(i) ` 15x^(2) - 8x + 1 = 0`
(ii) ` 6y^(2) - 19y + 10 = 0`

Options

A. ` x le y`
B. x = y or no relation can be established between x & y.
C. ` x gt y`
D. ` x lt y`

Explanation

To solve the equations given in the question, we will follow the steps below for each equation.

### Step-by-Step Solution

#### (i) Solve the equation: \( 15x^2 - 8x + 1 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 15 \), \( b = -8 \), and \( c = 1 \).

2. **Use the middle term splitting method**:
- We need to factor the quadratic. We look for two numbers that multiply to \( a \cdot c = 15 \cdot 1 = 15 \) and add up to \( b = -8 \).
- The numbers are -5 and -3 (since \(-5 \times -3 = 15\) and \(-5 + -3 = -8\)).

3. **Rewrite the equation**:
- Rewrite \( -8x \) as \( -5x - 3x \):
\[ 15x^2 - 5x - 3x + 1 = 0 \]

4. **Group the terms**:
- Group the first two and the last two terms:
\[ (15x^2 - 5x) + (-3x + 1) = 0 \]

5. **Factor by grouping**:
- Factor out common terms:
\[ 5x(3x - 1) - 1(3x - 1) = 0 \]
- This gives:
\[ (5x - 1)(3x - 1) = 0 \]

6. **Set each factor to zero**:
- Solve \( 5x - 1 = 0 \) and \( 3x - 1 = 0 \):
\[ 5x = 1 \implies x = \frac{1}{5} \]
\[ 3x = 1 \implies x = \frac{1}{3} \]

Thus, the values of \( x \) are \( \frac{1}{5} \) and \( \frac{1}{3} \).

#### (ii) Solve the equation: \( 6y^2 - 19y + 10 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 6 \), \( b = -19 \), and \( c = 10 \).

2. **Use the middle term splitting method**:
- We need to factor the quadratic. We look for two numbers that multiply to \( a \cdot c = 6 \cdot 10 = 60 \) and add up to \( b = -19 \).
- The numbers are -15 and -4 (since \(-15 \times -4 = 60\) and \(-15 + -4 = -19\)).

3. **Rewrite the equation**:
- Rewrite \( -19y \) as \( -15y - 4y \):
\[ 6y^2 - 15y - 4y + 10 = 0 \]

4. **Group the terms**:
- Group the first two and the last two terms:
\[ (6y^2 - 15y) + (-4y + 10) = 0 \]

5. **Factor by grouping**:
- Factor out common terms:
\[ 3y(2y - 5) - 2(2y - 5) = 0 \]
- This gives:
\[ (3y - 2)(2y - 5) = 0 \]

6. **Set each factor to zero**:
- Solve \( 3y - 2 = 0 \) and \( 2y - 5 = 0 \):
\[ 3y = 2 \implies y = \frac{2}{3} \]
\[ 2y = 5 \implies y = \frac{5}{2} \]

Thus, the values of \( y \) are \( \frac{2}{3} \) and \( \frac{5}{2} \).

### Summary of Values
- Values of \( x \): \( \frac{1}{5}, \frac{1}{3} \)
- Values of \( y \): \( \frac{2}{3}, \frac{5}{2} \)

### Comparison of \( x \) and \( y \)
- The denominators of \( x \) are 5 and 3, while the denominators of \( y \) are 3 and 2.
- Since the values of \( x \) and \( y \) cannot be directly compared due to different denominators, we conclude that there is no established relation between \( x \) and \( y \).

### Final Answer
The correct option is that no relation can be established between \( x \) and \( y \).