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Question
question
solve the following inequality algebraically.
|x - 8| ≤ 3
Step1: Recall absolute value inequality rule
For \(|A| \leq B\) (where \(B \geq 0\)), it is equivalent to \(-B \leq A \leq B\). Here, \(A = x - 8\) and \(B = 3\). So we get \(-3 \leq x - 8 \leq 3\).
Step2: Solve the compound inequality
Add 8 to all parts of the inequality.
Left part: \(-3 + 8 \leq x - 8 + 8\) which simplifies to \(5 \leq x\).
Right part: \(x - 8 + 8 \leq 3 + 8\) which simplifies to \(x \leq 11\).
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\(5 \leq x \leq 11\)