I. ` y^(2) + y - 1 = 4 - 2y - y^(2) `
II. ` x^(2)/2 - 3/2 x = x - 3`

Options

A. If ` x gt y`
B. If ` x lt y`
C. If ` x = y`
D. If ` x ge y`

Explanation

To solve the given equations step by step, let's start with the first equation involving \( y \):

### Step 1: Solve the first equation \( y^2 + y - 1 = 4 - 2y - y^2 \)

1. Rearrange the equation by moving all terms to one side:
\[ y^2 + y - 1 + 2y + y^2 - 4 = 0 \]
This simplifies to:
\[ 2y^2 + 3y - 5 = 0 \]

### Step 2: Factor the quadratic equation

2. To factor \( 2y^2 + 3y - 5 = 0 \), we can use the middle term splitting method. We need two numbers that multiply to \( 2 \times (-5) = -10 \) and add to \( 3 \). The numbers \( 5 \) and \( -2 \) work:
\[ 2y^2 + 5y - 2y - 5 = 0 \]
Grouping the terms:
\[ (2y^2 + 5y) + (-2y - 5) = 0 \]
Factoring by grouping:
\[ y(2y + 5) - 1(2y + 5) = 0 \]
This gives:
\[ (2y + 5)(y - 1) = 0 \]

### Step 3: Find the values of \( y \)

3. Set each factor to zero:
\[ 2y + 5 = 0 \quad \Rightarrow \quad y = -\frac{5}{2} \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]
Thus, the solutions for \( y \) are:
\[ y = -\frac{5}{2} \quad \text{and} \quad y = 1 \]

### Step 4: Solve the second equation \( \frac{x^2}{2} - \frac{3}{2}x = x - 3 \)

4. Rearranging gives:
\[ \frac{x^2}{2} - \frac{3}{2}x - x + 3 = 0 \]
Combine like terms:
\[ \frac{x^2}{2} - \frac{5}{2}x + 3 = 0 \]
Multiply through by \( 2 \) to eliminate the fraction:
\[ x^2 - 5x + 6 = 0 \]

### Step 5: Factor the quadratic equation

5. Factor \( x^2 - 5x + 6 = 0 \):
\[ (x - 2)(x - 3) = 0 \]

### Step 6: Find the values of \( x \)

6. Set each factor to zero:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Step 7: Compare the values of \( x \) and \( y \)

7. Now we have the values:
- For \( y \): \( -\frac{5}{2} \) and \( 1 \)
- For \( x \): \( 2 \) and \( 3 \)

Comparing these values:
- The maximum value of \( y \) is \( 1 \).
- The minimum value of \( x \) is \( 2 \).

### Conclusion

Since \( 2 > 1 \), we conclude that:
\[ x > y \]