You are choosing between two different cell phone plans. The first plan charges a rate of 24 cents per minute. The second plan charges a monthly fee of $49.95 plus 10 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

To determine how many minutes you would need to use for the second plan to be preferable, we need to compare the costs of both plans.

Let:

- $ x $ be the number of minutes used in a month.
- The cost of the first plan is $ 0.24x $, since it charges 24 cents per minute.
- The cost of the second plan is $ 49.95 + 0.10x $, which includes a fixed monthly fee of $49.95 and a rate of 10 cents per minute.

We want to find the number of minutes $ x $ where the second plan becomes cheaper than or equal to the first plan. This means solving the inequality:

\[
0.24x > 49.95 + 0.10x
\]

### Solving the inequality:

1. Subtract $ 0.10x $ from both sides:

\[
0.24x - 0.10x > 49.95
\]

\[
0.14x > 49.95
\]

2. Divide both sides by 0.14:

\[
x > \frac{49.95}{0.14}
\]

\[
x > 356.79
\]

### Conclusion:
You would need to use more than 356.79 minutes (approximately 357 minutes) in a month for the second plan to be preferable.

Would you like further details or have any questions?

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Here are 5 follow-up questions based on the problem:

1. How would the calculation change if the first plan charged 20 cents per minute?
2. What if the second plan charged a higher monthly fee, say $59.95, instead of $49.95?
3. How does the break-even point change if the cost per minute on both plans is adjusted by inflation?
4. What if there was an additional setup fee for both plans?
5. How would the analysis change if a third plan was introduced?

**Tip:** When comparing plans, always account for both fixed and variable costs to find the optimal choice.

Compare two cell phone plans using linear inequalities. Learn how to determine the number of minutes where the second plan becomes cheaper by solving 0.24x > 49.95 + 0.10x. This detailed step-by-step solution is ideal for students in Grades 8-10.

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