`I. X^(2) - 8x+15 = 0" "II. Y^(2) -12y + 36 = 0`

Options

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

Explanation

To solve the equations and determine the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( x^2 - 8x + 15 = 0 \)

1. We can factor the quadratic equation:
\[ x^2 - 8x + 15 = (x - 3)(x - 5) = 0 \]

2. Set each factor equal to zero:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]

Thus, the solutions for \( x \) are \( x = 3 \) and \( x = 5 \).

### Step 2: Solve the second equation \( y^2 - 12y + 36 = 0 \)

1. We can factor this quadratic equation:
\[ y^2 - 12y + 36 = (y - 6)(y - 6) = 0 \]

2. Set the factor equal to zero:
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]

Thus, the solution for \( y \) is \( y = 6 \).

### Step 3: Compare the values of \( x \) and \( y \)

Now we have:
- \( x = 3 \) or \( x = 5 \)
- \( y = 6 \)

We need to determine the relationship between \( x \) and \( y \):
- For \( x = 3 \): \( 3 < 6 \)
- For \( x = 5 \): \( 5 < 6 \)

In both cases, \( x < y \).

### Conclusion

The correct relationship is \( x < y \).