In each of the following questions, two equations are given. You have to solve them and
I. `6x^(2) + x-1 = 0` II. `8y^(2) + 10y + 3 = 0`

Options

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

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

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

1. **Rearranging the equation**: The equation is already in standard form \(ax^2 + bx + c = 0\) where \(a = 6\), \(b = 1\), and \(c = -1\).

2. **Factoring the quadratic**: We will factor the quadratic expression. We can rewrite the middle term:
\[ 6x^2 + 3x - 2x - 1 = 0 \]
Now, group the terms:
\[ (6x^2 + 3x) + (-2x - 1) = 0 \]
Factor out common terms:
\[ 3x(2x + 1) - 1(2x + 1) = 0 \]
This gives us:
\[ (3x - 1)(2x + 1) = 0 \]

3. **Finding the roots**: Set each factor to zero:
- \(3x - 1 = 0 \Rightarrow x = \frac{1}{3}\)
- \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)

Thus, the solutions for \(x\) are:
\[ x = \frac{1}{3} \quad \text{and} \quad x = -\frac{1}{2} \]

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

1. **Rearranging the equation**: The equation is already in standard form where \(a = 8\), \(b = 10\), and \(c = 3\).

2. **Factoring the quadratic**: We will factor the quadratic expression. We can rewrite the middle term:
\[ 8y^2 + 4y + 6y + 3 = 0 \]
Now, group the terms:
\[ (8y^2 + 4y) + (6y + 3) = 0 \]
Factor out common terms:
\[ 4y(2y + 1) + 3(2y + 1) = 0 \]
This gives us:
\[ (4y + 3)(2y + 1) = 0 \]

3. **Finding the roots**: Set each factor to zero:
- \(4y + 3 = 0 \Rightarrow y = -\frac{3}{4}\)
- \(2y + 1 = 0 \Rightarrow y = -\frac{1}{2}\)

Thus, the solutions for \(y\) are:
\[ y = -\frac{3}{4} \quad \text{and} \quad y = -\frac{1}{2} \]

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

Now we have the values:
- For \(x\): \(\frac{1}{3}\) and \(-\frac{1}{2}\)
- For \(y\): \(-\frac{3}{4}\) and \(-\frac{1}{2}\)

We will compare these values:
1. **Comparing \(x = \frac{1}{3}\) with \(y = -\frac{3}{4}\)**:
\[ \frac{1}{3} > -\frac{3}{4} \]

2. **Comparing \(x = \frac{1}{3}\) with \(y = -\frac{1}{2}\)**:
\[ \frac{1}{3} > -\frac{1}{2} \]

3. **Comparing \(x = -\frac{1}{2}\) with \(y = -\frac{3}{4}\)**:
\[ -\frac{1}{2} > -\frac{3}{4} \]

4. **Comparing \(x = -\frac{1}{2}\) with \(y = -\frac{1}{2}\)**:
\[ -\frac{1}{2} = -\frac{1}{2} \]

### Conclusion

From the comparisons, we can conclude that:
- \(x\) is always greater than or equal to \(y\).

Thus, the final answer is:
\[ \text{Option: } x \geq y \]