Question

when to use an and (intersection) or an or (union).

each absolute value inequality would turn in to a union of inequalities or an intersection of inequalities.

\\(|x + 2| < 1\\)

intersection

union

Explanation:

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<reasoning>

Analyze the inequality type

The given absolute value inequality is:
\[|x + 2| < 1\]
This inequality uses the "less than" (\(<\)) operator.

Apply the absolute value definition

Using Compound Inequalities, we know that an absolute value inequality of the form \(|u| < a\) (where \(a > 0\)) represents all numbers whose distance from zero is less than \(a\). This translates to:
\(-a < u < a\)
Which is a compound inequality joined by "AND":
\(u > -a\) AND \(u < a\)

Determine union or intersection

An "AND" compound inequality represents the overlap of the individual solution sets. In set theory, this overlap is called the intersection. Therefore, the inequality \(|x + 2| < 1\) turns into an intersection of inequalities:
\(-1 < x + 2 < 1\)
</reasoning>

<answer>
<mcq-correct>intersection</mcq-correct>
<mcq-option>union</mcq-option>
</answer>

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Answer:

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</pre_analysis>

<reasoning>

Analyze the inequality type

The given absolute value inequality is:
\[|x + 2| < 1\]
This inequality uses the "less than" (\(<\)) operator.

Apply the absolute value definition

Using Compound Inequalities, we know that an absolute value inequality of the form \(|u| < a\) (where \(a > 0\)) represents all numbers whose distance from zero is less than \(a\). This translates to:
\(-a < u < a\)
Which is a compound inequality joined by "AND":
\(u > -a\) AND \(u < a\)

Determine union or intersection

An "AND" compound inequality represents the overlap of the individual solution sets. In set theory, this overlap is called the intersection. Therefore, the inequality \(|x + 2| < 1\) turns into an intersection of inequalities:
\(-1 < x + 2 < 1\)
</reasoning>

<answer>
<mcq-correct>intersection</mcq-correct>
<mcq-option>union</mcq-option>
</answer>

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