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Sagot :
Let's solve the given system of linear equations and verify the solution using matrix methods:
The system of equations is:
[tex]\[ \begin{cases} 3x + 2y = 14 \\ 2x - 4y = 4 \end{cases} \][/tex]
### Step-by-Step Solution:
#### Part (a): Solve the System Using Substitution or Elimination Method
Choose the elimination method first to simplify both equations.
1. Equation 1:
[tex]\[ 3x + 2y = 14 \][/tex]
2. Equation 2:
[tex]\[ 2x - 4y = 4 \][/tex]
To eliminate one of the variables, let's manipulate the equations. We can achieve this by making the coefficients of [tex]\( y \)[/tex] in both equations the same.
Step 1: Multiply Equation 1 by 2:
[tex]\[ 2(3x + 2y) = 2(14) \\ 6x + 4y = 28 \][/tex]
Step 2: Write the modified system of equations:
[tex]\[ \begin{cases} 6x + 4y = 28 \\ 2x - 4y = 4 \end{cases} \][/tex]
Step 3: Add these two equations together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (6x + 4y) + (2x - 4y) = 28 + 4 \\ 8x = 32 \][/tex]
Step 4: Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{32}{8} \\ x = 4 \][/tex]
Step 5: Substitute [tex]\( x \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. Let's use Equation 1:
[tex]\[ 3(4) + 2y = 14 \\ 12 + 2y = 14 \\ 2y = 14 - 12 \\ 2y = 2 \\ y = 1 \][/tex]
So, the solution is [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex].
#### Part (b): Verify the Solution Using Matrix Methods
The system of equations can be written in matrix form [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] is the variable vector, and [tex]\( \mathbf{B} \)[/tex] is the constant vector.
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 2 & -4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 14 \\ 4 \end{pmatrix} \][/tex]
To solve for [tex]\( \mathbf{x} \)[/tex], we need the inverse of matrix [tex]\( A \)[/tex] denoted as [tex]\( A^{-1} \)[/tex].
Given [tex]\( A^{-1} \)[/tex] and result of [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.25 & 0.125 \\ 0.125 & -0.1875 \end{pmatrix} \][/tex]
The solution vector [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} = \begin{pmatrix} 0.25 & 0.125 \\ 0.125 & -0.1875 \end{pmatrix} \begin{pmatrix} 14 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} \][/tex]
Thus, the solution from matrix method confirms:
[tex]\[ x = 4, \quad y = 1 \][/tex]
To verify, we can substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex] back into the original equations:
1. For [tex]\( 3x + 2y = 14 \)[/tex]:
[tex]\[ 3(4) + 2(1) = 12 + 2 = 14 \][/tex]
2. For [tex]\( 2x - 4y = 4 \)[/tex]:
[tex]\[ 2(4) - 4(1) = 8 - 4 = 4 \][/tex]
Both equations are satisfied, thereby confirming our solution:
[tex]\[ x = 4, \quad y = 1 \][/tex]
The system of equations is:
[tex]\[ \begin{cases} 3x + 2y = 14 \\ 2x - 4y = 4 \end{cases} \][/tex]
### Step-by-Step Solution:
#### Part (a): Solve the System Using Substitution or Elimination Method
Choose the elimination method first to simplify both equations.
1. Equation 1:
[tex]\[ 3x + 2y = 14 \][/tex]
2. Equation 2:
[tex]\[ 2x - 4y = 4 \][/tex]
To eliminate one of the variables, let's manipulate the equations. We can achieve this by making the coefficients of [tex]\( y \)[/tex] in both equations the same.
Step 1: Multiply Equation 1 by 2:
[tex]\[ 2(3x + 2y) = 2(14) \\ 6x + 4y = 28 \][/tex]
Step 2: Write the modified system of equations:
[tex]\[ \begin{cases} 6x + 4y = 28 \\ 2x - 4y = 4 \end{cases} \][/tex]
Step 3: Add these two equations together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (6x + 4y) + (2x - 4y) = 28 + 4 \\ 8x = 32 \][/tex]
Step 4: Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{32}{8} \\ x = 4 \][/tex]
Step 5: Substitute [tex]\( x \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. Let's use Equation 1:
[tex]\[ 3(4) + 2y = 14 \\ 12 + 2y = 14 \\ 2y = 14 - 12 \\ 2y = 2 \\ y = 1 \][/tex]
So, the solution is [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex].
#### Part (b): Verify the Solution Using Matrix Methods
The system of equations can be written in matrix form [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] is the variable vector, and [tex]\( \mathbf{B} \)[/tex] is the constant vector.
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 2 & -4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 14 \\ 4 \end{pmatrix} \][/tex]
To solve for [tex]\( \mathbf{x} \)[/tex], we need the inverse of matrix [tex]\( A \)[/tex] denoted as [tex]\( A^{-1} \)[/tex].
Given [tex]\( A^{-1} \)[/tex] and result of [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.25 & 0.125 \\ 0.125 & -0.1875 \end{pmatrix} \][/tex]
The solution vector [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} = \begin{pmatrix} 0.25 & 0.125 \\ 0.125 & -0.1875 \end{pmatrix} \begin{pmatrix} 14 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} \][/tex]
Thus, the solution from matrix method confirms:
[tex]\[ x = 4, \quad y = 1 \][/tex]
To verify, we can substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex] back into the original equations:
1. For [tex]\( 3x + 2y = 14 \)[/tex]:
[tex]\[ 3(4) + 2(1) = 12 + 2 = 14 \][/tex]
2. For [tex]\( 2x - 4y = 4 \)[/tex]:
[tex]\[ 2(4) - 4(1) = 8 - 4 = 4 \][/tex]
Both equations are satisfied, thereby confirming our solution:
[tex]\[ x = 4, \quad y = 1 \][/tex]
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