In the following questions, two equations numbered I and II are given. You have to solve both the equations and
Give answer If
`2x^2+7x+5=0`
`5y^2+9y+4=0`

Options

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

Answer

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

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

1. **Identify the coefficients**:
- \(a = 2\), \(b = 7\), \(c = 5\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 5 = 49 - 40 = 9 \]

3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{9}}{2 \cdot 2} = \frac{-7 \pm 3}{4} \]

4. **Calculate the two possible values for \(x\)**:
- First root:
\[ x_1 = \frac{-7 + 3}{4} = \frac{-4}{4} = -1 \]
- Second root:
\[ x_2 = \frac{-7 - 3}{4} = \frac{-10}{4} = -2.5 \]

### Step 2: Solve the second equation \(5y^2 + 9y + 4 = 0\)

1. **Identify the coefficients**:
- \(a = 5\), \(b = 9\), \(c = 4\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 9^2 - 4 \cdot 5 \cdot 4 = 81 - 80 = 1 \]

3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-9 \pm \sqrt{1}}{2 \cdot 5} = \frac{-9 \pm 1}{10} \]

4. **Calculate the two possible values for \(y\)**:
- First root:
\[ y_1 = \frac{-9 + 1}{10} = \frac{-8}{10} = -0.8 \]
- Second root:
\[ y_2 = \frac{-9 - 1}{10} = \frac{-10}{10} = -1 \]

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

We have the values:
- For \(x\): \(-1\) and \(-2.5\)
- For \(y\): \(-0.8\) and \(-1\)

Now we will compare:
1. **Compare \(x_1 = -1\) with \(y_1 = -0.8\)**:
\(-1 < -0.8\) (so \(x_1 < y_1\))

2. **Compare \(x_1 = -1\) with \(y_2 = -1\)**:
\(-1 = -1\) (so \(x_1 = y_2\))

3. **Compare \(x_2 = -2.5\) with \(y_1 = -0.8\)**:
\(-2.5 < -0.8\) (so \(x_2 < y_1\))

4. **Compare \(x_2 = -2.5\) with \(y_2 = -1\)**:
\(-2.5 < -1\) (so \(x_2 < y_2\))

### Conclusion

From the comparisons, we can summarize:
- \(x_1 = -1\) is less than or equal to \(y_1 = -0.8\)
- \(x_2 = -2.5\) is less than both \(y_1\) and \(y_2\)

Thus, the relationship between \(x\) and \(y\) can be stated as:
\[ x \leq y \]

### Final Answer

The correct option is that \(x\) is less than or equal to \(y\).

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