One day a store sold 27 sweatshirts. White ones cost​ $11.95 and yellow ones cost $12.50. In​ all, ​$329.80 worth of sweatshirts were sold. How many of each color were​ sold?

Let the number of white sweatshirts sold be $ x $ and the number of yellow sweatshirts sold be $ y $. We know the following:

1. The total number of sweatshirts sold is 27, so:
   \[
   x + y = 27
   \]

2. The total revenue from selling the white and yellow sweatshirts is $329.80. The price of a white sweatshirt is $11.95, and the price of a yellow sweatshirt is $12.50, so:
   \[
   11.95x + 12.50y = 329.80
   \]

Now, we have the system of equations:
\[
x + y = 27
\]
\[
11.95x + 12.50y = 329.80
\]

### Step 1: Solve for $ y $ in terms of $ x $ from the first equation:
\[
y = 27 - x
\]

### Step 2: Substitute $ y = 27 - x $ into the second equation:
\[
11.95x + 12.50(27 - x) = 329.80
\]
Simplify and solve for $ x $:
\[
11.95x + 12.50 \times 27 - 12.50x = 329.80
\]
\[
11.95x + 337.50 - 12.50x = 329.80
\]
\[
(11.95 - 12.50)x = 329.80 - 337.50
\]
\[
-0.55x = -7.70
\]
\[
x = \frac{-7.70}{-0.55} = 14
\]

### Step 3: Substitute $ x = 14 $ back into the first equation to find $ y $:
\[
14 + y = 27
\]
\[
y = 27 - 14 = 13
\]

### Conclusion:
The store sold 14 white sweatshirts and 13 yellow sweatshirts.

Would you like more details on this solution? Here are some related questions:

1. How do you solve systems of linear equations with substitution?
2. What is the significance of the revenue equation in this problem?
3. Can we solve this system using the elimination method?
4. How does the price difference affect the solution?
5. What are some real-world applications of such systems of equations?

**Tip:** Always double-check your work by substituting your solution back into the original equations to ensure consistency.

Learn how to solve a system of equations to find out how many white and yellow sweatshirts were sold. The store sold 27 sweatshirts, earning a total of $329.80. White sweatshirts cost $11.95, and yellow ones cost $12.50. A detailed step-by-step solution shows how to use substitution to solve this algebraic problem. Ideal for Grades 8-10 students.

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