Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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### Solving the Compound Inequality

Solve the following inequality:

\[ 5|4x - 2| + 4 \leq 9 \]

**Explanation:**

1. **Start by isolating the absolute value expression:**

\[ 5|4x - 2| + 4 \leq 9 \]

Subtract 4 from both sides to isolate the absolute value term:

\[ 5|4x - 2| \leq 5 \]

2. **Divide both sides by 5:**

\[ |4x - 2| \leq 1 \]

3. **Solve the absolute value inequality:**

Given the absolute value inequality \( |A| \leq B \), this translates to:

\[ -B \leq A \leq B \]

For our inequality \( |4x - 2| \leq 1 \), this translates to:

\[ -1 \leq 4x - 2 \leq 1 \]

4. **Solve the compound inequality:**

First inequality:
\[ -1 \leq 4x - 2 \]

Add 2 to both sides:
\[ 1 \leq 4x \]

Divide by 4:
\[ \frac{1}{4} \leq x \]

Second inequality:
\[ 4x - 2 \leq 1 \]

Add 2 to both sides:
\[ 4x \leq 3 \]

Divide by 4:
\[ x \leq \frac{3}{4} \]

5. **Combine the solutions:**

\[ \frac{1}{4} \leq x \leq \frac{3}{4} \]

Thus, the solution to the inequality \( 5|4x - 2| + 4 \leq 9 \) is:

\[ \frac{1}{4} \leq x \leq \frac{3}{4} \]

This interval notation denotes that \( x \) is between \(\frac{1}{4}\) and \(\frac{3}{4}\) inclusive.
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Transcribed Image Text:### Solving the Compound Inequality Solve the following inequality: \[ 5|4x - 2| + 4 \leq 9 \] **Explanation:** 1. **Start by isolating the absolute value expression:** \[ 5|4x - 2| + 4 \leq 9 \] Subtract 4 from both sides to isolate the absolute value term: \[ 5|4x - 2| \leq 5 \] 2. **Divide both sides by 5:** \[ |4x - 2| \leq 1 \] 3. **Solve the absolute value inequality:** Given the absolute value inequality \( |A| \leq B \), this translates to: \[ -B \leq A \leq B \] For our inequality \( |4x - 2| \leq 1 \), this translates to: \[ -1 \leq 4x - 2 \leq 1 \] 4. **Solve the compound inequality:** First inequality: \[ -1 \leq 4x - 2 \] Add 2 to both sides: \[ 1 \leq 4x \] Divide by 4: \[ \frac{1}{4} \leq x \] Second inequality: \[ 4x - 2 \leq 1 \] Add 2 to both sides: \[ 4x \leq 3 \] Divide by 4: \[ x \leq \frac{3}{4} \] 5. **Combine the solutions:** \[ \frac{1}{4} \leq x \leq \frac{3}{4} \] Thus, the solution to the inequality \( 5|4x - 2| + 4 \leq 9 \) is: \[ \frac{1}{4} \leq x \leq \frac{3}{4} \] This interval notation denotes that \( x \) is between \(\frac{1}{4}\) and \(\frac{3}{4}\) inclusive.
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