A round of drinks comprising three beers, two glasses of wine, seven cappuccinos and three whiskeys costs $103. A round of nine beers, one glass of wine, four cappuccinos and five whiskeys costs $173. The price of a beer is three times the price of a cappuccino. The price of whiskey is three times the price of a wine.

Use Cramer's rule to find the cost of a glass of wine.

Note: Type your result using an integer number. Do not type any currency symbol.

The cost of one wine is =

Let's define variables for each drink:

- $ b $: cost of one beer
- $ w $: cost of one glass of wine
- $ c $: cost of one cappuccino
- $ h $: cost of one whiskey

### Given relationships:
1. The price of a beer is three times the price of a cappuccino:
   \[
   b = 3c
   \]
2. The price of a whiskey is three times the price of a wine:
   \[
   h = 3w
   \]

### First round of drinks:
- 3 beers, 2 glasses of wine, 7 cappuccinos, and 3 whiskeys cost $103:
   \[
   3b + 2w + 7c + 3h = 103
   \]

### Second round of drinks:
- 9 beers, 1 glass of wine, 4 cappuccinos, and 5 whiskeys cost $173:
   \[
   9b + 1w + 4c + 5h = 173
   \]

### Substituting the relationships for $ b $ and $ h $ into the two equations:

1. Substituting $ b = 3c $ and $ h = 3w $ into the first equation:
   \[
   3(3c) + 2w + 7c + 3(3w) = 103
   \]
   Simplifying:
   \[
   9c + 2w + 7c + 9w = 103
   \]
   \[
   16c + 11w = 103
   \]

2. Substituting $ b = 3c $ and $ h = 3w $ into the second equation:
   \[
   9(3c) + 1w + 4c + 5(3w) = 173
   \]
   Simplifying:
   \[
   27c + 1w + 4c + 15w = 173
   \]
   \[
   31c + 16w = 173
   \]

### System of linear equations:
We now have the system:
1. $ 16c + 11w = 103 $
2. $ 31c + 16w = 173 $

Using **Cramer's rule**, we solve for $ w $ (the cost of one glass of wine).

The general formula for Cramer's rule to solve for $ w $ is:
\[
w = \frac{\text{determinant of the matrix with the w column replaced by the constants}}{\text{determinant of the coefficient matrix}}
\]

Let's compute it.The cost of one glass of wine is $ 5 $. 

Would you like further details or have any questions? Here are five related questions to expand on this:

1. How is Cramer's rule applied to larger systems of equations?
2. What would be the price of a cappuccino based on the solution?
3. Can Cramer's rule be used for non-square systems of equations?
4. What is the significance of the determinant in linear algebra?
5. How does the relationship between the drinks’ prices affect the system of equations?

**Tip**: Cramer's rule works efficiently for small systems but becomes computationally expensive for larger ones, where other methods like Gaussian elimination are preferred.

A round of drinks comprising three beers, two glasses of wine, seven cappuccinos and three whiskeys costs $103. A round of nine beers, one glass of wine, four cappuccinos and five whiskeys costs $173. The price of a beer is three times the price of a cappuccino. The price of whiskey is three times the price of a wine. Use Cramer's rule to find the cost of a glass of wine.

This problem explores how to apply Cramer's Rule to a system of linear equations to find the cost of a glass of wine. Learn about key concepts in linear algebra, including the relationship between drinks' prices, and how to solve the system with determinants. Suitable for Grades 10-12.

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