Solve $3x + 8 > 2$ when $x$ is a real number.
A$(-\infty, -2)$
B$(2, \infty)$
C$[-2, \infty)$
D$(-2, \infty)$
Answer & Solution
Correct answer: D. $(-2, \infty)$
$3x > -6 \Rightarrow x > -2$.
In interval notation: $(-2, \infty)$ — open at $-2$ because strict inequality.
- **C** is closed at $-2$ — wrong for strict.
- **A** reverses the direction.
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