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solve for z. |z + 2| - 2 ≥ 80 write a compound inequality like 1 < x < …

Question

solve for z.
|z + 2| - 2 ≥ 80
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. use inte fractions, or improper fractions in simplest form.

Explanation:

Step1: Isolate the absolute value

Add 2 to both sides of the inequality \(|z + 2| - 2 \geq 80\) to isolate the absolute value expression.
\(|z + 2| - 2 + 2 \geq 80 + 2\)
\(|z + 2| \geq 82\)

Step2: Solve the absolute value inequality

The absolute value inequality \(|a| \geq b\) (where \(b>0\)) is equivalent to \(a \leq -b\) or \(a \geq b\). So for \(|z + 2| \geq 82\), we have two cases:

Case 1: \(z + 2 \leq -82\)

Subtract 2 from both sides:
\(z + 2 - 2 \leq -82 - 2\)
\(z \leq -84\)

Case 2: \(z + 2 \geq 82\)

Subtract 2 from both sides:
\(z + 2 - 2 \geq 82 - 2\)
\(z \geq 80\)

Answer:

\(z \leq -84\) or \(z \geq 80\)