In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
I`x^2-3x+2=0`
II `2y^2-7y+6=0`

Options

A. `xgey`
B. `xlty`
C. `xley`
D. x = y or relationship between x and y cannot be established.

Answer

To solve the given equations step by step, we will tackle each quadratic equation separately.

### Step 1: Solve the first equation \( I: x^2 - 3x + 2 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 1 \), \( b = -3 \), and \( c = 2 \).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( ac = 1 \times 2 = 2 \) and add up to \( b = -3 \).
- The numbers are \( -1 \) and \( -2 \).
- Therefore, we can factor the equation as:
\[ (x - 1)(x - 2) = 0 \]

3. **Set each factor to zero**:
- \( x - 1 = 0 \) → \( x = 1 \)
- \( x - 2 = 0 \) → \( x = 2 \)

4. **Solutions for x**:
- The solutions for the first equation are \( x = 1 \) and \( x = 2 \).

### Step 2: Solve the second equation \( II: 2y^2 - 7y + 6 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 2 \), \( b = -7 \), and \( c = 6 \).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( ac = 2 \times 6 = 12 \) and add up to \( b = -7 \).
- The numbers are \( -3 \) and \( -4 \).
- Therefore, we can rewrite the equation as:
\[ 2y^2 - 3y - 4y + 6 = 0 \]
- Now, factor by grouping:
\[ y(2y - 3) - 2(2y - 3) = 0 \]
\[ (2y - 3)(y - 2) = 0 \]

3. **Set each factor to zero**:
- \( 2y - 3 = 0 \) → \( y = \frac{3}{2} = 1.5 \)
- \( y - 2 = 0 \) → \( y = 2 \)

4. **Solutions for y**:
- The solutions for the second equation are \( y = 1.5 \) and \( y = 2 \).

### Step 3: Compare the solutions

- From the first equation, we have \( x = 1 \) and \( x = 2 \).
- From the second equation, we have \( y = 1.5 \) and \( y = 2 \).

### Step 4: Determine the relationship between x and y

- The possible pairs are:
- If \( x = 1 \), then \( y = 1.5 \) (1 < 1.5)
- If \( x = 2 \), then \( y = 2 \) (2 = 2)

Thus, we can conclude that \( y \) is greater than or equal to \( x \).

### Final Answer:
The correct relation is \( x \leq y \).

---