I. ` 15x^(2) - 29x - 14 = 0`
II. ` 6y^(2) - 5y - 25 = 0`

Options

A. `x gt y`
B. ` x ge y`
C. ` x lt y`
D. Relationship between x and y cannot be determined

Explanation

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation for \( x \)
The first equation is:
\[ 15x^2 - 29x - 14 = 0 \]

To solve this quadratic equation, we will use the middle-term splitting method. First, we need to find the product of the coefficient of \( x^2 \) (which is 15) and the constant term (which is -14):
\[ 15 \times -14 = -210 \]

Next, we need to find two numbers that multiply to -210 and add up to -29. The factors of -210 that satisfy this condition are -35 and 6, since:
\[ -35 + 6 = -29 \]

Now we can rewrite the equation:
\[ 15x^2 - 35x + 6x - 14 = 0 \]

### Step 2: Group the terms
Now, we group the terms:
\[ (15x^2 - 35x) + (6x - 14) = 0 \]

### Step 3: Factor by grouping
Factoring out the common terms in each group:
\[ 5x(3x - 7) + 2(3x - 7) = 0 \]

Now we can factor out \( (3x - 7) \):
\[ (3x - 7)(5x + 2) = 0 \]

### Step 4: Set each factor to zero
Now we set each factor to zero:
1. \( 3x - 7 = 0 \) → \( x = \frac{7}{3} \)
2. \( 5x + 2 = 0 \) → \( x = -\frac{2}{5} \)

Thus, the solutions for \( x \) are:
\[ x = \frac{7}{3} \quad \text{and} \quad x = -\frac{2}{5} \]

### Step 5: Solve the second equation for \( y \)
The second equation is:
\[ 6y^2 - 5y - 25 = 0 \]

Again, we will use the middle-term splitting method. First, we find the product of the coefficient of \( y^2 \) (which is 6) and the constant term (which is -25):
\[ 6 \times -25 = -150 \]

We need to find two numbers that multiply to -150 and add up to -5. The factors that satisfy this condition are -15 and 10, since:
\[ -15 + 10 = -5 \]

Now we can rewrite the equation:
\[ 6y^2 - 15y + 10y - 25 = 0 \]

### Step 6: Group the terms
Now, we group the terms:
\[ (6y^2 - 15y) + (10y - 25) = 0 \]

### Step 7: Factor by grouping
Factoring out the common terms in each group:
\[ 3y(2y - 5) + 5(2y - 5) = 0 \]

Now we can factor out \( (2y - 5) \):
\[ (2y - 5)(3y + 5) = 0 \]

### Step 8: Set each factor to zero
Now we set each factor to zero:
1. \( 2y - 5 = 0 \) → \( y = \frac{5}{2} \)
2. \( 3y + 5 = 0 \) → \( y = -\frac{5}{3} \)

Thus, the solutions for \( y \) are:
\[ y = \frac{5}{2} \quad \text{and} \quad y = -\frac{5}{3} \]

### Summary of Solutions
- For \( x \): \( x = \frac{7}{3} \) and \( x = -\frac{2}{5} \)
- For \( y \): \( y = \frac{5}{2} \) and \( y = -\frac{5}{3} \)

### Step 9: Establish the relationship between \( x \) and \( y \)
Now, we need to compare the values of \( x \) and \( y \):
1. \( \frac{7}{3} \approx 2.33 \) and \( \frac{5}{2} = 2.5 \)
2. \( -\frac{2}{5} = -0.4 \) and \( -\frac{5}{3} \approx -1.67 \)

From this, we can see:
- \( \frac{7}{3} > \frac{5}{2} \)
- \( -\frac{2}{5} > -\frac{5}{3} \)

### Conclusion
Thus, we cannot establish a definitive relationship between \( x \) and \( y \) since \( x \) can be greater than \( y \) in one case and less in another. Therefore, the relationship between \( x \) and \( y \) cannot be determined.

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