Directions:In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.
I. `6x^2-x-2=0` II. `5y^2-18y+9=0`

Options

A. if `x lt y`
B. if `x le y`
C. if `x gt y`
D. if x=y or the relationship can't be established.

Answer

To solve the given equations step by step, we'll start with each equation separately.

### Step 1: Solve Equation I: \( 6x^2 - x - 2 = 0 \)

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

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 6 \cdot (-2) = 1 + 48 = 49 \]

4. **Substitute into the quadratic formula**:
\[ x = \frac{-(-1) \pm \sqrt{49}}{2 \cdot 6} = \frac{1 \pm 7}{12} \]

5. **Calculate the two possible values for \( x \)**:
- First value:
\[ x_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \]
- Second value:
\[ x_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]

### Step 2: Solve Equation II: \( 5y^2 - 18y + 9 = 0 \)

1. **Identify the coefficients**:
- \( a = 5 \)
- \( b = -18 \)
- \( c = 9 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-18)^2 - 4 \cdot 5 \cdot 9 = 324 - 180 = 144 \]

4. **Substitute into the quadratic formula**:
\[ y = \frac{-(-18) \pm \sqrt{144}}{2 \cdot 5} = \frac{18 \pm 12}{10} \]

5. **Calculate the two possible values for \( y \)**:
- First value:
\[ y_1 = \frac{18 + 12}{10} = \frac{30}{10} = 3 \]
- Second value:
\[ y_2 = \frac{18 - 12}{10} = \frac{6}{10} = \frac{3}{5} \]

### Summary of Solutions:
- For Equation I: \( x = \frac{2}{3}, -\frac{1}{2} \)
- For Equation II: \( y = 3, \frac{3}{5} \)

### Final Comparison:
Now we compare the values of \( x \) and \( y \):
- \( x_1 = \frac{2}{3} \) and \( y_1 = 3 \) (Here \( x < y \))
- \( x_1 = \frac{2}{3} \) and \( y_2 = \frac{3}{5} \) (Here \( x > y \))
- \( x_2 = -\frac{1}{2} \) and \( y_1 = 3 \) (Here \( x < y \))
- \( x_2 = -\frac{1}{2} \) and \( y_2 = \frac{3}{5} \) (Here \( x < y \))

### Conclusion:
The relationships established are:
- \( x < y \) for all comparisons.