In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
I `2x^2-3x+1=0`
II `5y^2-6y-8=0`

Options

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

Answer

To solve the given equations step by step, we will first focus on the equation for \( x \) and then for \( y \).

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

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

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]

4. **Substitute values into the formula**:
\[ x = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 2} = \frac{3 \pm 1}{4} \]

5. **Calculate the two possible values for \( x \)**:
- \( x_1 = \frac{3 + 1}{4} = \frac{4}{4} = 1 \)
- \( x_2 = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2} \)

### Step 2: Solve the second equation \( 5y^2 - 6y - 8 = 0 \)

1. **Identify the coefficients**:
- \( a = 5 \)
- \( b = -6 \)
- \( c = -8 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 5 \cdot (-8) = 36 + 160 = 196 \]

4. **Substitute values into the formula**:
\[ y = \frac{-(-6) \pm \sqrt{196}}{2 \cdot 5} = \frac{6 \pm 14}{10} \]

5. **Calculate the two possible values for \( y \)**:
- \( y_1 = \frac{6 + 14}{10} = \frac{20}{10} = 2 \)
- \( y_2 = \frac{6 - 14}{10} = \frac{-8}{10} = -\frac{4}{5} \)

### Step 3: Compare the values of \( x \) and \( y \)

1. **Possible values for \( x \)**: \( 1 \) and \( \frac{1}{2} \)
2. **Possible values for \( y \)**: \( 2 \) and \( -\frac{4}{5} \)

### Step 4: Establish relationships

1. **First comparison**:
- \( x = \frac{1}{2} \) and \( y = 2 \):
- \( \frac{1}{2} < 2 \) (so \( x < y \))

2. **Second comparison**:
- \( x = 1 \) and \( y = -\frac{4}{5} \):
- \( 1 > -\frac{4}{5} \) (so \( x > y \))

### Conclusion

Since \( x \) can be both less than and greater than \( y \) depending on the values chosen, the relationship between \( x \) and \( y \) cannot be established.