`xgt0`
`{:("Quantity A","Quantity B"),((2+(2)/(3x))/(2),(3+(3)/(2x))/(3)):}`

Options

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the infromation given.

Answer

To solve the problem, we need to compare Quantity A and Quantity B given the condition \( x > 0 \).

### Step 1: Define Quantity A and Quantity B
- Quantity A: \( \frac{2 + \frac{2}{3x}}{2} \)
- Quantity B: \( \frac{3 + \frac{3}{2x}}{3} \)

### Step 2: Substitute \( x = 1 \)
Since \( x > 0 \), we can choose \( x = 1 \) for simplicity.
- Substitute \( x = 1 \) into Quantity A:
\[ \text{Quantity A} = \frac{2 + \frac{2}{3 \cdot 1}}{2} = \frac{2 + \frac{2}{3}}{2} \]

### Step 3: Simplify Quantity A
- Find a common denominator for the numerator:
\[ 2 = \frac{6}{3} \quad \Rightarrow \quad \text{Quantity A} = \frac{\frac{6}{3} + \frac{2}{3}}{2} = \frac{\frac{8}{3}}{2} \]
- Simplify:
\[ \text{Quantity A} = \frac{8}{3} \cdot \frac{1}{2} = \frac{8}{6} = \frac{4}{3} \approx 1.3333\ldots \]

### Step 4: Substitute \( x = 1 \) into Quantity B
- Substitute \( x = 1 \) into Quantity B:
\[ \text{Quantity B} = \frac{3 + \frac{3}{2 \cdot 1}}{3} = \frac{3 + \frac{3}{2}}{3} \]

### Step 5: Simplify Quantity B
- Find a common denominator for the numerator:
\[ 3 = \frac{6}{2} \quad \Rightarrow \quad \text{Quantity B} = \frac{\frac{6}{2} + \frac{3}{2}}{3} = \frac{\frac{9}{2}}{3} \]
- Simplify:
\[ \text{Quantity B} = \frac{9}{2} \cdot \frac{1}{3} = \frac{9}{6} = 1.5 \]

### Step 6: Compare Quantity A and Quantity B
- We have:
- Quantity A = \( \frac{4}{3} \approx 1.3333\ldots \)
- Quantity B = \( 1.5 \)
- Since \( 1.5 > 1.3333\ldots \), we conclude that:
\[ \text{Quantity B} > \text{Quantity A} \]

### Conclusion
Thus, the correct relationship is that Quantity B is greater than Quantity A.