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Sagot :
To determine which system the ordered pair [tex]\((3, -1)\)[/tex] is a solution of, we need to check each system of inequalities one by one.
1. System 1:
[tex]\[ \left\{ \begin{array}{l} y \geq x + 1 \\ y \leq -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y \geq x + 1 \implies -1 \geq 3 + 1 \implies -1 \geq 4 \quad \text{(False)} \][/tex]
[tex]\[ y \leq -2 \implies -1 \leq -2 \quad \text{(False)} \][/tex]
Since both conditions are false, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 1.
2. System 2:
[tex]\[ \left\{ \begin{array}{l} y \leq x + 1 \\ y \geq -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y \leq x + 1 \implies -1 \leq 3 + 1 \implies -1 \leq 4 \quad \text{(True)} \][/tex]
[tex]\[ y \geq -2 \implies -1 \geq -2 \quad \text{(True)} \][/tex]
Since both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is a solution to System 2.
3. System 3:
[tex]\[ \left\{ \begin{array}{l} y > x + 1 \\ y > -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y > x + 1 \implies -1 > 3 + 1 \implies -1 > 4 \quad \text{(False)} \][/tex]
[tex]\[ y > -2 \implies -1 > -2 \quad \text{(True)} \][/tex]
Since not both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 3.
4. System 4:
[tex]\[ \left\{ \begin{array}{l} y < x + 1 \\ y < -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y < x + 1 \implies -1 < 3 + 1 \implies -1 < 4 \quad \text{(True)} \][/tex]
[tex]\[ y < -2 \implies -1 < -2 \quad \text{(False)} \][/tex]
Since not both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 4.
After evaluating each system of inequalities, we find that the ordered pair [tex]\((3, -1)\)[/tex] satisfies System 2:
[tex]\[ \left\{ \begin{array}{l} y \leq x + 1 \\ y \geq -2 \end{array} \right. \][/tex]
1. System 1:
[tex]\[ \left\{ \begin{array}{l} y \geq x + 1 \\ y \leq -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y \geq x + 1 \implies -1 \geq 3 + 1 \implies -1 \geq 4 \quad \text{(False)} \][/tex]
[tex]\[ y \leq -2 \implies -1 \leq -2 \quad \text{(False)} \][/tex]
Since both conditions are false, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 1.
2. System 2:
[tex]\[ \left\{ \begin{array}{l} y \leq x + 1 \\ y \geq -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y \leq x + 1 \implies -1 \leq 3 + 1 \implies -1 \leq 4 \quad \text{(True)} \][/tex]
[tex]\[ y \geq -2 \implies -1 \geq -2 \quad \text{(True)} \][/tex]
Since both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is a solution to System 2.
3. System 3:
[tex]\[ \left\{ \begin{array}{l} y > x + 1 \\ y > -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y > x + 1 \implies -1 > 3 + 1 \implies -1 > 4 \quad \text{(False)} \][/tex]
[tex]\[ y > -2 \implies -1 > -2 \quad \text{(True)} \][/tex]
Since not both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 3.
4. System 4:
[tex]\[ \left\{ \begin{array}{l} y < x + 1 \\ y < -2 \end{array} \right. \][/tex]
Let's check each condition:
[tex]\[ y < x + 1 \implies -1 < 3 + 1 \implies -1 < 4 \quad \text{(True)} \][/tex]
[tex]\[ y < -2 \implies -1 < -2 \quad \text{(False)} \][/tex]
Since not both conditions are true, the ordered pair [tex]\((3, -1)\)[/tex] is not a solution to System 4.
After evaluating each system of inequalities, we find that the ordered pair [tex]\((3, -1)\)[/tex] satisfies System 2:
[tex]\[ \left\{ \begin{array}{l} y \leq x + 1 \\ y \geq -2 \end{array} \right. \][/tex]
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