Answer :
To determine which ordered pair satisfies both given inequalities, we can analyze each inequality step by step for the given ordered pairs. The inequalities are:
[tex]\[ \begin{array}{l} y \leq -x + 1 \\ y > x \end{array} \][/tex]
Let's check each ordered pair:
1. Ordered pair \((-1, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -(-1) + 1 \implies 0 \leq 1 + 1 \implies 0 \leq 2 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > -1 \quad (\text{True}) \][/tex]
Since this ordered pair satisfies both inequalities, \((-1, 0)\) is a valid solution.
2. Ordered pair \((0, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -0 + 1 \implies 0 \leq 1 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > 0 \quad (\text{False}) \][/tex]
Since the second inequality is not satisfied, \((0, 0)\) is not a valid solution.
3. Ordered pair \((1, 1)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 1 \leq -1 + 1 \implies 1 \leq 0 \quad (\text{False}) \][/tex]
- Since the first inequality is not satisfied, \((1, 1)\) is not a valid solution.
4. Ordered pair \((-2, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -(-2) + 1 \implies 0 \leq 2 + 1 \implies 0 \leq 3 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > -2 \quad (\text{True}) \][/tex]
Since this ordered pair satisfies both inequalities, \((-2, 0)\) is also a valid solution.
Therefore, based on the analysis of the given ordered pairs, the pair \((-1, 0)\) satisfies both inequalities:
[tex]\[ \begin{array}{l} y \leq -x + 1 \\ y > x \end{array} \][/tex]
Hence, the ordered pair [tex]\((-1, 0)\)[/tex] makes both inequalities true.
[tex]\[ \begin{array}{l} y \leq -x + 1 \\ y > x \end{array} \][/tex]
Let's check each ordered pair:
1. Ordered pair \((-1, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -(-1) + 1 \implies 0 \leq 1 + 1 \implies 0 \leq 2 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > -1 \quad (\text{True}) \][/tex]
Since this ordered pair satisfies both inequalities, \((-1, 0)\) is a valid solution.
2. Ordered pair \((0, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -0 + 1 \implies 0 \leq 1 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > 0 \quad (\text{False}) \][/tex]
Since the second inequality is not satisfied, \((0, 0)\) is not a valid solution.
3. Ordered pair \((1, 1)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 1 \leq -1 + 1 \implies 1 \leq 0 \quad (\text{False}) \][/tex]
- Since the first inequality is not satisfied, \((1, 1)\) is not a valid solution.
4. Ordered pair \((-2, 0)\):
- For the first inequality \(y \leq -x + 1\):
[tex]\[ 0 \leq -(-2) + 1 \implies 0 \leq 2 + 1 \implies 0 \leq 3 \quad (\text{True}) \][/tex]
- For the second inequality \(y > x\):
[tex]\[ 0 > -2 \quad (\text{True}) \][/tex]
Since this ordered pair satisfies both inequalities, \((-2, 0)\) is also a valid solution.
Therefore, based on the analysis of the given ordered pairs, the pair \((-1, 0)\) satisfies both inequalities:
[tex]\[ \begin{array}{l} y \leq -x + 1 \\ y > x \end{array} \][/tex]
Hence, the ordered pair [tex]\((-1, 0)\)[/tex] makes both inequalities true.
