Solve the following compound inequality. Write the solution set using interval notation and graph it. \( \frac{1}{3}(x+2)>2 \) or \( 0<3-x \)

First, let's solve each part of the compound inequality separately. 1. For the inequality \( \frac{1}{3}(x+2) > 2 \): Multiply both sides by 3: \( x + 2 > 6 \). Subtract 2 from both sides: \( x > 4 \). 2. For the inequality \( 0 < 3 - x \): Rearranging gives: \( x < 3 \). Now, we combine the solutions from both inequalities: - From the first inequality, we have \( x > 4 \). - From the second inequality, we have \( x < 3 \). Since it’s an "or" condition, the solution includes all values that satisfy either condition. Thus, the solution set in interval notation is: \( (-\infty, 3) \cup (4, \infty) \). To graph it, draw a number line: - Shade the area to the left of 3 (but not including 3) and the area to the right of 4 (but not including 4), indicating the two intervals where the values of \( x \) are valid solutions.