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Chapter 1: Problem 92

Solve each absolute value inequality. $$1<\left|x-\frac{11}{3}\right|+\frac{7}{3}$$

Short Answer

Expert verified
The solution to the absolute value inequality is every real number less than 4 or greater than 3. In interval notation, this is written as \( (-\infty, 4) \cup (3, \infty) \).

Step by step solution

01

Isolate the Absolute Value

Subtract \( \frac{7}{3} \) from both sides of the inequality to isolate the absolute value on one side. This will give: \( \frac{-2}{3} < \left|x - \frac{11}{3}\right| \)
02

Split into Two Inequalities

The next step is to split the original inequality into two separate inequalities, based on the property of absolute value. One of the inequalities will be: \( x - \frac{11}{3} > \frac{-2}{3} \), and the other one will be: \( -\left(x - \frac{11}{3}\right) > \frac{-2}{3} \)
03

Solve Each Inequality

The inequality \( x - \frac{11}{3} > \frac{-2}{3} \) can be solved step by step by adding \( \frac{11}{3} \) to both sides to get \( x > \frac{9}{3} \) , which simplifies to \( x > 3 \). The other inequality \( -\left(x - \frac{11}{3}\right) > \frac{-2}{3} \) can be solved by first multiplying both sides by -1 to reverse the inequality sign to '<', giving \( x - \frac{11}{3} < \frac{2}{3} \). Adding \( \frac{11}{3} \) to both sides gives \( x < \frac{11}{3} + \frac{2}{3} = 4 \). This gets the solution \( x < 4 \)

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