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Sagot :
To solve the system of equations:
[tex]\[ \begin{aligned} 9x - 2y &= 2 \\ -63x + 14y &= -14 \end{aligned} \][/tex]
Follow the detailed steps below.
### Step 1: Write the equations in standard form.
The equations are already provided in standard form:
1. [tex]\(9x - 2y = 2\)[/tex]
2. [tex]\(-63x + 14y = -14\)[/tex]
### Step 2: Form the augmented matrix.
We represent the system of equations as an augmented matrix:
[tex]\[ \begin{bmatrix} 9 & -2 & | & 2 \\ -63 & 14 & | & -14 \end{bmatrix} \][/tex]
### Step 3: Simplify the second equation if possible.
Notice that the second row is a multiple of the first row. Specifically, if we multiply the first row by [tex]\(-7\)[/tex], we get the second row:
[tex]\[ -7 \cdot (9x - 2y) = -7 \cdot 2 \][/tex]
which simplifies to:
[tex]\[ -63x + 14y = -14 \][/tex]
This means the second equation is not independent, and confirms that the system may not have a unique solution because it represents the same line.
### Step 4: Solve the first equation for one of the variables.
Let's solve the first equation for [tex]\(x\)[/tex]:
[tex]\[ 9x - 2y = 2 \implies 9x = 2y + 2 \implies x = \frac{2y + 2}{9} \][/tex]
### Step 5: Substitute [tex]\(x\)[/tex] into the second equation (if needed, but already confirmed linear dependency).
To verify, substitute [tex]\(x = \frac{2y + 2}{9}\)[/tex] into the second equation:
[tex]\[ -63 \left(\frac{2y + 2}{9}\right) + 14y = -14 \][/tex]
By simplifying:
[tex]\[ -63 \cdot \frac{2y + 2}{9} + 14y = -14 \implies -7(2y + 2) + 14y = -14 \implies -14y - 14 + 14y = -14 \implies -14 = -14 \][/tex]
This validates that the second equation does not provide new information, confirming the two lines are dependent.
### Step 6: Express the solution.
Since the equations are dependent, we have infinite solutions along the line described by the first (or second) equation. The solutions can be expressed in terms of [tex]\(y\)[/tex].
If we let [tex]\(y = t\)[/tex] (a parameter), then:
[tex]\[ x = \frac{2t + 2}{9} \][/tex]
The solution set can be represented as:
[tex]\[ \{ (x, y) \mid x = \frac{2y + 2}{9}, y \in \mathbb{R} \} \][/tex]
### Conclusion:
The system of equations does not have a unique solution because the equations are not independent; they represent the same line. Therefore, the solutions are infinitely many and can be expressed parametrically as:
[tex]\[ x = \frac{2y + 2}{9} \][/tex]
where [tex]\(y\)[/tex] is any real number.
[tex]\[ \begin{aligned} 9x - 2y &= 2 \\ -63x + 14y &= -14 \end{aligned} \][/tex]
Follow the detailed steps below.
### Step 1: Write the equations in standard form.
The equations are already provided in standard form:
1. [tex]\(9x - 2y = 2\)[/tex]
2. [tex]\(-63x + 14y = -14\)[/tex]
### Step 2: Form the augmented matrix.
We represent the system of equations as an augmented matrix:
[tex]\[ \begin{bmatrix} 9 & -2 & | & 2 \\ -63 & 14 & | & -14 \end{bmatrix} \][/tex]
### Step 3: Simplify the second equation if possible.
Notice that the second row is a multiple of the first row. Specifically, if we multiply the first row by [tex]\(-7\)[/tex], we get the second row:
[tex]\[ -7 \cdot (9x - 2y) = -7 \cdot 2 \][/tex]
which simplifies to:
[tex]\[ -63x + 14y = -14 \][/tex]
This means the second equation is not independent, and confirms that the system may not have a unique solution because it represents the same line.
### Step 4: Solve the first equation for one of the variables.
Let's solve the first equation for [tex]\(x\)[/tex]:
[tex]\[ 9x - 2y = 2 \implies 9x = 2y + 2 \implies x = \frac{2y + 2}{9} \][/tex]
### Step 5: Substitute [tex]\(x\)[/tex] into the second equation (if needed, but already confirmed linear dependency).
To verify, substitute [tex]\(x = \frac{2y + 2}{9}\)[/tex] into the second equation:
[tex]\[ -63 \left(\frac{2y + 2}{9}\right) + 14y = -14 \][/tex]
By simplifying:
[tex]\[ -63 \cdot \frac{2y + 2}{9} + 14y = -14 \implies -7(2y + 2) + 14y = -14 \implies -14y - 14 + 14y = -14 \implies -14 = -14 \][/tex]
This validates that the second equation does not provide new information, confirming the two lines are dependent.
### Step 6: Express the solution.
Since the equations are dependent, we have infinite solutions along the line described by the first (or second) equation. The solutions can be expressed in terms of [tex]\(y\)[/tex].
If we let [tex]\(y = t\)[/tex] (a parameter), then:
[tex]\[ x = \frac{2t + 2}{9} \][/tex]
The solution set can be represented as:
[tex]\[ \{ (x, y) \mid x = \frac{2y + 2}{9}, y \in \mathbb{R} \} \][/tex]
### Conclusion:
The system of equations does not have a unique solution because the equations are not independent; they represent the same line. Therefore, the solutions are infinitely many and can be expressed parametrically as:
[tex]\[ x = \frac{2y + 2}{9} \][/tex]
where [tex]\(y\)[/tex] is any real number.
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