = large cars, v = vans

Step 2 — Write constraint equations

l = s + 2 (2 more large than small)
v = s (same number of vans as small cars)

Step 3 — Write capacity equation

4s + 6l + 8v = 66

Step 4 — Substitute both relationships

4s + 6(s + 2) + 8s = 66
4s + 6s + 12 + 8s = 66
18s = 54
s = 3

Step 5 — Find other vehicles

l = s + 2 = 3 + 2 = 5
v = s = 3

Verification

3(4) + 5(6) + 3(8) = 12 + 30 + 24 = 66 ✓

Answer

3 small cars, 5 large cars, and 3 vans

3
Reverse the Unknown
A group rents 5 small cars and 8 large cars for their trip. Each small car holds the same number of people, each large car holds 2 more people than a small car, and the total capacity is 86 people. How many people does each small car hold?
Step 1 — Define variables

Let x = people per small car
Then x + 2 = people per large car

Step 2 — Set up capacity equation

5x + 8(x + 2) = 86

Step 3 — Solve for x

5x + 8x + 16 = 86
13x + 16 = 86
13x = 70
x = 70/13 ≈ 5.38

Step 4 — Check for whole number

Since we need whole people, let's verify: if small cars hold 5 people and large cars hold 7 people:
5(5) + 8(7) = 25 + 56 = 81 ≠ 86

Step 5 — Try 6 people per small car

If small cars hold 6 people, large cars hold 8 people:
5(6) + 8(8) = 30 + 64 = 94 ≠ 86

Mathematical Answer

Each small car holds 70/13 ≈ 5.4 people (The problem numbers don't yield a clean integer solution)

4
Add Cost Constraint
A group rents small cars ($40/day, 4 people) and large cars ($60/day, 6 people). They rent 1 more large car than small cars. The total capacity is 38 people and total cost is $420. How many of each car type did they rent?
Step 1 — Set up the system

Let s = small cars, l = large cars
Constraint: l = s + 1
Capacity: 4s + 6l = 38
Cost: 40s + 60l = 420

Step 2 — Use capacity equation first

4s + 6(s + 1) = 38
4s + 6s + 6 = 38
10s = 32
s = 3.2

Step 3 — Check with cost equation

If s = 3.2, then l = 4.2
Cost check: 40(3.2) + 60(4.2) = 128 + 252 = 380 ≠ 420

Step 4 — Try systematic approach

Let's try s = 3, l = 4:
Capacity: 4(3) + 6(4) = 12 + 24 = 36 ≠ 38
Try s = 4, l = 5:
Capacity: 4(4) + 6(5) = 16 + 30 = 46 ≠ 38

Step 5 — Conclusion

The three constraints (relationship, capacity, and cost) are inconsistent with these particular numbers. In real problems, you'd need to adjust one of the constraints.

Mathematical Note

This system has no solution with the given constraints — a reminder that not all word problems have neat answers!

Frequently Asked Questions

How do you set up a system of equations for capacity problems? +
Write one equation for the constraint relationship and one for the total capacity. In this problem, if small cars = s and large cars = l, then l = s + 3 (constraint) and 4s + 6l = 58 (capacity). The key is identifying what each vehicle type contributes to the total.
What's the easiest way to solve a two-variable system? +
Use substitution when one equation easily expresses one variable in terms of the other. Here, l = s + 3 substitutes cleanly into 4s + 6l = 58, giving 4s + 6(s + 3) = 58, which simplifies to one variable.
Why do capacity problems often have clean integer answers? +
Because you can't rent part of a car or transport part of a person. Real-world constraints force the mathematical solution to be whole numbers, which is why these problems are designed with numbers that work out evenly.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-11-09

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