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

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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.
All real numbers $$ b $$ such that $$ b \leq -6 $$.

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

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