In the following questions, there are two equations in x and y . You have to solve both the equations and give answer
I. ` x^(2) = 4`
II. ` y^(2) - 6 y + 8 = 0`

Options

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

Answer

To solve the given equations step by step, we will work through each equation separately and then compare the results.

### Step 1: Solve the first equation \( x^2 = 4 \)

To solve for \( x \), we can take the square root of both sides:

\[ x = \pm \sqrt{4} \]

This gives us:

\[ x = 2 \quad \text{or} \quad x = -2 \]

### Step 2: Solve the second equation \( y^2 - 6y + 8 = 0 \)

We can factor this quadratic equation. We need two numbers that multiply to \( 8 \) (the constant term) and add up to \( -6 \) (the coefficient of \( y \)). The numbers that satisfy this are \( -4 \) and \( -2 \):

\[ y^2 - 4y - 2y + 8 = 0 \]

Now, we can group the terms:

\[ (y - 4)(y - 2) = 0 \]

Setting each factor to zero gives us:

\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]
\[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \]

### Step 3: List the solutions

From the first equation, we have:
- \( x = 2 \)
- \( x = -2 \)

From the second equation, we have:
- \( y = 4 \)
- \( y = 2 \)

### Step 4: Compare the values of \( x \) and \( y \)

Now we will compare the values of \( x \) and \( y \):

1. For \( x = 2 \):
- \( 2 \leq 4 \) (True)
- \( 2 \leq 2 \) (True)

2. For \( x = -2 \):
- \( -2 \leq 4 \) (True)
- \( -2 \leq 2 \) (True)

### Conclusion

From the comparisons, we can conclude that:

\[ x \leq y \]

Thus, the final answer is:

\[ \text{The relation is } x \leq y \]

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