Question

question
solve the following inequality algebraically.
|x - 3| ≤ 8

Explanation:

Step1: Recall absolute value inequality rule

For \(|a| \leq b\) (where \(b\geq0\)), it is equivalent to \(-b \leq a \leq b\). Here, \(a = x - 3\) and \(b = 8\) (and \(8\geq0\) holds), so we can rewrite the inequality \(|x - 3| \leq 8\) as \(-8 \leq x - 3 \leq 8\).

Step2: Solve the compound inequality

To solve \(-8 \leq x - 3 \leq 8\), we add 3 to all parts of the compound inequality.
For the left part: \(-8 + 3 \leq x - 3 + 3\), which simplifies to \(-5 \leq x\).
For the right part: \(x - 3 + 3 \leq 8 + 3\), which simplifies to \(x \leq 11\).
Combining these two results, we get \(-5 \leq x \leq 11\).

Answer:

The solution to the inequality \(|x - 3| \leq 8\) is \(\boldsymbol{-5 \leq x \leq 11}\) (or in interval notation, \([-5, 11]\)).