In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answers
I `x^2+8x+15=0`
II `2y^2+7y+3=0`

Options

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

Answer

To solve the given equations step by step, we will first solve each quadratic equation separately.

### Step 1: Solve Equation I: \( x^2 + 8x + 15 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 1 \), \( b = 8 \), and \( c = 15 \).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( c \) (15) and add up to \( b \) (8).
- The numbers 3 and 5 satisfy this condition because \( 3 \times 5 = 15 \) and \( 3 + 5 = 8 \).

3. **Rewrite the equation**:
- We can factor the equation as:
\[ (x + 3)(x + 5) = 0 \]

4. **Set each factor to zero**:
- \( x + 3 = 0 \) or \( x + 5 = 0 \)

5. **Solve for \( x \)**:
- From \( x + 3 = 0 \), we get \( x = -3 \).
- From \( x + 5 = 0 \), we get \( x = -5 \).

Thus, the solutions for Equation I are:
\[ x = -3 \quad \text{and} \quad x = -5 \]

### Step 2: Solve Equation II: \( 2y^2 + 7y + 3 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 2 \), \( b = 7 \), and \( c = 3 \).

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( a \cdot c = 2 \cdot 3 = 6 \) and add up to \( b \) (7).
- The numbers 6 and 1 satisfy this condition because \( 6 \times 1 = 6 \) and \( 6 + 1 = 7 \).

3. **Rewrite the equation**:
- We can rewrite the middle term:
\[ 2y^2 + 6y + 1y + 3 = 0 \]
- Grouping gives us:
\[ 2y(y + 3) + 1(y + 3) = 0 \]
- Factoring out the common term:
\[ (y + 3)(2y + 1) = 0 \]

4. **Set each factor to zero**:
- \( y + 3 = 0 \) or \( 2y + 1 = 0 \)

5. **Solve for \( y \)**:
- From \( y + 3 = 0 \), we get \( y = -3 \).
- From \( 2y + 1 = 0 \), we get \( y = -\frac{1}{2} \).

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

### Step 3: Compare the solutions

- The solutions for \( x \) are \( -3 \) and \( -5 \).
- The solutions for \( y \) are \( -3 \) and \( -\frac{1}{2} \).

### Conclusion:
- The common value is \( -3 \), which means \( x = y \) for \( x = -3 \).
- For the other values, \( -5 < -\frac{1}{2} \).

Thus, we can conclude:
\[ x \leq y \]

### Final Answer:
The correct option is \( x \leq y \).

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