In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answers
I `2x^2+3x+1=0`
II `8y^2+22y+15=0`

Options

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

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 + 3x + 1 = 0 \]

1. **Factor the equation**: We can rewrite the middle term (3x) as the sum of two terms that multiply to \(2 \times 1 = 2\) and add up to \(3\). We can split it into \(2x + x\).
\[ 2x^2 + 2x + x + 1 = 0 \]

2. **Group the terms**: Now, we can group the first two terms and the last two terms:
\[ (2x^2 + 2x) + (x + 1) = 0 \]

3. **Factor out the common terms**:
\[ 2x(x + 1) + 1(x + 1) = 0 \]

4. **Factor out the common binomial**:
\[ (x + 1)(2x + 1) = 0 \]

5. **Set each factor to zero**:
- From \(x + 1 = 0\), we get:
\[ x = -1 \]
- From \(2x + 1 = 0\), we get:
\[ 2x = -1 \implies x = -\frac{1}{2} \]

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

### Step 2: Solve Equation II
The second equation is:
\[ 8y^2 + 22y + 15 = 0 \]

1. **Factor the equation**: We need two numbers that multiply to \(8 \times 15 = 120\) and add up to \(22\). These numbers are \(12\) and \(10\).
\[ 8y^2 + 12y + 10y + 15 = 0 \]

2. **Group the terms**:
\[ (8y^2 + 12y) + (10y + 15) = 0 \]

3. **Factor out the common terms**:
\[ 4y(2y + 3) + 5(2y + 3) = 0 \]

4. **Factor out the common binomial**:
\[ (2y + 3)(4y + 5) = 0 \]

5. **Set each factor to zero**:
- From \(2y + 3 = 0\), we get:
\[ 2y = -3 \implies y = -\frac{3}{2} \]
- From \(4y + 5 = 0\), we get:
\[ 4y = -5 \implies y = -\frac{5}{4} \]

Thus, the solutions for \(y\) are:
\[ y = -\frac{3}{2} \quad \text{and} \quad y = -\frac{5}{4} \]

### Step 3: Compare the Values of \(x\) and \(y\)
Now we have:
- For \(x\): \( -1 \) and \( -\frac{1}{2} \)
- For \(y\): \( -\frac{3}{2} \) and \( -\frac{5}{4} \)

To compare:
- \( -1 > -\frac{3}{2} \) (which is -1.5)
- \( -\frac{1}{2} > -\frac{5}{4} \) (which is -1.25)

Both values of \(x\) are greater than both values of \(y\).

### Final Conclusion
Thus, the correct relation is:
\[ x > y \]