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To determine which ordered pairs satisfy the given system of inequalities, we will evaluate each pair by substituting [tex]\( x = 6 \)[/tex] into the inequalities and check the corresponding [tex]\( y \)[/tex]-values.
The system of inequalities is:
[tex]\[ \begin{array}{l} y \leq \frac{2}{3} x + 1 \\ y > -\frac{1}{4} x + 2 \end{array} \][/tex]
First, let's substitute [tex]\( x = 6 \)[/tex] into the inequalities to simplify them:
1. For the first inequality, [tex]\( y \leq \frac{2}{3} x + 1 \)[/tex]:
[tex]\[ y \leq \frac{2}{3} \cdot 6 + 1 = 4 + 1 = 5 \][/tex]
2. For the second inequality, [tex]\( y > -\frac{1}{4} x + 2 \)[/tex]:
[tex]\[ y > -\frac{1}{4} \cdot 6 + 2 = -1.5 + 2 = 0.5 \][/tex]
So, the simplified system of inequalities when [tex]\( x = 6 \)[/tex] is:
[tex]\[ \begin{array}{l} y \leq 5 \\ y > 0.5 \end{array} \][/tex]
Now, let's evaluate each ordered pair to see if they satisfy both conditions.
1. For the ordered pair [tex]\( (6, -2) \)[/tex]:
- Check the first inequality: [tex]\( -2 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( -2 > 0.5 \)[/tex] (False)
- Since one of the inequalities is false, [tex]\( (6, -2) \)[/tex] is not a solution.
2. For the ordered pair [tex]\( (6, 0.5) \)[/tex]:
- Check the first inequality: [tex]\( 0.5 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( 0.5 > 0.5 \)[/tex] (False)
- Since one of the inequalities is false, [tex]\( (6, 0.5) \)[/tex] is not a solution.
3. For the ordered pair [tex]\( (6, 5) \)[/tex]:
- Check the first inequality: [tex]\( 5 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( 5 > 0.5 \)[/tex] (True)
- Since both inequalities are true, [tex]\( (6, 5) \)[/tex] is a solution.
4. For the ordered pair [tex]\( (6, 8) \)[/tex]:
- Check the first inequality: [tex]\( 8 \leq 5 \)[/tex] (False)
- Check the second inequality: [tex]\( 8 > 0.5 \)[/tex] (True)
- Since one of the inequalities is false, [tex]\( (6, 8) \)[/tex] is not a solution.
The ordered pair that satisfies both inequalities in the system is [tex]\( (6, 5) \)[/tex].
The system of inequalities is:
[tex]\[ \begin{array}{l} y \leq \frac{2}{3} x + 1 \\ y > -\frac{1}{4} x + 2 \end{array} \][/tex]
First, let's substitute [tex]\( x = 6 \)[/tex] into the inequalities to simplify them:
1. For the first inequality, [tex]\( y \leq \frac{2}{3} x + 1 \)[/tex]:
[tex]\[ y \leq \frac{2}{3} \cdot 6 + 1 = 4 + 1 = 5 \][/tex]
2. For the second inequality, [tex]\( y > -\frac{1}{4} x + 2 \)[/tex]:
[tex]\[ y > -\frac{1}{4} \cdot 6 + 2 = -1.5 + 2 = 0.5 \][/tex]
So, the simplified system of inequalities when [tex]\( x = 6 \)[/tex] is:
[tex]\[ \begin{array}{l} y \leq 5 \\ y > 0.5 \end{array} \][/tex]
Now, let's evaluate each ordered pair to see if they satisfy both conditions.
1. For the ordered pair [tex]\( (6, -2) \)[/tex]:
- Check the first inequality: [tex]\( -2 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( -2 > 0.5 \)[/tex] (False)
- Since one of the inequalities is false, [tex]\( (6, -2) \)[/tex] is not a solution.
2. For the ordered pair [tex]\( (6, 0.5) \)[/tex]:
- Check the first inequality: [tex]\( 0.5 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( 0.5 > 0.5 \)[/tex] (False)
- Since one of the inequalities is false, [tex]\( (6, 0.5) \)[/tex] is not a solution.
3. For the ordered pair [tex]\( (6, 5) \)[/tex]:
- Check the first inequality: [tex]\( 5 \leq 5 \)[/tex] (True)
- Check the second inequality: [tex]\( 5 > 0.5 \)[/tex] (True)
- Since both inequalities are true, [tex]\( (6, 5) \)[/tex] is a solution.
4. For the ordered pair [tex]\( (6, 8) \)[/tex]:
- Check the first inequality: [tex]\( 8 \leq 5 \)[/tex] (False)
- Check the second inequality: [tex]\( 8 > 0.5 \)[/tex] (True)
- Since one of the inequalities is false, [tex]\( (6, 8) \)[/tex] is not a solution.
The ordered pair that satisfies both inequalities in the system is [tex]\( (6, 5) \)[/tex].
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