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The problem asks for a linear equation that models the total cost $ C $ for $ L $ tennis lessons, given that:

- The cost for 7 lessons is $82.
- The cost for 11 lessons is $122.

To solve this, we need to derive the equation of the line in the form $ C = mL + b $, where $ m $ is the slope (cost per lesson) and $ b $ is the y-intercept (fixed cost).

### Step 1: Calculate the slope
The slope $ m $ is given by the change in cost divided by the change in the number of lessons:
\[
m = \frac{122 - 82}{11 - 7} = \frac{40}{4} = 10
\]
Thus, the slope is $ m = 10 $, meaning each lesson costs $10.

### Step 2: Find the y-intercept
We can now use one of the points (7, 82) or (11, 122) to find the y-intercept $ b $. Let's use the point (7, 82):
\[
C = mL + b \implies 82 = 10 \times 7 + b
\]
\[
82 = 70 + b \implies b = 82 - 70 = 12
\]
So, the y-intercept is $ b = 12 $.

### Step 3: Write the equation
The equation for the total cost $ C $ as a function of the number of lessons $ L $ is:
\[
C = 10L + 12
\]

### Step 4: Check the answer choices
From the options provided:
- A. $ C = 10L - 12 $
- B. $ C = -10L - 12 $
- C. $ C = -20L + 14 $
- D. $ C = 10L + 12 $

The correct equation is **D**.

Would you like more details or have any questions about this solution?

### Follow-up Questions:
1. How do you determine the slope in a linear equation problem?
2. Why do we need two points to determine a linear equation?
3. What role does the y-intercept play in the equation?
4. How can we check if the derived equation fits the given data?
5. What happens if the slope is negative in a cost problem?

### Tip:
When solving for the y-intercept, always ensure to substitute values for both variables (cost and lessons) to avoid simple mistakes.

The cost for 7 tennis lessons is $82. The cost for 11 tennis lessons is $122. Which of the following linear equations model the total cost C for L lessons?

Learn how to model the total cost for tennis lessons using a linear equation. This problem involves finding the slope and y-intercept based on the cost of 7 and 11 lessons. Suitable for students in Grades 7-10.

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