In each of the following questions two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. Give answer
I `x^2+7x+12=0`
II `y^2`+5y+6=0

Options

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

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation \( x^2 + 7x + 12 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 7 \), and \( c = 12 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( c \) (12) and add up to \( b \) (7). The numbers are 3 and 4.
3. **Write the factored form**:
\[ (x + 3)(x + 4) = 0 \]
4. **Set each factor to zero**:
- \( x + 3 = 0 \) ⇒ \( x = -3 \)
- \( x + 4 = 0 \) ⇒ \( x = -4 \)

So, the solutions for \( x \) are \( x = -3 \) and \( x = -4 \).

### Step 2: Solve the second equation \( y^2 + 5y + 6 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 5 \), and \( c = 6 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( c \) (6) and add up to \( b \) (5). The numbers are 2 and 3.
3. **Write the factored form**:
\[ (y + 2)(y + 3) = 0 \]
4. **Set each factor to zero**:
- \( y + 2 = 0 \) ⇒ \( y = -2 \)
- \( y + 3 = 0 \) ⇒ \( y = -3 \)

So, the solutions for \( y \) are \( y = -2 \) and \( y = -3 \).

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

Now we have the following values:
- For \( x \): \( -3, -4 \)
- For \( y \): \( -2, -3 \)

We will compare these values:

1. **Compare \( x = -3 \) and \( y = -2 \)**:
- \( -3 < -2 \) (So, \( x < y \))

2. **Compare \( x = -3 \) and \( y = -3 \)**:
- \( -3 = -3 \) (So, \( x = y \))

3. **Compare \( x = -4 \) and \( y = -2 \)**:
- \( -4 < -2 \) (So, \( x < y \))

4. **Compare \( x = -4 \) and \( y = -3 \)**:
- \( -4 < -3 \) (So, \( x < y \))

### Conclusion

From our comparisons, we can conclude that:
- \( x \) is either less than or equal to \( y \).

Thus, the final answer is that \( x \) is either less than or equal to \( y \).