I. `6x^(2)+x-1=0`
II. `8y^(2)+10y+3=0`

Options

A. `xley`
B. `xgty`
C. `xgey`
D. `xlty`

Answer

To solve the given quadratic equations and find the relation between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first quadratic equation \( 6x^2 + x - 1 = 0 \)

We will use the method of factoring. We need to find two numbers that multiply to \( 6 \times -1 = -6 \) and add up to \( 1 \) (the coefficient of \( x \)).

The numbers that satisfy this are \( 3 \) and \( -2 \) because:
- \( 3 + (-2) = 1 \)
- \( 3 \times (-2) = -6 \)

Now we can rewrite the equation as:
\[ 6x^2 + 3x - 2x - 1 = 0 \]

### Step 2: Factor the equation

Grouping the terms:
\[ (6x^2 + 3x) + (-2x - 1) = 0 \]
Factoring out common terms:
\[ 3x(2x + 1) - 1(2x + 1) = 0 \]
Now, factor by grouping:
\[ (3x - 1)(2x + 1) = 0 \]

### Step 3: Find the roots for \( x \)

Setting each factor to zero gives:
1. \( 3x - 1 = 0 \) → \( x = \frac{1}{3} \)
2. \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \)

So the roots for \( x \) are \( x_1 = \frac{1}{3} \) and \( x_2 = -\frac{1}{2} \).

### Step 4: Solve the second quadratic equation \( 8y^2 + 10y + 3 = 0 \)

Again, we will factor this equation. We need two numbers that multiply to \( 8 \times 3 = 24 \) and add up to \( 10 \).

The numbers that satisfy this are \( 6 \) and \( 4 \) because:
- \( 6 + 4 = 10 \)
- \( 6 \times 4 = 24 \)

Now we can rewrite the equation as:
\[ 8y^2 + 6y + 4y + 3 = 0 \]

### Step 5: Factor the equation

Grouping the terms:
\[ (8y^2 + 6y) + (4y + 3) = 0 \]
Factoring out common terms:
\[ 2y(4y + 3) + 1(4y + 3) = 0 \]
Now, factor by grouping:
\[ (2y + 1)(4y + 3) = 0 \]

### Step 6: Find the roots for \( y \)

Setting each factor to zero gives:
1. \( 2y + 1 = 0 \) → \( y = -\frac{1}{2} \)
2. \( 4y + 3 = 0 \) → \( y = -\frac{3}{4} \)

So the roots for \( y \) are \( y_1 = -\frac{1}{2} \) and \( y_2 = -\frac{3}{4} \).

### Step 7: Compare the roots of \( x \) and \( y \)

Now we have:
- \( x_1 = \frac{1}{3} \), \( x_2 = -\frac{1}{2} \)
- \( y_1 = -\frac{1}{2} \), \( y_2 = -\frac{3}{4} \)

Now we compare:
1. \( x_2 = -\frac{1}{2} \) and \( y_1 = -\frac{1}{2} \): \( x_2 = y_1 \)
2. \( x_2 = -\frac{1}{2} \) and \( y_2 = -\frac{3}{4} \): \( x_2 > y_2 \)
3. \( x_1 = \frac{1}{3} \) and \( y_1 = -\frac{1}{2} \): \( x_1 > y_1 \)
4. \( x_1 = \frac{1}{3} \) and \( y_2 = -\frac{3}{4} \): \( x_1 > y_2 \)

### Conclusion

From the comparisons, we can conclude that:
\[ x > y \]