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Let's break down the problem:

We are given two rental plans, and we need to find the number of miles at which both plans will cost the same.

- **Plan 1**:  
  Initial fee = \$61.98  
  Additional cost per mile = \$0.07  

- **Plan 2**:  
  Initial fee = \$51.98  
  Additional cost per mile = \$0.12  

Let the number of miles driven be $ x $. The total cost for each plan can be expressed as:

- **Cost of Plan 1**:  
  \[ C_1 = 61.98 + 0.07x \]

- **Cost of Plan 2**:  
  \[ C_2 = 51.98 + 0.12x \]

To find the number of miles $ x $ at which both plans cost the same, we need to set the two costs equal:

\[
61.98 + 0.07x = 51.98 + 0.12x
\]

Now, let's solve for $ x $.Salma would need to drive **200 miles** for the two rental plans to cost the same.

Would you like any further details or have any questions?

Here are five related questions to consider:
1. What is the cost for each plan if Salma drives 100 miles?
2. What if the cost per mile for one of the plans changes? How would the comparison change?
3. How would a 10% discount on the initial fee of both plans affect the breakeven point?
4. What if Salma plans to drive only 50 miles? Which plan would be cheaper?
5. How many miles should Salma drive to make Plan 2 the more economical choice?

**Tip:** Always compare both fixed and variable costs when deciding between two pricing plans!

Salma will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $61.98 and costs an additional $0.07 per mile driven. The second plan has an initial fee of $51.98 and costs an additional $0.12 per mile driven. How many miles would Salma need to drive for the two plans to cost the same?

Learn how to compare two car rental plans by solving linear equations to find out how many miles Salma should drive for both plans to cost the same. This problem involves cost comparison using basic algebraic equations, ideal for Grades 6-8.

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