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Question
- solve the inequality: $-3(n + 8) > 36$
a. $n < -20$
b. $n < 9$
c. $n > -20$
d. $n < 3$
Step1: Divide both sides by -3 (reverse inequality)
When dividing an inequality by a negative number, the inequality sign flips. So, divide both sides of \(-3(n + 8) > 36\) by \(-3\):
\(n + 8 < \frac{36}{-3}\)
\(n + 8 < -12\)
Step2: Subtract 8 from both sides
Subtract 8 from both sides to solve for \(n\):
\(n + 8 - 8 < -12 - 8\)
\(n < -20\)
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A. \(n < -20\)