Problem 92 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

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 1: 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: \( \frac{-2}{3} < \left|x - \frac{11}{3}\right| \)

Step 2: 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: \( -\left(x - \frac{11}{3}\right) > \frac{-2}{3} \)

Step 3: 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 \( x < 4 \)