In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I `2x^2-9x+9=0`
II `y^2-7y+12=0`

Options

A. `xlty`
B. `xley`
C. `xgty`
D. x = y or relation can’t be established

Answer

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

### Step 1: Solve Equation I: \(2x^2 - 9x + 9 = 0\)

1. **Identify the coefficients**:
- \(a = 2\), \(b = -9\), \(c = 9\)

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-9)^2 - 4 \cdot 2 \cdot 9 = 81 - 72 = 9 \]

4. **Substitute values into the formula**:
\[ x = \frac{-(-9) \pm \sqrt{9}}{2 \cdot 2} = \frac{9 \pm 3}{4} \]

5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{9 + 3}{4} = \frac{12}{4} = 3 \]
- Second value:
\[ x_2 = \frac{9 - 3}{4} = \frac{6}{4} = \frac{3}{2} \]

### Step 2: Solve Equation II: \(y^2 - 7y + 12 = 0\)

1. **Identify the coefficients**:
- \(a = 1\), \(b = -7\), \(c = 12\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1 \]

4. **Substitute values into the formula**:
\[ y = \frac{-(-7) \pm \sqrt{1}}{2 \cdot 1} = \frac{7 \pm 1}{2} \]

5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{7 + 1}{2} = \frac{8}{2} = 4 \]
- Second value:
\[ y_2 = \frac{7 - 1}{2} = \frac{6}{2} = 3 \]

### Summary of Solutions:
- From Equation I: \(x = 3\) or \(x = \frac{3}{2}\)
- From Equation II: \(y = 4\) or \(y = 3\)

### Conclusion:
We have the pairs of solutions:
1. \(x = 3\), \(y = 4\)
2. \(x = 3\), \(y = 3\)
3. \(x = \frac{3}{2}\), \(y = 4\)
4. \(x = \frac{3}{2}\), \(y = 3\)

### Answer:
The appropriate answer based on the conditions \(x \leq y\) is:
- \(x = 3\) and \(y = 4\) (Option 2: \(x \leq y\))