In the following questions, two equations numbered I and II are given. You have to solve both the equations and
I. `2x^2-17x+36=0 `
II. `y^2-7y+12=0`

Options

A. `xlty`
B. `xgty`
C. `x le y`
D. `x ge y`

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second one.

### Step 1: Solve the first equation \( 2x^2 - 17x + 36 = 0 \)

1. **Rearranging the equation**: The equation is already in standard form.
\[ 2x^2 - 17x + 36 = 0 \]

2. **Factoring the quadratic**: We look for two numbers that multiply to \( 2 \times 36 = 72 \) and add to \( -17 \). The numbers are \( -8 \) and \( -9 \).
\[ 2x^2 - 8x - 9x + 36 = 0 \]

3. **Grouping**: We can group the terms:
\[ (2x^2 - 8x) + (-9x + 36) = 0 \]

4. **Factoring by grouping**:
\[ 2x(x - 4) - 9(x - 4) = 0 \]
\[ (2x - 9)(x - 4) = 0 \]

5. **Setting each factor to zero**:
\[ 2x - 9 = 0 \quad \text{or} \quad x - 4 = 0 \]

6. **Solving for \( x \)**:
\[ 2x - 9 = 0 \implies 2x = 9 \implies x = \frac{9}{2} = 4.5 \]
\[ x - 4 = 0 \implies x = 4 \]

So, the solutions for \( x \) are \( x = 4 \) and \( x = 4.5 \).

### Step 2: Solve the second equation \( y^2 - 7y + 12 = 0 \)

1. **Rearranging the equation**: The equation is already in standard form.
\[ y^2 - 7y + 12 = 0 \]

2. **Factoring the quadratic**: We look for two numbers that multiply to \( 12 \) and add to \( -7 \). The numbers are \( -3 \) and \( -4 \).
\[ y^2 - 3y - 4y + 12 = 0 \]

3. **Grouping**: We can group the terms:
\[ (y^2 - 3y) + (-4y + 12) = 0 \]

4. **Factoring by grouping**:
\[ y(y - 3) - 4(y - 3) = 0 \]
\[ (y - 3)(y - 4) = 0 \]

5. **Setting each factor to zero**:
\[ y - 3 = 0 \quad \text{or} \quad y - 4 = 0 \]

6. **Solving for \( y \)**:
\[ y - 3 = 0 \implies y = 3 \]
\[ y - 4 = 0 \implies y = 4 \]

So, the solutions for \( y \) are \( y = 3 \) and \( y = 4 \).

### Summary of Solutions
- The values of \( x \) are \( 4 \) and \( 4.5 \).
- The values of \( y \) are \( 3 \) and \( 4 \).

### Conclusion
Now, we can compare the values of \( x \) and \( y \):
- The maximum value of \( x \) is \( 4.5 \) and the maximum value of \( y \) is \( 4 \).

Thus, we conclude that \( x \geq y \).