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Sagot :
Sure! To solve the given system of equations, let's follow these steps in detail:
The system of equations is:
[tex]\[ \begin{cases} 5x + 2y = 7 \\ -2x + 6y = 9 \end{cases} \][/tex]
### Step 1: Write the system of equations in matrix form
The system can be represented in matrix form as \( A \vec{x} = \vec{B} \), where:
[tex]\[ A = \begin{pmatrix} 5 & 2 \\ -2 & 6 \end{pmatrix}, \quad \vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} 7 \\ 9 \end{pmatrix} \][/tex]
### Step 2: Calculate the determinant of matrix \( A \)
The determinant of \( A \) is:
[tex]\[ \text{det}(A) = (5 \cdot 6) - (2 \cdot -2) = 30 + 4 = 34 \][/tex]
Since the determinant is non-zero, the matrix \( A \) is invertible, and the system has a unique solution.
### Step 3: Find the inverse of matrix \( A \)
The inverse \( A^{-1} \) is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} 6 & -2 \\ 2 & 5 \end{pmatrix} = \frac{1}{34} \begin{pmatrix} 6 & -2 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} \frac{6}{34} & \frac{-2}{34} \\ \frac{2}{34} & \frac{5}{34} \end{pmatrix} \][/tex]
### Step 4: Multiply the inverse of \( A \) with \( B \) to find \( \vec{x} \)
[tex]\[ \vec{x} = A^{-1} \vec{B} = \begin{pmatrix} \frac{6}{34} & \frac{-2}{34} \\ \frac{2}{34} & \frac{5}{34} \end{pmatrix} \begin{pmatrix} 7 \\ 9 \end{pmatrix} \][/tex]
### Step 5: Perform the matrix multiplication
[tex]\[ x = \left( \frac{6}{34} \cdot 7 + \frac{-2}{34} \cdot 9 \right) = \left( \frac{42}{34} - \frac{18}{34} \right) = \frac{24}{34} = \frac{12}{17} \approx 0.705882 \][/tex]
[tex]\[ y = \left( \frac{2}{34} \cdot 7 + \frac{5}{34} \cdot 9 \right) = \left( \frac{14}{34} + \frac{45}{34} \right) = \frac{59}{34} \approx 1.735294 \][/tex]
### Step 6: Extract the y-coordinate and round to the nearest tenth
The calculated value for \( y \) is approximately \( 1.735294 \).
Rounding \( 1.735294 \) to the nearest tenth gives \( 1.7 \).
So, the y-coordinate of the solution, rounded to the nearest tenth, is:
[tex]\[ \boxed{1.7} \][/tex]
The system of equations is:
[tex]\[ \begin{cases} 5x + 2y = 7 \\ -2x + 6y = 9 \end{cases} \][/tex]
### Step 1: Write the system of equations in matrix form
The system can be represented in matrix form as \( A \vec{x} = \vec{B} \), where:
[tex]\[ A = \begin{pmatrix} 5 & 2 \\ -2 & 6 \end{pmatrix}, \quad \vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} 7 \\ 9 \end{pmatrix} \][/tex]
### Step 2: Calculate the determinant of matrix \( A \)
The determinant of \( A \) is:
[tex]\[ \text{det}(A) = (5 \cdot 6) - (2 \cdot -2) = 30 + 4 = 34 \][/tex]
Since the determinant is non-zero, the matrix \( A \) is invertible, and the system has a unique solution.
### Step 3: Find the inverse of matrix \( A \)
The inverse \( A^{-1} \) is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} 6 & -2 \\ 2 & 5 \end{pmatrix} = \frac{1}{34} \begin{pmatrix} 6 & -2 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} \frac{6}{34} & \frac{-2}{34} \\ \frac{2}{34} & \frac{5}{34} \end{pmatrix} \][/tex]
### Step 4: Multiply the inverse of \( A \) with \( B \) to find \( \vec{x} \)
[tex]\[ \vec{x} = A^{-1} \vec{B} = \begin{pmatrix} \frac{6}{34} & \frac{-2}{34} \\ \frac{2}{34} & \frac{5}{34} \end{pmatrix} \begin{pmatrix} 7 \\ 9 \end{pmatrix} \][/tex]
### Step 5: Perform the matrix multiplication
[tex]\[ x = \left( \frac{6}{34} \cdot 7 + \frac{-2}{34} \cdot 9 \right) = \left( \frac{42}{34} - \frac{18}{34} \right) = \frac{24}{34} = \frac{12}{17} \approx 0.705882 \][/tex]
[tex]\[ y = \left( \frac{2}{34} \cdot 7 + \frac{5}{34} \cdot 9 \right) = \left( \frac{14}{34} + \frac{45}{34} \right) = \frac{59}{34} \approx 1.735294 \][/tex]
### Step 6: Extract the y-coordinate and round to the nearest tenth
The calculated value for \( y \) is approximately \( 1.735294 \).
Rounding \( 1.735294 \) to the nearest tenth gives \( 1.7 \).
So, the y-coordinate of the solution, rounded to the nearest tenth, is:
[tex]\[ \boxed{1.7} \][/tex]
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