Break-Even: When Are Health Club Costs the Same?

Systems & Applications 9th-10th Grade
PROBLEM
You are choosing between two health clubs. Club A offers membership for a fee of $11 plus a monthly fee of $17. Club B offers membership for a fee of $25 plus a monthly fee of $10. After how many months will the total cost of each health club be the same? What will be the total cost for each club?

What You Will Learn

  • Setting up linear equations from real-world cost scenarios
  • Solving systems of equations by substitution and elimination
  • Interpreting break-even points in business contexts
  • Understanding the relationship between fixed and variable costs
  • Verifying solutions by checking both original equations

What This Looks Like

Let's visualize how the costs grow over time for each club:

MonthsClub A CostClub B CostDifference
0$11$25Club A cheaper by $14
1$28$35Club A cheaper by $7
2$45$45Same cost!
3$62$55Club B cheaper by $7
4$79$65Club B cheaper by $14

Solution: Method 1 — The Equation Setup

The most natural approach is to write cost equations for each club and find where they're equal.

Step 1 — Define the variable

Let m = number of months of membership.

Step 2 — Write cost equations for each club

Each club has a one-time initial fee plus monthly fees that accumulate:

Club A total cost: C₁ = 11 + 17m
Club B total cost: C₂ = 25 + 10m

Step 3 — Set the costs equal

We want to find when the total costs are the same, so:

11 + 17m = 25 + 10m

Step 4 — Solve for m

Subtract 10m from both sides:

11 + 7m = 25

Subtract 11 from both sides:

7m = 14

Divide by 7:

m = 2

Step 5 — Find the total cost at break-even

Substitute m = 2 into either cost equation:

Club A: 11 + 17(2) = 11 + 34 = 45
Club B: 25 + 10(2) = 25 + 20 = 45

Solution: Method 2 — The Cost Difference Analysis

Instead of setting equations equal, we can think about how the cost difference changes over time.

Step 1 — Analyze the initial difference

At the start (month 0), Club B costs more:

Initial difference = $25 - $11 = $14

Club B starts $14 ahead in cost.

Step 2 — Find how the gap closes each month

Each month, Club A charges $17 while Club B charges $10:

Monthly difference = $17 - $10 = $7

Club A gains $7 on Club B each month.

Step 3 — Calculate when the gap closes

Club A needs to make up the $14 initial disadvantage at $7 per month:

Months needed = $14 ÷ $7 = 2 months

Step 4 — Find the total cost

After 2 months, both clubs will cost:

Club A: $11 + 2($17) = $45
Club B: $25 + 2($10) = $45
After 2 months, both health clubs will cost the same amount: $45.

Verification

Let's check our answer by substituting m = 2 back into both original cost equations:

Club A at 2 months: 11 + 17(2) = 11 + 34 = 45 ✓
Club B at 2 months: 25 + 10(2) = 25 + 20 = 45 ✓

Perfect! Both clubs cost exactly $45 after 2 months.

We can also check our logic by testing nearby values:

  • At 1 month: Club A costs $28, Club B costs $35 (Club A is cheaper)
  • At 3 months: Club A costs $62, Club B costs $55 (Club B is cheaper)

The costs cross over exactly at 2 months, confirming our answer.

Watch Out For These

✗ Mixing up initial and monthly fees
Writing Club A as "17 + 11m" instead of "11 + 17m". The $11 is the one-time initial fee, not the monthly rate. Always check which number goes with which type of fee.
✗ Forgetting to add the initial fee
Setting up the equation as just "17m = 10m" and getting m = 0. This ignores the initial membership fees entirely. The full cost includes both the initial fee AND the accumulated monthly fees.
✗ Solving for total cost instead of months
Setting "11 + 17 = 25 + 10" and getting confused. We need to set up equations with the variable m representing months, not solve for a specific month's cost directly.

Does This Seem Reasonable?

Our answer says the clubs cost the same after 2 months. Let's see if this makes intuitive sense:

Initial advantage: Club A starts $14 cheaper ($11 vs $25 initial fee).

Monthly disadvantage: Club A costs $7 more per month ($17 vs $10 monthly fee).

Break-even timing: It should take about $14 ÷ $7 = 2 months for Club A's higher monthly fees to catch up to Club B's higher initial fee.

This matches our calculated answer perfectly. For short-term memberships (under 2 months), choose Club A. For long-term memberships (over 2 months), choose Club B.

