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We are given the following system of equations from the problem:

\[
c - p = 9 \quad \text{(1)}
\]
\[
8c + 14p = 402 \quad \text{(2)}
\]

Where:
- $ c $ is the price of a cake.
- $ p $ is the price of a pie.

We are tasked with finding the price of a cake, $ c $.

### Step 1: Express $ c $ in terms of $ p $
From equation (1), we can express $ c $ as:
\[
c = p + 9
\]

### Step 2: Substitute into the second equation
Now, substitute $ c = p + 9 $ into equation (2):
\[
8(p + 9) + 14p = 402
\]
Simplify this:
\[
8p + 72 + 14p = 402
\]
Combine like terms:
\[
22p + 72 = 402
\]
Subtract 72 from both sides:
\[
22p = 330
\]
Solve for $ p $:
\[
p = \frac{330}{22} = 15
\]

### Step 3: Solve for $ c $
Now that we know $ p = 15 $, substitute this value back into the equation $ c = p + 9 $:
\[
c = 15 + 9 = 24
\]

Thus, the price of a cake is **$24**, which corresponds to option C.

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How would the result change if the price difference between a cake and a pie were different?
2. What if the bakery had sold more cakes than pies, how would that affect the system of equations?
3. How can this system of equations be solved graphically?
4. How does the total revenue equation (the second equation) change if discounts are applied?
5. Can you apply this method to other scenarios, such as solving for two unknown prices in a shopping bill?

**Tip:** Always double-check your substitutions to avoid small mistakes when solving systems of equations.

At a bakery, the price of a cake is $9 more than the price of a pie. One day, the bakery sold 8 cakes and 14 pies for a total of $402. The find c, the cost in dollars of a cake, and p, the cost in dollars of a pie. What is the price of a cake?

Learn how to solve a system of linear equations to find the price of a cake and pie when given their total sales. This step-by-step solution uses substitution to solve for the unknowns. Suitable for students in Grades 8-10.

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