In each of questions two eqautions numbered Iand II are given you have to solve both equations and give answer
I.`2x^2+7x+5=0`
II.`3y^2+5y+2=0`

Options

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

Answer

To solve the given equations step by step, we will first solve each equation separately and then analyze the results.

### Step 1: Solve the first equation \(2x^2 + 7x + 5 = 0\)

1. **Identify coefficients**: Here, \(a = 2\), \(b = 7\), and \(c = 5\).
2. **Use the quadratic formula**: The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 5 = 49 - 40 = 9 \]
4. **Substitute values into the formula**:
\[ x = \frac{-7 \pm \sqrt{9}}{2 \cdot 2} = \frac{-7 \pm 3}{4} \]
5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{-7 + 3}{4} = \frac{-4}{4} = -1 \]
- Second value:
\[ x_2 = \frac{-7 - 3}{4} = \frac{-10}{4} = -\frac{5}{2} = -2.5 \]

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

1. **Identify coefficients**: Here, \(a = 3\), \(b = 5\), and \(c = 2\).
2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1 \]
4. **Substitute values into the formula**:
\[ y = \frac{-5 \pm \sqrt{1}}{2 \cdot 3} = \frac{-5 \pm 1}{6} \]
5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3} \approx -0.67 \]
- Second value:
\[ y_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1 \]

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

We have the following values:
- From the first equation: \(x_1 = -1\) and \(x_2 = -2.5\)
- From the second equation: \(y_1 = -\frac{2}{3} \approx -0.67\) and \(y_2 = -1\)

Now, we will compare each pair of values:

1. **For \(x_1 = -1\) and \(y_1 = -\frac{2}{3}\)**:
- \(-\frac{2}{3} \approx -0.67\) is greater than \(-1\). Thus, \(x_1 < y_1\).

2. **For \(x_1 = -1\) and \(y_2 = -1\)**:
- Here, \(x_1 = y_2\). Thus, \(x_1 = y_2\).

3. **For \(x_2 = -2.5\) and \(y_1 = -\frac{2}{3}\)**:
- \(-2.5 < -0.67\). Thus, \(x_2 < y_1\).

4. **For \(x_2 = -2.5\) and \(y_2 = -1\)**:
- \(-2.5 < -1\). Thus, \(x_2 < y_2\).

### Conclusion

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

### Final Answer

The answer is: \(x \leq y\).