I.` 15x^(2) - 2x-1=0`
II.` y^(2) - 9y+14 = 0`

Options

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

Explanation

To solve the given quadratic equations step by step, we will start with the first equation and then proceed to the second one.

### Step 1: Solve the first equation \( 15x^2 - 2x - 1 = 0 \)

To factor the quadratic equation, we need to find two numbers that multiply to \( 15 \times (-1) = -15 \) and add up to \( -2 \).

1. The numbers that satisfy this condition are \( -5 \) and \( 3 \).
2. Rewrite the equation by splitting the middle term:
\[ 15x^2 - 5x + 3x - 1 = 0 \]
3. Group the terms:
\[ (15x^2 - 5x) + (3x - 1) = 0 \]
4. Factor by grouping:
\[ 5x(3x - 1) + 1(3x - 1) = 0 \]
5. Factor out the common term:
\[ (3x - 1)(5x + 1) = 0 \]
6. Set each factor to zero:
\[ 3x - 1 = 0 \quad \text{or} \quad 5x + 1 = 0 \]
7. Solve for \( x \):
\[ 3x - 1 = 0 \implies x = \frac{1}{3} \]
\[ 5x + 1 = 0 \implies x = -\frac{1}{5} \]

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

### Step 2: Solve the second equation \( y^2 - 9y + 14 = 0 \)

1. We need to find two numbers that multiply to \( 14 \) and add up to \( -9 \).
2. The numbers that satisfy this condition are \( -7 \) and \( -2 \).
3. Rewrite the equation by splitting the middle term:
\[ y^2 - 7y - 2y + 14 = 0 \]
4. Group the terms:
\[ (y^2 - 7y) + (-2y + 14) = 0 \]
5. Factor by grouping:
\[ y(y - 7) - 2(y - 7) = 0 \]
6. Factor out the common term:
\[ (y - 7)(y - 2) = 0 \]
7. Set each factor to zero:
\[ y - 7 = 0 \quad \text{or} \quad y - 2 = 0 \]
8. Solve for \( y \):
\[ y - 7 = 0 \implies y = 7 \]
\[ y - 2 = 0 \implies y = 2 \]

Thus, the values of \( y \) are \( y = 7 \) and \( y = 2 \).

### Step 3: Determine the relationship between \( x \) and \( y \)

Now we have the values:
- \( x = \frac{1}{3}, -\frac{1}{5} \)
- \( y = 7, 2 \)

Since \( \frac{1}{3} \) and \( -\frac{1}{5} \) are both less than both \( 7 \) and \( 2 \), we can conclude that \( x < y \).

### Final Answer:
The values of \( x \) are \( \frac{1}{3} \) and \( -\frac{1}{5} \), and the values of \( y \) are \( 7 \) and \( 2 \). The relationship is \( x < y \).

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