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Sagot :
Certainly! Let's work through the problem step by step.
The given system of linear equations is:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \quad \text{(Equation 1)} \\ 4x - 14y = -68 \quad \text{(Equation 2)} \end{array} \][/tex]
We want to replace one of these equations by the sum of that equation and a multiple of the other. We'll accomplish this by eliminating one variable through algebraic manipulation, thus ensuring that the new system has the same solutions as the original system.
### Step 1: Multiply Equation 2 to Facilitate Addition
First, we will multiply Equation 2 by 2 to align it for addition with Equation 1. This is done because multiplying by 2 will make the coefficients of [tex]\(x\)[/tex] in both equations the same, which simplifies the elimination process.
[tex]\[ 2 \cdot (4x - 14y) = 2 \cdot (-68) \][/tex]
This gives us:
[tex]\[ 8x - 28y = -136 \quad \text{(Modified Equation 2)} \][/tex]
### Step 2: Add Equation 1 and Modified Equation 2
Now we will add Equation 1 and Modified Equation 2 to eliminate [tex]\(x\)[/tex]:
[tex]\[ (8x + 7y) + (8x - 28y) = 39 + (-136) \][/tex]
Simplifying this:
[tex]\[ 8x + 8x + 7y - 28y = 39 - 136 \][/tex]
Combine like terms:
[tex]\[ 16x - 21y = -97 \quad \text{(New Equation)} \][/tex]
So the new system of equations is:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \quad \text{(Original Equation)} \\ 16x - 21y = -97 \quad \text{(New Equation)} \end{array} \][/tex]
### Step 3: Verify Solutions of Both Systems
The key here is to ensure both systems have the same solutions.
1. Original System:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \\ 4x - 14y = -68 \end{array} \][/tex]
Solving these equations, you find:
[tex]\[ x = \frac{1}{2}, \quad y = 5 \][/tex]
2. Modified System:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \\ 16x - 21y = -97 \end{array} \][/tex]
Solving these equations, you will find the same solutions:
[tex]\[ x = \frac{1}{2}, \quad y = 5 \][/tex]
### Conclusion:
Adding a multiple of one equation to another results in a new equation that still maintains the relationships and solutions of the original system. Hence, replacing one equation in the system with the sum of that equation and a multiple of the other does indeed produce a system with the same solutions.
Thus, we have confirmed that the new system:
[tex]\[\begin{array}{l} 8x + 7y = 39 \\ 16x - 21y = -97 \end{array}\][/tex]
has the same solution [tex]\((x, y) = \left(\frac{1}{2}, 5\right)\)[/tex] as the original system:
[tex]\[\begin{array}{l} 8x + 7y = 39 \\ 4x - 14y = -68 \end{array}\][/tex]
The given system of linear equations is:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \quad \text{(Equation 1)} \\ 4x - 14y = -68 \quad \text{(Equation 2)} \end{array} \][/tex]
We want to replace one of these equations by the sum of that equation and a multiple of the other. We'll accomplish this by eliminating one variable through algebraic manipulation, thus ensuring that the new system has the same solutions as the original system.
### Step 1: Multiply Equation 2 to Facilitate Addition
First, we will multiply Equation 2 by 2 to align it for addition with Equation 1. This is done because multiplying by 2 will make the coefficients of [tex]\(x\)[/tex] in both equations the same, which simplifies the elimination process.
[tex]\[ 2 \cdot (4x - 14y) = 2 \cdot (-68) \][/tex]
This gives us:
[tex]\[ 8x - 28y = -136 \quad \text{(Modified Equation 2)} \][/tex]
### Step 2: Add Equation 1 and Modified Equation 2
Now we will add Equation 1 and Modified Equation 2 to eliminate [tex]\(x\)[/tex]:
[tex]\[ (8x + 7y) + (8x - 28y) = 39 + (-136) \][/tex]
Simplifying this:
[tex]\[ 8x + 8x + 7y - 28y = 39 - 136 \][/tex]
Combine like terms:
[tex]\[ 16x - 21y = -97 \quad \text{(New Equation)} \][/tex]
So the new system of equations is:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \quad \text{(Original Equation)} \\ 16x - 21y = -97 \quad \text{(New Equation)} \end{array} \][/tex]
### Step 3: Verify Solutions of Both Systems
The key here is to ensure both systems have the same solutions.
1. Original System:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \\ 4x - 14y = -68 \end{array} \][/tex]
Solving these equations, you find:
[tex]\[ x = \frac{1}{2}, \quad y = 5 \][/tex]
2. Modified System:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \\ 16x - 21y = -97 \end{array} \][/tex]
Solving these equations, you will find the same solutions:
[tex]\[ x = \frac{1}{2}, \quad y = 5 \][/tex]
### Conclusion:
Adding a multiple of one equation to another results in a new equation that still maintains the relationships and solutions of the original system. Hence, replacing one equation in the system with the sum of that equation and a multiple of the other does indeed produce a system with the same solutions.
Thus, we have confirmed that the new system:
[tex]\[\begin{array}{l} 8x + 7y = 39 \\ 16x - 21y = -97 \end{array}\][/tex]
has the same solution [tex]\((x, y) = \left(\frac{1}{2}, 5\right)\)[/tex] as the original system:
[tex]\[\begin{array}{l} 8x + 7y = 39 \\ 4x - 14y = -68 \end{array}\][/tex]
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