In the following questions, there are two equations in x and y . You have to solve both the equations and give answer
I.` 2 x^(2) - 3 x + 1 = 0`
II. ` 2 y^(2) - 5y + 3 = 0 `

Options

A. If ` x gt y `
B. If ` x lt y `
C. If ` x ge y `
D. If ` x le y `

Answer

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

### 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 quadratic formula**:
\[ x = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 2} = \frac{3 \pm 1}{4} \]

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

Thus, the solutions for \( x \) are:
\[ x = 1 \quad \text{and} \quad x = \frac{1}{2} \]

### Step 2: Solve the second equation \( 2y^2 - 5y + 3 = 0 \)

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = -5 \)
- \( c = 3 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 2} = \frac{5 \pm 1}{4} \]

5. **Calculate the two possible values for \( y \)**:
- \( y_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2} \)
- \( y_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1 \)

Thus, the solutions for \( y \) are:
\[ y = \frac{3}{2} \quad \text{and} \quad y = 1 \]

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

- The values of \( x \) are \( 1 \) and \( \frac{1}{2} \).
- The values of \( y \) are \( 1 \) and \( \frac{3}{2} \).

Now, we compare:
1. \( \frac{1}{2} < 1 \)
2. \( 1 = 1 \)
3. \( 1 < \frac{3}{2} \)

### Conclusion

From the comparisons, we can conclude:
- \( x \) can be less than or equal to \( y \).
- Therefore, the final relation is:
\[ x \leq y \]