In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option
I. ` x^(2) - 25 x + 156 = 0 `
II. ` y^(2) - 32 y +255 = 0 `

Options

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

Answer

To solve the given equations step by step, we will follow these procedures for both equations.

### Step 1: Solve the first equation
The first equation is:
\[ x^2 - 25x + 156 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \( 156 \) (the constant term) and add up to \( -25 \) (the coefficient of \( x \)).

1. The numbers that satisfy these conditions are \( -13 \) and \( -12 \).
2. We can rewrite the equation as:
\[ (x - 13)(x - 12) = 0 \]

Now, we can set each factor to zero:
- \( x - 13 = 0 \) → \( x = 13 \)
- \( x - 12 = 0 \) → \( x = 12 \)

### Step 2: Solve the second equation
The second equation is:
\[ y^2 - 32y + 255 = 0 \]

Similarly, we will factor this quadratic equation. We need to find two numbers that multiply to \( 255 \) and add up to \( -32 \).

1. The numbers that satisfy these conditions are \( -17 \) and \( -15 \).
2. We can rewrite the equation as:
\[ (y - 17)(y - 15) = 0 \]

Now, we can set each factor to zero:
- \( y - 17 = 0 \) → \( y = 17 \)
- \( y - 15 = 0 \) → \( y = 15 \)

### Step 3: Compare the values of \( x \) and \( y \)
From the solutions, we have:
- Values of \( x \): \( 12, 13 \)
- Values of \( y \): \( 15, 17 \)

Now, we will compare the values:
- For \( x = 12 \), \( y = 15 \) → \( x < y \)
- For \( x = 13 \), \( y = 15 \) → \( x < y \)
- For \( x = 12 \), \( y = 17 \) → \( x < y \)
- For \( x = 13 \), \( y = 17 \) → \( x < y \)

In all cases, we find that \( x < y \).

### Final Conclusion
The relation we have established is:
\[ x < y \]