https://file.aiquickdraw.com/mathbot/file/17293055104417m1vke72.jpg

To solve this problem, we can define variables and set up equations for the two plans.

Let:
- $ x $ be the number of minutes of calls.

### Plan 1:
- Monthly fee: $24
- Additional cost per minute: $0.09

Total cost for Plan 1: 
\[
C_1 = 24 + 0.09x
\]

### Plan 2:
- Monthly fee: $18
- Additional cost per minute: $0.14

Total cost for Plan 2: 
\[
C_2 = 18 + 0.14x
\]

We are asked to find when the costs of the two plans are equal, so we set the two cost equations equal to each other:
\[
24 + 0.09x = 18 + 0.14x
\]

### Solving the equation:
1. Subtract 18 from both sides:
   \[
   24 - 18 = 0.14x - 0.09x
   \]
   \[
   6 = 0.05x
   \]

2. Solve for $ x $ by dividing both sides by 0.05:
   \[
   x = \frac{6}{0.05} = 120
   \]

So, the costs of the two plans will be equal when the customer uses **120 minutes** of calls.

Would you like more details on this solution or further clarification?

### 5 Related Questions:
1. What is the cost of each plan if the customer uses 150 minutes?
2. How much will the customer pay for 200 minutes on Plan 2?
3. For what number of minutes is Plan 1 cheaper than Plan 2?
4. How does the cost difference between the two plans change as the number of minutes increases?
5. How can you graph the two cost functions to visually represent the point where the costs are equal?

### Tip:
Always set up your equations carefully by identifying fixed and variable costs. Then, solve by equating the total costs when comparing multiple options.

Customers of a phone company can choose between two service plans for long distance calls. The first plan has a $24 monthly fee and charges an additional $0.09 for each minute of calls. The second plan has an $18 monthly fee and charges an additional $0.14 for each minute of calls. For how many minutes of calls will the costs of the two plans be equal?

Solve a cost comparison problem between two phone plans with fixed and variable costs. Learn how to use linear equations to determine the breakeven point at 120 minutes. This solution covers essential algebraic concepts and is suitable for students in Grades 8-10.

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