The General Pattern

This is a classic linear break-even problem. The general structure is:

Option 1: C₁ = a₁ + b₁t
Option 2: C₂ = a₂ + b₂t

Break-even when: a₁ + b₁t = a₂ + b₂t
Solving: t = (a₂ - a₁) ÷ (b₁ - b₂)

Where:

  • a₁, a₂ are the initial costs
  • b₁, b₂ are the rates per time period
  • t is the break-even time

This formula only works when b₁ ≠ b₂ (different rates). If the monthly fees were the same, the cheaper initial option would always remain cheaper.

Real Applications

  • Cell phone plans: Comparing unlimited vs pay-per-use plans with different activation fees and monthly rates
  • Software subscriptions: Annual vs monthly billing with different upfront discounts
  • Equipment leasing: Lease vs buy decisions where lease has lower upfront cost but higher monthly payments

What If?

1
Change a Monthly Fee
Club A: $15 initial fee + $12/month. Club B: $35 initial fee + $8/month. After how many months will the costs be equal?
Step 1 — Set up cost equations

Club A: C₁ = 15 + 12m
Club B: C₂ = 35 + 8m

Step 2 — Set costs equal

15 + 12m = 35 + 8m

Step 3 — Solve for m

Subtract 8m: 15 + 4m = 35
Subtract 15: 4m = 20
Divide by 4: m = 5

Step 4 — Find total cost

At 5 months: 15 + 12(5) = 75
Check: 35 + 8(5) = 75

Answer: 5 months, $75 total cost

2
Reverse the Question
Club A: $20 initial + $15/month. Club B: $50 initial + unknown monthly fee. If the costs are equal after 6 months, what is Club B's monthly fee?
Step 1 — Find Club A's cost at 6 months

Club A cost = 20 + 15(6) = 20 + 90 = 110

Step 2 — Set up Club B equation

Let x = Club B's monthly fee
Club B cost = 50 + x(6) = 50 + 6x

Step 3 — Set costs equal

50 + 6x = 110

Step 4 — Solve for x

Subtract 50: 6x = 60
Divide by 6: x = 10

Answer: Club B's monthly fee is $10

3
Add a Third Option
Club A: $10 initial + $20/month. Club B: $40 initial + $12/month. Club C: $0 initial + $24/month. Find when Club A equals Club C, and when Club B equals Club C.
Step 1 — Club A vs Club C

10 + 20m = 0 + 24m
10 = 4m
m = 2.5 months

Step 2 — Club B vs Club C

40 + 12m = 0 + 24m
40 = 12m
m = 40/12 = 10/3 ≈ 3.33 months

Step 3 — Verify costs

At 2.5 months: A = C = 24(2.5) = $60
At 3.33 months: B = C = 24(10/3) = $80

Answer: A=C at 2.5 months ($60), B=C at 3⅓ months ($80)

4
Add a Constraint
Same clubs as original problem, but Club A offers a promotion: first month is free (you still pay the $11 initial fee). After how many months will the costs be equal now?
Step 1 — Adjust Club A's cost equation

With first month free: Club A = 11 + 17(m-1) for m ≥ 1
Simplified: Club A = 11 + 17m - 17 = 17m - 6

Step 2 — Set equal to Club B

17m - 6 = 25 + 10m

Step 3 — Solve for m

Subtract 10m: 7m - 6 = 25
Add 6: 7m = 31
m = 31/7 ≈ 4.43 months

Step 4 — Find total cost

At 31/7 months: 17(31/7) - 6 = 74.43 - 6 = $68.43

Answer: About 4.43 months, costing $68.43 each

Frequently Asked Questions

How do you solve break-even problems with different membership fees?+
Set up cost equations for each option, then find where they're equal. In this problem, Club A costs $11 + $17m and Club B costs $25 + $10m. Setting 17m + 11 = 10m + 25 gives us m = 2 months as the break-even point.
What's the difference between initial fees and monthly fees in cost comparison?+
Initial fees are one-time payments at signup, while monthly fees repeat each month. Your total cost is initial fee plus (monthly fee × number of months). Higher initial fees can be offset by lower monthly fees over time.
When should you choose the option with higher monthly fees?+
Choose higher monthly fees if you plan to stay for less time than the break-even point. Here, Club A has higher monthly fees but lower initial cost, so it's better for memberships shorter than 2 months.
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.

2026-08-28