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Sagot :
To determine the costs of the Ferris wheel, water slide, and merry-go-round tickets given the expenses and the purchases by Ryan, Michelle, and Erwin, we need to use a system of linear equations. Let's set up the equations and the corresponding matrices to solve for [tex]\(x\)[/tex] (cost of the Ferris wheel ticket), [tex]\(y\)[/tex] (cost of the water slide ticket), and [tex]\(z\)[/tex] (cost of the merry-go-round ticket).
Given:
- Ryan spent [tex]\( \$ 13.50 \)[/tex].
- Michelle spent [tex]\( \$ 16.50 \)[/tex].
- Erwin spent [tex]\( \$ 14 \)[/tex].
Ryan bought:
- 3 Ferris wheel tickets
- 2 water slide tickets
- 0 merry-go-round tickets
Michelle bought:
- 1 Ferris wheel ticket
- 0 water slide tickets
- 4 merry-go-round tickets
Erwin bought:
- 3 Ferris wheel tickets
- 1 water slide ticket
- 1 merry-go-round ticket
We can form the following system of linear equations:
1. [tex]\( 3x + 2y + 0z = 13.50 \)[/tex]
2. [tex]\( 1x + 0y + 4z = 16.50 \)[/tex]
3. [tex]\( 3x + 1y + 1z = 14 \)[/tex]
These equations can be written in matrix form as:
[tex]\[ A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]
Where:
[tex]\[ A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 0 & 4 \\ 3 & 1 & 1 \end{bmatrix} \][/tex]
[tex]\[ B = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]
To solve for [tex]\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)[/tex], we need to find the inverse of matrix [tex]\( A \)[/tex] and multiply it by matrix [tex]\( B \)[/tex]:
[tex]\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} B \][/tex]
Given the final results from solving this equation:
[tex]\[ X = \begin{bmatrix} 2.5 \\ 3.0 \\ 3.5 \end{bmatrix} \][/tex]
Therefore, the costs of the tickets are:
- Ferris wheel ticket cost, [tex]\( x \)[/tex]: [tex]\( \$ 2.50 \)[/tex]
- Water slide ticket cost, [tex]\( y \)[/tex]: [tex]\( \$ 3.00 \)[/tex]
- Merry-go-round ticket cost, [tex]\( z \)[/tex]: [tex]\( \$ 3.50 \)[/tex]
Thus, the values for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] are:
[tex]\[ x = 2.5, \quad y = 3.0, \quad z = 3.5 \][/tex]
Given:
- Ryan spent [tex]\( \$ 13.50 \)[/tex].
- Michelle spent [tex]\( \$ 16.50 \)[/tex].
- Erwin spent [tex]\( \$ 14 \)[/tex].
Ryan bought:
- 3 Ferris wheel tickets
- 2 water slide tickets
- 0 merry-go-round tickets
Michelle bought:
- 1 Ferris wheel ticket
- 0 water slide tickets
- 4 merry-go-round tickets
Erwin bought:
- 3 Ferris wheel tickets
- 1 water slide ticket
- 1 merry-go-round ticket
We can form the following system of linear equations:
1. [tex]\( 3x + 2y + 0z = 13.50 \)[/tex]
2. [tex]\( 1x + 0y + 4z = 16.50 \)[/tex]
3. [tex]\( 3x + 1y + 1z = 14 \)[/tex]
These equations can be written in matrix form as:
[tex]\[ A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]
Where:
[tex]\[ A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 0 & 4 \\ 3 & 1 & 1 \end{bmatrix} \][/tex]
[tex]\[ B = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]
To solve for [tex]\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)[/tex], we need to find the inverse of matrix [tex]\( A \)[/tex] and multiply it by matrix [tex]\( B \)[/tex]:
[tex]\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} B \][/tex]
Given the final results from solving this equation:
[tex]\[ X = \begin{bmatrix} 2.5 \\ 3.0 \\ 3.5 \end{bmatrix} \][/tex]
Therefore, the costs of the tickets are:
- Ferris wheel ticket cost, [tex]\( x \)[/tex]: [tex]\( \$ 2.50 \)[/tex]
- Water slide ticket cost, [tex]\( y \)[/tex]: [tex]\( \$ 3.00 \)[/tex]
- Merry-go-round ticket cost, [tex]\( z \)[/tex]: [tex]\( \$ 3.50 \)[/tex]
Thus, the values for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] are:
[tex]\[ x = 2.5, \quad y = 3.0, \quad z = 3.5 \][/tex]
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