In each of these questions, two equations I. and II. are give. you have to solve both the equations and give answer
I. ` x^(2) - 11 x + 30 = 0`
II. ` 56 y^(2) - 151 y + 99 = 0 `

Options

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

Answer

To solve the equations provided in the question, we will proceed step by step for each equation.

### Step 1: Solve the first equation \( x^2 - 11x + 30 = 0 \)

1. **Factor the quadratic equation**: We need to find two numbers that multiply to \( 30 \) (the constant term) and add up to \( -11 \) (the coefficient of \( x \)).
- The numbers are \( -6 \) and \( -5 \).

Therefore, we can rewrite the equation as:
\[ (x - 6)(x - 5) = 0 \]

2. **Set each factor to zero**:
\[ x - 6 = 0 \quad \text{or} \quad x - 5 = 0 \]

3. **Solve for \( x \)**:
- From \( x - 6 = 0 \), we get \( x = 6 \).
- From \( x - 5 = 0 \), we get \( x = 5 \).

Thus, the solutions for the first equation are:
\[ x = 6 \quad \text{and} \quad x = 5 \]

### Step 2: Solve the second equation \( 56y^2 - 151y + 99 = 0 \)

1. **Factor the quadratic equation**: We need to find two numbers that multiply to \( 56 \times 99 = 5544 \) and add up to \( -151 \).
- The numbers are \( -88 \) and \( -63 \).

Therefore, we can rewrite the equation as:
\[ 56y^2 - 88y - 63y + 99 = 0 \]

2. **Group the terms**:
\[ (56y^2 - 88y) + (-63y + 99) = 0 \]

3. **Factor by grouping**:
- From the first group \( 56y^2 - 88y \), we can factor out \( 8y \):
\[ 8y(7y - 11) \]
- From the second group \( -63y + 99 \), we can factor out \( -9 \):
\[ -9(7y - 11) \]

Thus, we can rewrite the equation as:
\[ (8y - 9)(7y - 11) = 0 \]

4. **Set each factor to zero**:
\[ 8y - 9 = 0 \quad \text{or} \quad 7y - 11 = 0 \]

5. **Solve for \( y \)**:
- From \( 8y - 9 = 0 \), we get \( y = \frac{9}{8} \).
- From \( 7y - 11 = 0 \), we get \( y = \frac{11}{7} \).

Thus, the solutions for the second equation are:
\[ y = \frac{9}{8} \quad \text{and} \quad y = \frac{11}{7} \]

### Summary of Solutions
- From the first equation: \( x = 6 \) or \( x = 5 \)
- From the second equation: \( y = \frac{9}{8} \) or \( y = \frac{11}{7} \)

### Step 3: Compare \( x \) and \( y \)

Now we need to compare the values of \( x \) and \( y \):
- For \( x = 6 \) and \( y = \frac{11}{7} \approx 1.57 \), we have \( 6 > \frac{11}{7} \).
- For \( x = 5 \) and \( y = \frac{9}{8} = 1.125 \), we have \( 5 > \frac{9}{8} \).

### Conclusion
In both cases, \( x \) is greater than \( y \). Therefore, we conclude that:
\[ x > y \]