In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct options
I. ` 6 x^(2) + 5 x + 1 = 0 `
II. ` 2 y^(2) + 5y + 3 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If ` x le y`

Answer

To solve the given equations and determine the relationship between the values of \(x\) and \(y\), we will follow these steps:

### Step 1: Solve the first equation \(6x^2 + 5x + 1 = 0\)

We will factor the quadratic equation. We need to find two numbers that multiply to \(6 \times 1 = 6\) and add up to \(5\).

1. The numbers \(3\) and \(2\) satisfy this condition.
2. Rewrite \(5x\) as \(3x + 2x\):
\[ 6x^2 + 3x + 2x + 1 = 0 \]
3. Group the terms:
\[ (6x^2 + 3x) + (2x + 1) = 0 \]
4. Factor by grouping:
\[ 3x(2x + 1) + 1(2x + 1) = 0 \]
5. Factor out the common term \((2x + 1)\):
\[ (2x + 1)(3x + 1) = 0 \]

### Step 2: Find the values of \(x\)

Set each factor equal to zero:
1. \(2x + 1 = 0\) gives:
\[ x = -\frac{1}{2} \]
2. \(3x + 1 = 0\) gives:
\[ x = -\frac{1}{3} \]

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

Similarly, we will factor this quadratic equation. We need two numbers that multiply to \(2 \times 3 = 6\) and add up to \(5\).

1. The numbers \(2\) and \(3\) satisfy this condition.
2. Rewrite \(5y\) as \(2y + 3y\):
\[ 2y^2 + 2y + 3y + 3 = 0 \]
3. Group the terms:
\[ (2y^2 + 2y) + (3y + 3) = 0 \]
4. Factor by grouping:
\[ 2y(y + 1) + 3(y + 1) = 0 \]
5. Factor out the common term \((y + 1)\):
\[ (y + 1)(2y + 3) = 0 \]

### Step 4: Find the values of \(y\)

Set each factor equal to zero:
1. \(y + 1 = 0\) gives:
\[ y = -1 \]
2. \(2y + 3 = 0\) gives:
\[ y = -\frac{3}{2} \]

### Step 5: Compare the values of \(x\) and \(y\)

Now we have:
- Values of \(x\): \(-\frac{1}{2}\) and \(-\frac{1}{3}\)
- Values of \(y\): \(-1\) and \(-\frac{3}{2}\)

We will compare these values:
1. Compare \(x = -\frac{1}{2}\) with \(y = -1\):
\[ -\frac{1}{2} > -1 \quad \text{(True)} \]
2. Compare \(x = -\frac{1}{2}\) with \(y = -\frac{3}{2}\):
\[ -\frac{1}{2} > -\frac{3}{2} \quad \text{(True)} \]
3. Compare \(x = -\frac{1}{3}\) with \(y = -1\):
\[ -\frac{1}{3} > -1 \quad \text{(True)} \]
4. Compare \(x = -\frac{1}{3}\) with \(y = -\frac{3}{2}\):
\[ -\frac{1}{3} > -\frac{3}{2} \quad \text{(True)} \]

### Conclusion

In all comparisons, \(x\) is greater than \(y\). Therefore, the correct relationship is:
\[ x > y \]