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This problem involves setting up a system of equations to solve for the fee $ x $ (fixed amount for the oil change) and the cost per quart of oil $ y $.

The two pieces of information we have are:

1. Customer A: 5 quarts of oil, total cost $37.45.
2. Customer B: 7 quarts of oil, total cost $46.45.

The relationship between the cost and the variables can be modeled by the following system of equations:

1. For Customer A: 
\[
x + 5y = 37.45
\]
2. For Customer B:
\[
x + 7y = 46.45
\]

### Solving the system of equations:
We can subtract the first equation from the second to eliminate $ x $:

\[
(x + 7y) - (x + 5y) = 46.45 - 37.45
\]
\[
2y = 9
\]
\[
y = 4.50
\]

Now, substitute $ y = 4.50 $ into the first equation:

\[
x + 5(4.50) = 37.45
\]
\[
x + 22.50 = 37.45
\]
\[
x = 37.45 - 22.50 = 14.95
\]

### Final Answer:
- The fee $ x $ is $14.95.
- The cost per quart of oil $ y $ is $4.50.

Would you like further clarification on this solution?

Here are 5 related questions to extend the concept:
1. What if Customer C receives 8 quarts of oil—how much would they pay?
2. How would the total cost change if the service center increases the fee by $5?
3. Can this model be extended to include a discount for customers who buy more than 6 quarts of oil?
4. What would happen to the system of equations if there was no fixed fee, only a cost per quart of oil?
5. How can this problem be solved graphically?

**Tip:** Always check your system of equations by substituting the values back into the original equations to ensure accuracy!

A service center charges a fee of x dollars for an oil change plus y dollars per quart of oil used. Customer A receives 5 quarts of oil and pays a total of $37.45. Customer B receives 7 quarts of oil and pays a total of $46.45. Find the fee and cost per quart of oil.

Learn how to solve a system of linear equations involving a fixed service fee and a variable cost per quart of oil. This real-life problem applies algebraic concepts to find the fee and cost per quart using substitution or elimination methods. Suitable for students in Grades 8-10.

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