Question

question 8
solve: $4|x - 4| + 2 > 8$
the answer has the form:
$\circ (a, b)$
$\circ (-infty, a) \cup (b, infty)$
state your solution using interval notation like above:

question help: message instructor

Explanation:

Step1: Isolate the absolute value

Subtract 2 from both sides: \(4|x - 4| + 2 - 2 > 8 - 2\)
Simplify: \(4|x - 4| > 6\)
Divide both sides by 4: \(\frac{4|x - 4|}{4} > \frac{6}{4}\)
Simplify: \(|x - 4| > \frac{3}{2}\)

Step2: Solve the absolute value inequality

For \(|x - 4| > \frac{3}{2}\), we have two cases:
Case 1: \(x - 4 > \frac{3}{2}\)
Add 4 to both sides: \(x - 4 + 4 > \frac{3}{2} + 4\)
Simplify: \(x > \frac{3}{2} + \frac{8}{2} = \frac{11}{2}\)

Case 2: \(x - 4 < -\frac{3}{2}\)
Add 4 to both sides: \(x - 4 + 4 < -\frac{3}{2} + 4\)
Simplify: \(x < -\frac{3}{2} + \frac{8}{2} = \frac{5}{2}\)

Answer:

\((-\infty, \frac{5}{2}) \cup (\frac{11}{2}, \infty)\)