`I. x^(2) -7x + 12 = 0" "II. 3y^(2) - 11y + 10 = 0`

Options

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

Explanation

To solve the given equations step by step, we will factorize each quadratic equation and find the values of \( x \) and \( y \).

### Step 1: Solve the first equation \( x^2 - 7x + 12 = 0 \)

1. **Factor the equation**:
We need to find two numbers that multiply to \( 12 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)).
The numbers that satisfy this are \( -3 \) and \( -4 \).

Therefore, we can rewrite the equation as:
\[ (x - 3)(x - 4) = 0 \]

2. **Set each factor to zero**:
\[ x - 3 = 0 \quad \text{or} \quad x - 4 = 0 \]

3. **Solve for \( x \)**:
\[ x = 3 \quad \text{or} \quad x = 4 \]

### Step 2: Solve the second equation \( 3y^2 - 11y + 10 = 0 \)

1. **Factor the equation**:
We need to find two numbers that multiply to \( 3 \times 10 = 30 \) and add up to \( -11 \).
The numbers that satisfy this are \( -6 \) and \( -5 \).

Therefore, we can rewrite the equation as:
\[ 3y^2 - 6y - 5y + 10 = 0 \]
Grouping gives:
\[ 3y(y - 2) - 5(y - 2) = 0 \]
Factoring out \( (y - 2) \):
\[ (3y - 5)(y - 2) = 0 \]

2. **Set each factor to zero**:
\[ 3y - 5 = 0 \quad \text{or} \quad y - 2 = 0 \]

3. **Solve for \( y \)**:
\[ 3y = 5 \quad \Rightarrow \quad y = \frac{5}{3} \]
\[ y = 2 \]

### Summary of Solutions
- The solutions for \( x \) are \( x = 3 \) and \( x = 4 \).
- The solutions for \( y \) are \( y = \frac{5}{3} \) and \( y = 2 \).