A round of drinks comprising four beers, two glasses of wine, seven cappuccinos and three whiskeys costs $85. A round of nine beers, one glass of wine, four cappuccinos and five whiskeys costs $121. The price of a beer is double the price of a cappuccino. The price of whiskey is double the price of a wine.



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

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

The cost of one whiskey is = Blank 1

To solve the problem using Cramer's rule, we'll follow these steps:

### Step 1: Define the variables
Let's denote the prices as follows:
- Let $ b $ be the price of one beer.
- Let $ w $ be the price of one glass of wine.
- Let $ c $ be the price of one cappuccino.
- Let $ h $ be the price of one whiskey.

### Step 2: Set up the equations
From the problem statement, we have the following information:
1. $ b = 2c $ (The price of a beer is double the price of a cappuccino)
2. $ h = 2w $ (The price of whiskey is double the price of wine)

We also have the cost equations:
\[
4b + 2w + 7c + 3h = 85 \quad \text{(Equation 1)}
\]
\[
9b + 1w + 4c + 5h = 121 \quad \text{(Equation 2)}
\]

### Step 3: Substitute the relationships into the equations
Using $ b = 2c $ and $ h = 2w $, we substitute into the two equations:

**Equation 1:**
\[
4(2c) + 2w + 7c + 3(2w) = 85
\]
\[
8c + 2w + 7c + 6w = 85
\]
\[
15c + 8w = 85 \quad \text{(Equation 3)}
\]

**Equation 2:**
\[
9(2c) + 1w + 4c + 5(2w) = 121
\]
\[
18c + w + 4c + 10w = 121
\]
\[
22c + 11w = 121 \quad \text{(Equation 4)}
\]

### Step 4: Set up the system of linear equations
Now we have the system of equations:
\[
\begin{aligned}
15c + 8w &= 85 \quad \text{(Equation 3)} \\
22c + 11w &= 121 \quad \text{(Equation 4)}
\end{aligned}
\]

### Step 5: Solve the system using Cramer's rule
The system can be written in matrix form $ A \mathbf{x} = \mathbf{b} $, where:
\[
A = \begin{pmatrix} 15 & 8 \\ 22 & 11 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} c \\ w \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 85 \\ 121 \end{pmatrix}
\]

Using Cramer's rule:
\[
c = \frac{\text{det}(A_c)}{\text{det}(A)}, \quad w = \frac{\text{det}(A_w)}{\text{det}(A)}
\]
where:
\[
A_c = \begin{pmatrix} 85 & 8 \\ 121 & 11 \end{pmatrix}, \quad A_w = \begin{pmatrix} 15 & 85 \\ 22 & 121 \end{pmatrix}
\]

First, we calculate $ \text{det}(A) $:
\[
\text{det}(A) = 15(11) - 8(22) = 165 - 176 = -11
\]

Next, we calculate $ \text{det}(A_w) $:
\[
\text{det}(A_w) = 15(121) - 22(85) = 1815 - 1870 = -55
\]

Finally, using Cramer's rule:
\[
w = \frac{-55}{-11} = 5
\]

Since $ h = 2w $, the cost of one whiskey $ h $ is:
\[
h = 2(5) = 10
\]

### Final Answer
The cost of one whiskey is **10**.

A round of drinks comprising four beers, two glasses of wine, seven cappuccinos and three whiskeys costs $85. A round of nine beers, one glass of wine, four cappuccinos and five whiskeys costs $121. The price of a beer is double the price of a cappuccino. The price of whiskey is double the price of a wine. Use Cramer's rule to find the cost of a glass of whiskey.

Learn how to apply Cramer's rule to find the cost of a whiskey from a system of linear equations involving beers, wine, cappuccinos, and whiskey prices. This problem explores key concepts of linear algebra, suitable for Grades 10-12.

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