In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct options
I. ` x^(2) - 11 x + 30 = 0 `
II. ` y^(2) - 15 y + 56 = 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 start with the first equation and then move on to the second 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 that satisfy this are \(-5\) and \(-6\).

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

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

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

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

### Step 2: Solve the second equation \( y^2 - 15y + 56 = 0 \)

1. **Factor the quadratic equation**: We need to find two numbers that multiply to \(56\) (the constant term) and add up to \(-15\) (the coefficient of \(y\)).
- The numbers that satisfy this are \(-7\) and \(-8\).

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

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

3. **Solve for \(y\)**:
- From \(y - 7 = 0\), we get \(y = 7\).
- From \(y - 8 = 0\), we get \(y = 8\).

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

### Final Solutions
- The solutions for the first equation \(x^2 - 11x + 30 = 0\) are \(x = 5\) and \(x = 6\).
- The solutions for the second equation \(y^2 - 15y + 56 = 0\) are \(y = 7\) and \(y = 8\).