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To solve the system of equations
[tex]\[ \begin{array}{l} 3x + 4y = 5 \\ 6x - y = 1 \end{array} \][/tex]
we can use the method of determinants, also known as Cramer's Rule. Here are the steps to find the solution [tex]\( (x, y) \)[/tex]:
### Step 1: Set up the equations
We start by setting up the two linear equations:
1. [tex]\( 3x + 4y = 5 \)[/tex]
2. [tex]\( 6x - y = 1 \)[/tex]
### Step 2: Formulate the coefficient matrix and the constant matrix
We identify the coefficients from the given equations:
For the equation [tex]\( 3x + 4y = 5 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] (a₁) = 3
- Coefficient of [tex]\( y \)[/tex] (b₁) = 4
- Constant term (c₁) = 5
For the equation [tex]\( 6x - y = 1 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] (a₂) = 6
- Coefficient of [tex]\( y \)[/tex] (b₂) = -1
- Constant term (c₂) = 1
### Step 3: Calculate the determinant of the coefficient matrix ([tex]\( \Delta \)[/tex])
The determinant [tex]\(\Delta\)[/tex] of the coefficient matrix is given by:
[tex]\[ \Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1 \][/tex]
Substituting the values:
[tex]\[ \Delta = (3 \cdot -1) - (6 \cdot 4) = -3 - 24 = -27 \][/tex]
### Step 4: Calculate the determinant for [tex]\( x \)[/tex] ([tex]\( \Delta_x \)[/tex])
The determinant [tex]\( \Delta_x \)[/tex] is obtained by replacing the first column of the coefficient matrix with the constants from the right-hand side:
[tex]\[ \Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1 \][/tex]
Substituting the values:
[tex]\[ \Delta_x = (5 \cdot -1) - (1 \cdot 4) = -5 - 4 = -9 \][/tex]
### Step 5: Calculate the determinant for [tex]\( y \)[/tex] ([tex]\( \Delta_y \)[/tex])
The determinant [tex]\( \Delta_y \)[/tex] is obtained by replacing the second column of the coefficient matrix with the constants from the right-hand side:
[tex]\[ \Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1 \][/tex]
Substituting the values:
[tex]\[ \Delta_y = (3 \cdot 1) - (6 \cdot 5) = 3 - 30 = -27 \][/tex]
### Step 6: Solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
Using Cramer's Rule:
[tex]\[ x = \frac{\Delta_x}{\Delta} \][/tex]
[tex]\[ y = \frac{\Delta_y}{\Delta} \][/tex]
Substituting the values:
[tex]\[ x = \frac{-9}{-27} = \frac{1}{3} \quad \text{or} \quad 0.3333 \][/tex]
[tex]\[ y = \frac{-27}{-27} = 1 \][/tex]
### Step 7: State the solution
The solution to the system of equations is:
[tex]\[ x = \frac{1}{3}, \quad y = 1 \][/tex]
Or, in decimal form:
[tex]\[ x \approx 0.3333, \quad y = 1 \][/tex]
Therefore, the values that simultaneously satisfy both equations are [tex]\( x \approx 0.3333 \)[/tex] and [tex]\( y = 1 \)[/tex].
[tex]\[ \begin{array}{l} 3x + 4y = 5 \\ 6x - y = 1 \end{array} \][/tex]
we can use the method of determinants, also known as Cramer's Rule. Here are the steps to find the solution [tex]\( (x, y) \)[/tex]:
### Step 1: Set up the equations
We start by setting up the two linear equations:
1. [tex]\( 3x + 4y = 5 \)[/tex]
2. [tex]\( 6x - y = 1 \)[/tex]
### Step 2: Formulate the coefficient matrix and the constant matrix
We identify the coefficients from the given equations:
For the equation [tex]\( 3x + 4y = 5 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] (a₁) = 3
- Coefficient of [tex]\( y \)[/tex] (b₁) = 4
- Constant term (c₁) = 5
For the equation [tex]\( 6x - y = 1 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] (a₂) = 6
- Coefficient of [tex]\( y \)[/tex] (b₂) = -1
- Constant term (c₂) = 1
### Step 3: Calculate the determinant of the coefficient matrix ([tex]\( \Delta \)[/tex])
The determinant [tex]\(\Delta\)[/tex] of the coefficient matrix is given by:
[tex]\[ \Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1 \][/tex]
Substituting the values:
[tex]\[ \Delta = (3 \cdot -1) - (6 \cdot 4) = -3 - 24 = -27 \][/tex]
### Step 4: Calculate the determinant for [tex]\( x \)[/tex] ([tex]\( \Delta_x \)[/tex])
The determinant [tex]\( \Delta_x \)[/tex] is obtained by replacing the first column of the coefficient matrix with the constants from the right-hand side:
[tex]\[ \Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1 \][/tex]
Substituting the values:
[tex]\[ \Delta_x = (5 \cdot -1) - (1 \cdot 4) = -5 - 4 = -9 \][/tex]
### Step 5: Calculate the determinant for [tex]\( y \)[/tex] ([tex]\( \Delta_y \)[/tex])
The determinant [tex]\( \Delta_y \)[/tex] is obtained by replacing the second column of the coefficient matrix with the constants from the right-hand side:
[tex]\[ \Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1 \][/tex]
Substituting the values:
[tex]\[ \Delta_y = (3 \cdot 1) - (6 \cdot 5) = 3 - 30 = -27 \][/tex]
### Step 6: Solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
Using Cramer's Rule:
[tex]\[ x = \frac{\Delta_x}{\Delta} \][/tex]
[tex]\[ y = \frac{\Delta_y}{\Delta} \][/tex]
Substituting the values:
[tex]\[ x = \frac{-9}{-27} = \frac{1}{3} \quad \text{or} \quad 0.3333 \][/tex]
[tex]\[ y = \frac{-27}{-27} = 1 \][/tex]
### Step 7: State the solution
The solution to the system of equations is:
[tex]\[ x = \frac{1}{3}, \quad y = 1 \][/tex]
Or, in decimal form:
[tex]\[ x \approx 0.3333, \quad y = 1 \][/tex]
Therefore, the values that simultaneously satisfy both equations are [tex]\( x \approx 0.3333 \)[/tex] and [tex]\( y = 1 \)[/tex].
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