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To determine the values of [tex]\(x\)[/tex] that make [tex]\(A \cup B = \varnothing\)[/tex], we need to first solve the inequalities for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] and find their union.
Let's start with the inequality for set [tex]\(A\)[/tex]:
[tex]\[ 3x + 4 \geq 13 \][/tex]
Subtract 4 from both sides:
[tex]\[ 3x \geq 9 \][/tex]
Divide both sides by 3:
[tex]\[ x \geq 3 \][/tex]
So, the solutions to the inequality [tex]\(3x + 4 \geq 13\)[/tex] are:
[tex]\[ A = \{x \mid x \geq 3\} \][/tex]
Next, solve the inequality for set [tex]\(B\)[/tex]:
[tex]\[ \frac{1}{2}x + 3 \leq 4 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{1}{2}x \leq 1 \][/tex]
Multiply both sides by 2:
[tex]\[ x \leq 2 \][/tex]
So, the solutions to the inequality [tex]\(\frac{1}{2}x + 3 \leq 4\)[/tex] are:
[tex]\[ B = \{x \mid x \leq 2\} \][/tex]
Now, let's consider the union of sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A \cup B = \{x \mid x \geq 3\} \cup \{x \mid x \leq 2\} \][/tex]
To see if [tex]\(A \cup B = \varnothing\)[/tex], we need to check if there is any overlap or if the sets together cover all possible real values, but do not make a situation where there is no possible value for [tex]\(x\)[/tex].
Here, [tex]\(A = \{x \geq 3\}\)[/tex] and [tex]\(B = \{x \leq 2\}\)[/tex].
- [tex]\( x \geq 3 \)[/tex] means [tex]\(x\)[/tex] values start from 3 and go to [tex]\(\infty\)[/tex].
- [tex]\( x \leq 2 \)[/tex] means [tex]\(x\)[/tex] values start from [tex]\(-\infty\)[/tex] to 2.
Since there is no [tex]\(x\)[/tex] that can simultaneously satisfy [tex]\(x \geq 3\)[/tex] and [tex]\(x \leq 2\)[/tex], [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not overlap.
Thus, [tex]\(A \cup B = \varnothing\)[/tex] will happen only if the sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not include any [tex]\(x\)[/tex] values between 2 and 3 (as they do not overlap).
Considering the provided options:
- [tex]\(2 < x < 3\)[/tex]: This set includes values between 2 and 3, but no value from here satisfies either [tex]\(x \geq 3\)[/tex] or [tex]\(x \leq 2\)[/tex].
- [tex]\(2 \leq x \leq 3\)[/tex]: This set has the bounds included, but those explicitly do not satisfy both sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
- [tex]\(x \leq 2\)[/tex] and [tex]\(x \geq 3\)[/tex]: This is the correct interpretation of [tex]\(A \cup B = \varnothing\)[/tex] since these boundaries refer to sets [tex]\(\{x | x \leq 2\}\)[/tex] and [tex]\(\{x | x \geq 3\}\)[/tex] which is the only situation for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] not to overlap.
- [tex]\(x < 2\)[/tex] and [tex]\(x > 3\)[/tex]: This would imply no valid overlapping range as these do not satisfy the bounds of [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
Thus, to satisfy [tex]\(A \cup B = \varnothing\)[/tex] the correct answer is:
[tex]\[ x \leq 2 \text{ and } x \geq 3 \][/tex]
Let's start with the inequality for set [tex]\(A\)[/tex]:
[tex]\[ 3x + 4 \geq 13 \][/tex]
Subtract 4 from both sides:
[tex]\[ 3x \geq 9 \][/tex]
Divide both sides by 3:
[tex]\[ x \geq 3 \][/tex]
So, the solutions to the inequality [tex]\(3x + 4 \geq 13\)[/tex] are:
[tex]\[ A = \{x \mid x \geq 3\} \][/tex]
Next, solve the inequality for set [tex]\(B\)[/tex]:
[tex]\[ \frac{1}{2}x + 3 \leq 4 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{1}{2}x \leq 1 \][/tex]
Multiply both sides by 2:
[tex]\[ x \leq 2 \][/tex]
So, the solutions to the inequality [tex]\(\frac{1}{2}x + 3 \leq 4\)[/tex] are:
[tex]\[ B = \{x \mid x \leq 2\} \][/tex]
Now, let's consider the union of sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A \cup B = \{x \mid x \geq 3\} \cup \{x \mid x \leq 2\} \][/tex]
To see if [tex]\(A \cup B = \varnothing\)[/tex], we need to check if there is any overlap or if the sets together cover all possible real values, but do not make a situation where there is no possible value for [tex]\(x\)[/tex].
Here, [tex]\(A = \{x \geq 3\}\)[/tex] and [tex]\(B = \{x \leq 2\}\)[/tex].
- [tex]\( x \geq 3 \)[/tex] means [tex]\(x\)[/tex] values start from 3 and go to [tex]\(\infty\)[/tex].
- [tex]\( x \leq 2 \)[/tex] means [tex]\(x\)[/tex] values start from [tex]\(-\infty\)[/tex] to 2.
Since there is no [tex]\(x\)[/tex] that can simultaneously satisfy [tex]\(x \geq 3\)[/tex] and [tex]\(x \leq 2\)[/tex], [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not overlap.
Thus, [tex]\(A \cup B = \varnothing\)[/tex] will happen only if the sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not include any [tex]\(x\)[/tex] values between 2 and 3 (as they do not overlap).
Considering the provided options:
- [tex]\(2 < x < 3\)[/tex]: This set includes values between 2 and 3, but no value from here satisfies either [tex]\(x \geq 3\)[/tex] or [tex]\(x \leq 2\)[/tex].
- [tex]\(2 \leq x \leq 3\)[/tex]: This set has the bounds included, but those explicitly do not satisfy both sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
- [tex]\(x \leq 2\)[/tex] and [tex]\(x \geq 3\)[/tex]: This is the correct interpretation of [tex]\(A \cup B = \varnothing\)[/tex] since these boundaries refer to sets [tex]\(\{x | x \leq 2\}\)[/tex] and [tex]\(\{x | x \geq 3\}\)[/tex] which is the only situation for [tex]\(A\)[/tex] and [tex]\(B\)[/tex] not to overlap.
- [tex]\(x < 2\)[/tex] and [tex]\(x > 3\)[/tex]: This would imply no valid overlapping range as these do not satisfy the bounds of [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
Thus, to satisfy [tex]\(A \cup B = \varnothing\)[/tex] the correct answer is:
[tex]\[ x \leq 2 \text{ and } x \geq 3 \][/tex]
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