b) \( \frac{b}{3} \geqslant \frac{b}{2}+1 \geqslant b-1 \)

To solve the compound inequality: \[ \frac{b}{3} \geq \frac{b}{2} + 1 \geq b - 1 \] we'll break it down into two separate inequalities and solve each step by step. Then, we'll determine the values of \( b \) that satisfy both conditions simultaneously. ### 1. First Inequality: \[ \frac{b}{3} \geq \frac{b}{2} + 1 \] **Step 1:** Subtract \(\frac{b}{2}\) from both sides to isolate terms involving \( b \). \[ \frac{b}{3} - \frac{b}{2} \geq 1 \] **Step 2:** Find a common denominator to combine the fractions. The least common denominator between 3 and 2 is 6. \[ \frac{2b}{6} - \frac{3b}{6} \geq 1 \\ \frac{-b}{6} \geq 1 \] **Step 3:** Multiply both sides by 6 to eliminate the denominator. Since we're multiplying by a positive number, the inequality sign remains the same. \[ -b \geq 6 \] **Step 4:** Multiply both sides by \(-1\) to solve for \( b \). Remember to reverse the inequality sign when multiplying or dividing by a negative number. \[ b \leq -6 \] ### 2. Second Inequality: \[ \frac{b}{2} + 1 \geq b - 1 \] **Step 1:** Subtract \(\frac{b}{2}\) from both sides to isolate terms involving \( b \). \[ 1 \geq b - 1 - \frac{b}{2} \] **Step 2:** Combine like terms. Notice that \( b - \frac{b}{2} = \frac{b}{2} \). \[ 1 \geq \frac{b}{2} - 1 \] **Step 3:** Add 1 to both sides to further isolate \( \frac{b}{2} \). \[ 2 \geq \frac{b}{2} \] **Step 4:** Multiply both sides by 2 to solve for \( b \). \[ 4 \geq b \\ \text{or equivalently} \\ b \leq 4 \] ### 3. Combine Both Inequalities: From the first inequality, we have: \[ b \leq -6 \] From the second inequality, we have: \[ b \leq 4 \] To satisfy **both** conditions simultaneously, \( b \) must satisfy the more restrictive condition. Therefore, the solution is: \[ b \leq -6 \] ### **Solution:** All real numbers b with b at most –6. In other words, the solution is b ≤ –6.