In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
I `3x^2+4x+1=0`
II `y^2+5y+6=0`

Options

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

Answer

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

### Step 1: Solve Equation I: \(3x^2 + 4x + 1 = 0\)

1. **Identify the coefficients**:
- Here, \(a = 3\), \(b = 4\), and \(c = 1\).

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 1 = 16 - 12 = 4 \]
Since the discriminant is positive, we will have two real solutions.

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-4 \pm \sqrt{4}}{2 \cdot 3} = \frac{-4 \pm 2}{6} \]

5. **Calculate the two possible values for \(x\)**:
- First solution:
\[ x_1 = \frac{-4 + 2}{6} = \frac{-2}{6} = -\frac{1}{3} \]
- Second solution:
\[ x_2 = \frac{-4 - 2}{6} = \frac{-6}{6} = -1 \]

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

### Step 2: Solve Equation II: \(y^2 + 5y + 6 = 0\)

1. **Identify the coefficients**:
- Here, \(a = 1\), \(b = 5\), and \(c = 6\).

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]
Since the discriminant is positive, we will have two real solutions.

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-5 \pm \sqrt{1}}{2 \cdot 1} = \frac{-5 \pm 1}{2} \]

5. **Calculate the two possible values for \(y\)**:
- First solution:
\[ y_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \]
- Second solution:
\[ y_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \]

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

### Conclusion

The solutions to the equations are:
- From Equation I: \(x = -1\) and \(x = -\frac{1}{3}\)
- From Equation II: \(y = -2\) and \(y = -3\)

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

- Compare \(x\) values with \(y\) values:
- For \(x = -1\) and \(y = -2\): \(-1 > -2\)
- For \(x = -\frac{1}{3}\) and \(y = -3\): \(-\frac{1}{3} > -3\)

Both comparisons show that \(x\) is greater than \(y\).

### Final Answer

The relation is \(x > y\).

---