In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer
I.` 2 x ^(2) - 9 x + 9 = 0`
II. ` 6 y ^(2) - 17 y + 12 = 0 `

Options

A. If x= y or no relation
B. If ` x gt y `
C. If ` x lt y `
D. If `x ge y`

Answer

To solve the given equations step by step, we will start with equation (I) and then move to equation (II).

### Step 1: Solve Equation (I)
The first equation is:
\[ 2x^2 - 9x + 9 = 0 \]

We will use the factorization method to solve this quadratic equation.

1. **Multiply the coefficient of \(x^2\) (which is 2) with the constant term (which is 9)**:
\[ 2 \times 9 = 18 \]

2. **We need to find two numbers that multiply to 18 and add up to -9**. The numbers that satisfy this condition are -6 and -3.

3. **Rewrite the equation using these factors**:
\[ 2x^2 - 6x - 3x + 9 = 0 \]

4. **Group the terms**:
\[ (2x^2 - 6x) + (-3x + 9) = 0 \]

5. **Factor out the common terms from each group**:
\[ 2x(x - 3) - 3(x - 3) = 0 \]

6. **Factor out the common binomial factor**:
\[ (2x - 3)(x - 3) = 0 \]

7. **Set each factor to zero**:
- For \(2x - 3 = 0\):
\[ 2x = 3 \]
\[ x = \frac{3}{2} = 1.5 \]

- For \(x - 3 = 0\):
\[ x = 3 \]

Thus, the solutions for \(x\) are:
\[ x = 1.5 \quad \text{and} \quad x = 3 \]

### Step 2: Solve Equation (II)
The second equation is:
\[ 6y^2 - 17y + 12 = 0 \]

We will also use the factorization method for this quadratic equation.

1. **Multiply the coefficient of \(y^2\) (which is 6) with the constant term (which is 12)**:
\[ 6 \times 12 = 72 \]

2. **We need to find two numbers that multiply to 72 and add up to -17**. The numbers that satisfy this condition are -9 and -8.

3. **Rewrite the equation using these factors**:
\[ 6y^2 - 9y - 8y + 12 = 0 \]

4. **Group the terms**:
\[ (6y^2 - 9y) + (-8y + 12) = 0 \]

5. **Factor out the common terms from each group**:
\[ 3y(2y - 3) - 4(2y - 3) = 0 \]

6. **Factor out the common binomial factor**:
\[ (2y - 3)(3y - 4) = 0 \]

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

- For \(3y - 4 = 0\):
\[ 3y = 4 \]
\[ y = \frac{4}{3} \approx 1.33 \]

Thus, the solutions for \(y\) are:
\[ y = 1.5 \quad \text{and} \quad y \approx 1.33 \]

### Step 3: Compare Values of \(x\) and \(y\)
Now we have:
- \(x = 1.5\) and \(x = 3\)
- \(y = 1.5\) and \(y \approx 1.33\)

To compare:
- For \(x = 3\): \(3 > 1.5\) and \(3 > 1.33\)
- For \(x = 1.5\): \(1.5 = 1.5\) and \(1.5 > 1.33\)

### Conclusion
From the comparisons, we can conclude that:
- \(x\) can be greater than \(y\) or equal to \(y\) depending on the values chosen.

### Final Answer
Thus, the relationship can be summarized as:
\[ x \geq y \]