I. ` x^(2)-7x+10= 0`
II. `y^(2)+11y+10 = 0`

Options

A. `x gt y `
B. `x ge y`
C. ` x lt y`
D. ` x le y`

Explanation

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation \( x^2 - 7x + 10 = 0 \)

1. **Identify the equation**: The equation is \( x^2 - 7x + 10 = 0 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 10 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)).
- The numbers are \( -5 \) and \( -2 \).
3. **Write the factored form**: The equation can be factored as:
\[ (x - 5)(x - 2) = 0 \]
4. **Set each factor to zero**:
- \( x - 5 = 0 \) leads to \( x = 5 \)
- \( x - 2 = 0 \) leads to \( x = 2 \)

**Values of \( x \)**: \( x = 5 \) and \( x = 2 \)

### Step 2: Solve the second equation \( y^2 + 11y + 10 = 0 \)

1. **Identify the equation**: The equation is \( y^2 + 11y + 10 = 0 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 10 \) and add up to \( 11 \).
- The numbers are \( 1 \) and \( 10 \).
3. **Write the factored form**: The equation can be factored as:
\[ (y + 1)(y + 10) = 0 \]
4. **Set each factor to zero**:
- \( y + 1 = 0 \) leads to \( y = -1 \)
- \( y + 10 = 0 \) leads to \( y = -10 \)

**Values of \( y \)**: \( y = -1 \) and \( y = -10 \)

### Step 3: Establish a relation between \( x \) and \( y \)

1. **List the values**:
- \( x \) can be \( 5 \) or \( 2 \)
- \( y \) can be \( -1 \) or \( -10 \)
2. **Compare the values**:
- For \( x = 5 \): \( y = -1 \) and \( y = -10 \) → both values of \( y \) are less than \( x \).
- For \( x = 2 \): \( y = -1 \) and \( y = -10 \) → both values of \( y \) are less than \( x \).

### Conclusion

From the above analysis, we can conclude that \( y < x \) for all values of \( x \) and \( y \) derived from the equations.

### Final Relation

Thus, we can express the relationship as:
\[ x > y \]