In each of these questions, two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` x^(2) + 20 = 9 x `
II. ` y^(2) + 42 = 13 y `

Options

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

Answer

To solve the given equations step by step, we will follow the process of rearranging, factoring, and finding the roots for both equations.

### Step 1: Solve the first equation \( x^2 + 20 = 9x \)

1. Rearrange the equation:
\[ x^2 - 9x + 20 = 0 \]

2. Factor the quadratic equation:
- We need two numbers that multiply to \( 20 \) and add up to \( -9 \). The numbers are \( -5 \) and \( -4 \).
\[ (x - 5)(x - 4) = 0 \]

3. Set each factor to zero:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Step 2: Solve the second equation \( y^2 + 42 = 13y \)

1. Rearrange the equation:
\[ y^2 - 13y + 42 = 0 \]

2. Factor the quadratic equation:
- We need two numbers that multiply to \( 42 \) and add up to \( -13 \). The numbers are \( -6 \) and \( -7 \).
\[ (y - 6)(y - 7) = 0 \]

3. Set each factor to zero:
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]
\[ y - 7 = 0 \quad \Rightarrow \quad y = 7 \]

### Step 3: Summary of the solutions

- From the first equation, we found:
- \( x = 4 \) and \( x = 5 \)

- From the second equation, we found:
- \( y = 6 \) and \( y = 7 \)

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

- For \( x = 4 \):
- \( y \) can be \( 6 \) or \( 7 \) (both are greater than \( 4 \)).

- For \( x = 5 \):
- \( y \) can be \( 6 \) or \( 7 \) (both are greater than \( 5 \)).

Thus, in both cases, we have \( x < y \).

### Final Answer
The relationship established is:
\[ x < y \]