Question

(a) |x + 1| - 2 = 5
the solution set is (-4, 6).
enter your answer in roster form.
part 2 of 3
(b) |x + 1| - 2 ≤ 5
the solution set is
enter your answer in interval notation.

Explanation:

Part (a)

Step1: Solve the equation \(|x + 1| - 2 = 5\)

First, isolate the absolute value expression. Add 2 to both sides of the equation:
\(|x + 1| = 5 + 2\)
\(|x + 1| = 7\)

Step2: Solve the absolute value equation

The absolute value equation \(|A| = B\) (where \(B\geq0\)) has solutions \(A = B\) or \(A = -B\). So for \(|x + 1| = 7\), we have two cases:

• Case 1: \(x + 1 = 7\)

Subtract 1 from both sides: \(x = 7 - 1 = 6\)

• Case 2: \(x + 1 = -7\)

Subtract 1 from both sides: \(x = -7 - 1 = -8\) Wait, the original solution set was given as \((-4, 6)\) which seems incorrect. Wait, let's re - solve:

Wait, original equation \(|x + 1|-2 = 5\)
Add 2: \(|x + 1|=7\)
So \(x + 1 = 7\) gives \(x = 6\); \(x + 1=-7\) gives \(x=-8\). But the user's initial input had a wrong solution set. Let's do it correctly.

Wait, maybe the user made a mistake in the initial solution. Let's solve it properly:

Starting with \(|x + 1|-2 = 5\)

Step 1: Add 2 to both sides: \(|x + 1|=5 + 2=7\)

Step 2: By the definition of absolute value, \(x + 1 = 7\) or \(x + 1=-7\)

For \(x + 1 = 7\), \(x=7 - 1 = 6\)

For \(x + 1=-7\), \(x=-7 - 1=-8\)

So the solution set in roster form is \(\{-8, 6\}\)

Part (b)

Step1: Isolate the absolute value expression

Start with \(|x + 1|-2\leq5\)

Add 2 to both sides of the inequality: \(|x + 1|\leq5 + 2\)

\(|x + 1|\leq7\)

Step2: Solve the absolute value inequality

The absolute value inequality \(|A|\leq B\) (where \(B\geq0\)) is equivalent to \(-B\leq A\leq B\). So for \(|x + 1|\leq7\), we have:

\(-7\leq x + 1\leq7\)

Step3: Solve the compound inequality

Subtract 1 from all parts of the compound inequality:

\(-7-1\leq x+1 - 1\leq7 - 1\)

\(-8\leq x\leq6\)

In interval notation, this is \([-8, 6]\)

Part (a) Answer: \(\{-8, 6\}\)

Part (b) Answer: \([-8, 6]\)

Answer:

Step1: Isolate the absolute value expression

Start with \(|x + 1|-2\leq5\)

Add 2 to both sides of the inequality: \(|x + 1|\leq5 + 2\)

\(|x + 1|\leq7\)

Step2: Solve the absolute value inequality

The absolute value inequality \(|A|\leq B\) (where \(B\geq0\)) is equivalent to \(-B\leq A\leq B\). So for \(|x + 1|\leq7\), we have:

\(-7\leq x + 1\leq7\)

Step3: Solve the compound inequality

Subtract 1 from all parts of the compound inequality:

\(-7-1\leq x+1 - 1\leq7 - 1\)

\(-8\leq x\leq6\)

In interval notation, this is \([-8, 6]\)

Part (a) Answer: \(\{-8, 6\}\)

Part (b) Answer: \([-8, 6]\)