In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer
I. ` x^(2) = 256 `
II. ` y^(2) - 17 y + 16 = 0`

Options

A. A)If x= y or no relation can be established
B. B)If ` x gt y `
C. C)If ` x lt y `
D. D)If `x le y`

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve the first equation \( x^2 = 256 \)

1. **Rearranging the equation**:
\[ x^2 - 256 = 0 \]

2. **Recognizing the difference of squares**:
\[ x^2 - 16^2 = 0 \]

3. **Factoring the equation**:
\[ (x - 16)(x + 16) = 0 \]

4. **Finding the values of \( x \)**:
- From \( x - 16 = 0 \), we get \( x = 16 \).
- From \( x + 16 = 0 \), we get \( x = -16 \).

Thus, the solutions for the first equation are:
\[ x = 16 \quad \text{or} \quad x = -16 \]

### Step 2: Solve the second equation \( y^2 - 17y + 16 = 0 \)

1. **Factoring the quadratic equation**:
- We need two numbers that multiply to \( 16 \) (the constant term) and add to \( -17 \) (the coefficient of \( y \)).
- The numbers are \( -16 \) and \( -1 \).

2. **Writing the factored form**:
\[ (y - 16)(y - 1) = 0 \]

3. **Finding the values of \( y \)**:
- From \( y - 16 = 0 \), we get \( y = 16 \).
- From \( y - 1 = 0 \), we get \( y = 1 \).

Thus, the solutions for the second equation are:
\[ y = 16 \quad \text{or} \quad y = 1 \]

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

Now we have the following values:
- From the first equation: \( x = 16 \) or \( x = -16 \)
- From the second equation: \( y = 16 \) or \( y = 1 \)

### Step 4: Establish the relationship between \( x \) and \( y \)

1. **Comparing values**:
- If \( x = 16 \), then \( y = 16 \) gives \( x = y \).
- If \( x = -16 \), then \( y = 1 \) gives \( x < y \).

### Conclusion

The relationship between \( x \) and \( y \) can be summarized as:
- \( x \) can be equal to \( y \) (when both are 16).
- \( x \) can also be less than \( y \) (when \( x = -16 \) and \( y = 1 \)).

Thus, the final answer is:
\[ x \leq y \]