In each of the following questions, two equations are given. You have to solve them and
I. `8x^(2) + 10x +3 = 0` II. `2y^(2) + 5y + 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 one.

### Step 1: Solve the first equation \(8x^2 + 10x + 3 = 0\)

We can use the factorization method here.

1. **Rearranging the equation**:
We can rewrite the equation as:
\[ 8x^2 + 4x + 6x + 3 = 0 \]

2. **Grouping terms**:
Now we can group the terms:
\[ (8x^2 + 4x) + (6x + 3) = 0 \]

3. **Factoring out common terms**:
From the first group, we can factor out \(4x\):
\[ 4x(2x + 1) + 3(2x + 1) = 0 \]

4. **Factoring the entire equation**:
Now we can factor out \((2x + 1)\):
\[ (2x + 1)(4x + 3) = 0 \]

5. **Finding the roots**:
Setting each factor to zero gives:
\[ 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]
\[ 4x + 3 = 0 \quad \Rightarrow \quad x = -\frac{3}{4} \]

### Step 2: Solve the second equation \(2y^2 + 5y + 3 = 0\)

Similarly, we will factor this equation.

1. **Rearranging the equation**:
We can rewrite the equation as:
\[ 2y^2 + 3y + 2y + 3 = 0 \]

2. **Grouping terms**:
Now we can group the terms:
\[ (2y^2 + 3y) + (2y + 3) = 0 \]

3. **Factoring out common terms**:
From the first group, we can factor out \(y\):
\[ y(2y + 3) + 1(2y + 3) = 0 \]

4. **Factoring the entire equation**:
Now we can factor out \((2y + 3)\):
\[ (y + 1)(2y + 3) = 0 \]

5. **Finding the roots**:
Setting each factor to zero gives:
\[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]
\[ 2y + 3 = 0 \quad \Rightarrow \quad y = -\frac{3}{2} \]

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

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

### Step 4: Determine the relationship between \(x\) and \(y\)

1. **Comparing \(-\frac{1}{2}\) and \(-1\)**:
\(-\frac{1}{2} > -1\)

2. **Comparing \(-\frac{1}{2}\) and \(-\frac{3}{2}\)**:
\(-\frac{1}{2} > -\frac{3}{2}\)

3. **Comparing \(-\frac{3}{4}\) and \(-1\)**:
\(-\frac{3}{4} > -1\)

4. **Comparing \(-\frac{3}{4}\) and \(-\frac{3}{2}\)**:
\(-\frac{3}{4} > -\frac{3}{2}\)

### Conclusion

In all cases, the values of \(x\) are greater than the values of \(y\). Therefore, we conclude that:

\[ x > y \]

### Final Answer
The answer is \(x > y\).

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