I. `2x^(2) + 17x + 35 = 0`
II. `8y^(2) - 14y-15 = 0`

Options

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

Explanation

To solve the equations provided, we will follow these steps:

### Step 1: Solve the first equation \(2x^2 + 17x + 35 = 0\)

1. **Identify the coefficients**:
- \(a = 2\), \(b = 17\), and \(c = 35\).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \(a \cdot c = 2 \cdot 35 = 70\) and add to \(b = 17\).
- The numbers are \(10\) and \(7\).
- Rewrite the equation:
\[ 2x^2 + 10x + 7x + 35 = 0 \]
- Group the terms:
\[ (2x^2 + 10x) + (7x + 35) = 0 \]
- Factor by grouping:
\[ 2x(x + 5) + 7(x + 5) = 0 \]
- Factor out the common term:
\[ (2x + 7)(x + 5) = 0 \]

3. **Find the values of \(x\)**:
- Set each factor to zero:
\[ 2x + 7 = 0 \quad \Rightarrow \quad x = -\frac{7}{2} \]
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]

### Step 2: Solve the second equation \(8y^2 - 14y - 15 = 0\)

1. **Identify the coefficients**:
- \(a = 8\), \(b = -14\), and \(c = -15\).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \(a \cdot c = 8 \cdot (-15) = -120\) and add to \(b = -14\).
- The numbers are \(-20\) and \(6\).
- Rewrite the equation:
\[ 8y^2 - 20y + 6y - 15 = 0 \]
- Group the terms:
\[ (8y^2 - 20y) + (6y - 15) = 0 \]
- Factor by grouping:
\[ 4y(2y - 5) + 3(2y - 5) = 0 \]
- Factor out the common term:
\[ (4y + 3)(2y - 5) = 0 \]

3. **Find the values of \(y\)**:
- Set each factor to zero:
\[ 4y + 3 = 0 \quad \Rightarrow \quad y = -\frac{3}{4} \]
\[ 2y - 5 = 0 \quad \Rightarrow \quad y = \frac{5}{2} \]

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

- The values of \(x\) are \(x = -\frac{7}{2}\) and \(x = -5\).
- The values of \(y\) are \(y = -\frac{3}{4}\) and \(y = \frac{5}{2}\).

### Conclusion

- Since \(y = \frac{5}{2}\) is greater than both values of \(x\) (as \(-\frac{3}{4} > -5\) and \(-\frac{3}{4} > -\frac{7}{2}\)), we conclude that \(y > x\).

### Final Answer

The values of \(x\) are \(-\frac{7}{2}\) and \(-5\), and the values of \(y\) are \(-\frac{3}{4}\) and \(\frac{5}{2}\). The relation is \(y > x\).

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