I. ` 3x^(2) + 7x = 6 " "` II. ` 6(2y^(2)+1) = 17 y`

Options

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

Explanation

To solve the equations step by step, we will tackle each equation separately and find the values of \( x \) and \( y \).

### Step 1: Solve the first equation \( 3x^2 + 7x = 6 \)

1. **Rearranging the equation**:
\[ 3x^2 + 7x - 6 = 0 \]

2. **Factoring the quadratic equation**:
We need to factor \( 3x^2 + 7x - 6 \). We look for two numbers that multiply to \( 3 \times (-6) = -18 \) and add to \( 7 \). The numbers \( 9 \) and \( -2 \) work.
\[ 3x^2 + 9x - 2x - 6 = 0 \]
Grouping terms:
\[ (3x^2 + 9x) + (-2x - 6) = 0 \]
Factoring by grouping:
\[ 3x(x + 3) - 2(x + 3) = 0 \]
Factor out \( (x + 3) \):
\[ (x + 3)(3x - 2) = 0 \]

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

### Step 2: Solve the second equation \( 6(2y^2 + 1) = 17y \)

1. **Rearranging the equation**:
\[ 12y^2 + 6 - 17y = 0 \quad \Rightarrow \quad 12y^2 - 17y + 6 = 0 \]

2. **Factoring the quadratic equation**:
We need to find two numbers that multiply to \( 12 \times 6 = 72 \) and add to \( -17 \). The numbers \( -9 \) and \( -8 \) work.
\[ 12y^2 - 9y - 8y + 6 = 0 \]
Grouping terms:
\[ (12y^2 - 9y) + (-8y + 6) = 0 \]
Factoring by grouping:
\[ 3y(4y - 3) - 2(4y - 3) = 0 \]
Factor out \( (4y - 3) \):
\[ (4y - 3)(3y - 2) = 0 \]

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

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

We found:
- Values of \( x \): \( -3 \) and \( \frac{2}{3} \)
- Values of \( y \): \( \frac{3}{4} \) and \( \frac{2}{3} \)

Now, we compare these values:
- \( \frac{2}{3} \) is greater than \( -3 \) and less than \( \frac{3}{4} \).
- Therefore, \( x \) can be less than or equal to \( y \).

### Final Conclusion:
The relationship between \( x \) and \( y \) is:
\[ x \leq y \]