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To determine which of the ordered pairs [tex]\((6, -2)\)[/tex], [tex]\((6, 0.5)\)[/tex], [tex]\((6, 5)\)[/tex], or [tex]\((6, 8)\)[/tex] are solutions to the given system of inequalities, we need to check each pair against both inequalities.
The system of inequalities is:
1. [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]
2. [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]
Let's check each ordered pair:
1. For the pair [tex]\((6, -2)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ -2 \leq \frac{2}{3}(6) + 1 \implies -2 \leq 4 + 1 \implies -2 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ -2 > -\frac{1}{4}(6) + 2 \implies -2 > -1.5 + 2 \implies -2 > 0.5 \quad \text{(False)} \][/tex]
Since one of the inequalities is false, [tex]\((6, -2)\)[/tex] is not a solution.
2. For the pair [tex]\((6, 0.5)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 0.5 \leq \frac{2}{3}(6) + 1 \implies 0.5 \leq 4 + 1 \implies 0.5 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ 0.5 > -\frac{1}{4}(6) + 2 \implies 0.5 > -1.5 + 2 \implies 0.5 > 0.5 \quad \text{(False)} \][/tex]
Since one of the inequalities is false, [tex]\((6, 0.5)\)[/tex] is not a solution.
3. For the pair [tex]\((6, 5)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 5 \leq \frac{2}{3}(6) + 1 \implies 5 \leq 4 + 1 \implies 5 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ 5 > -\frac{1}{4}(6) + 2 \implies 5 > -1.5 + 2 \implies 5 > 0.5 \quad \text{(True)} \][/tex]
Since both inequalities are true, [tex]\((6, 5)\)[/tex] is a solution.
4. For the pair [tex]\((6, 8)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 8 \leq \frac{2}{3}(6) + 1 \implies 8 \leq 4 + 1 \implies 8 \leq 5 \quad \text{(False)} \][/tex]
- Since the first inequality is false, [tex]\((6, 8)\)[/tex] is not a solution.
Therefore, the ordered pair that is included in the solution to the system is [tex]\((6, 5)\)[/tex].
The system of inequalities is:
1. [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]
2. [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]
Let's check each ordered pair:
1. For the pair [tex]\((6, -2)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ -2 \leq \frac{2}{3}(6) + 1 \implies -2 \leq 4 + 1 \implies -2 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ -2 > -\frac{1}{4}(6) + 2 \implies -2 > -1.5 + 2 \implies -2 > 0.5 \quad \text{(False)} \][/tex]
Since one of the inequalities is false, [tex]\((6, -2)\)[/tex] is not a solution.
2. For the pair [tex]\((6, 0.5)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 0.5 \leq \frac{2}{3}(6) + 1 \implies 0.5 \leq 4 + 1 \implies 0.5 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ 0.5 > -\frac{1}{4}(6) + 2 \implies 0.5 > -1.5 + 2 \implies 0.5 > 0.5 \quad \text{(False)} \][/tex]
Since one of the inequalities is false, [tex]\((6, 0.5)\)[/tex] is not a solution.
3. For the pair [tex]\((6, 5)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 5 \leq \frac{2}{3}(6) + 1 \implies 5 \leq 4 + 1 \implies 5 \leq 5 \quad \text{(True)} \][/tex]
- Check inequality [tex]\( y > -\frac{1}{4}x + 2 \)[/tex]:
[tex]\[ 5 > -\frac{1}{4}(6) + 2 \implies 5 > -1.5 + 2 \implies 5 > 0.5 \quad \text{(True)} \][/tex]
Since both inequalities are true, [tex]\((6, 5)\)[/tex] is a solution.
4. For the pair [tex]\((6, 8)\)[/tex]:
- Check inequality [tex]\( y \leq \frac{2}{3}x + 1 \)[/tex]:
[tex]\[ 8 \leq \frac{2}{3}(6) + 1 \implies 8 \leq 4 + 1 \implies 8 \leq 5 \quad \text{(False)} \][/tex]
- Since the first inequality is false, [tex]\((6, 8)\)[/tex] is not a solution.
Therefore, the ordered pair that is included in the solution to the system is [tex]\((6, 5)\)[/tex].
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