In these questions two equations numbered I and II are given. You have to solve both the 'equations and mark the appropriate option. Give answer
`6x^2+5x+1=0`
`6y^2+7y+2=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 \(6x^2 + 5x + 1 = 0\)

To solve this quadratic equation, we will use the factorization method.

1. Identify \(a = 6\), \(b = 5\), and \(c = 1\).
2. We need to find two numbers that multiply to \(a \cdot c = 6 \cdot 1 = 6\) and add up to \(b = 5\). The numbers are \(3\) and \(2\).
3. Rewrite the equation as:
\[ 6x^2 + 3x + 2x + 1 = 0 \]
4. Group the terms:
\[ (6x^2 + 3x) + (2x + 1) = 0 \]
5. Factor by grouping:
\[ 3x(2x + 1) + 1(2x + 1) = 0 \]
6. Factor out the common term:
\[ (3x + 1)(2x + 1) = 0 \]
7. Set each factor to zero:
- \(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\)
- \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)

### Step 2: Solve the second equation \(6y^2 + 7y + 2 = 0\)

Now, we will solve the second quadratic equation using the same method.

1. Identify \(a = 6\), \(b = 7\), and \(c = 2\).
2. We need to find two numbers that multiply to \(a \cdot c = 6 \cdot 2 = 12\) and add up to \(b = 7\). The numbers are \(4\) and \(3\).
3. Rewrite the equation as:
\[ 6y^2 + 4y + 3y + 2 = 0 \]
4. Group the terms:
\[ (6y^2 + 4y) + (3y + 2) = 0 \]
5. Factor by grouping:
\[ 2y(3y + 2) + 1(3y + 2) = 0 \]
6. Factor out the common term:
\[ (2y + 1)(3y + 2) = 0 \]
7. Set each factor to zero:
- \(2y + 1 = 0 \Rightarrow y = -\frac{1}{2}\)
- \(3y + 2 = 0 \Rightarrow y = -\frac{2}{3}\)

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

Now we have the values:
- For \(x\): \(-\frac{1}{3} \approx -0.33\) and \(-\frac{1}{2} = -0.5\)
- For \(y\): \(-\frac{1}{2} = -0.5\) and \(-\frac{2}{3} \approx -0.66\)

Now we will compare these values:

1. Compare \(x = -\frac{1}{3}\) with \(y = -\frac{1}{2}\):
- \(-0.33 > -0.5\) (So, \(x > y\))

2. Compare \(x = -\frac{1}{3}\) with \(y = -\frac{2}{3}\):
- \(-0.33 > -0.66\) (So, \(x > y\))

3. Compare \(x = -\frac{1}{2}\) with \(y = -\frac{1}{2}\):
- \(-0.5 = -0.5\) (So, \(x = y\))

4. Compare \(x = -\frac{1}{2}\) with \(y = -\frac{2}{3}\):
- \(-0.5 > -0.66\) (So, \(x > y\))

### Conclusion

From the comparisons, we can conclude that \(x\) is greater than or equal to \(y\).

### Final Answer
The correct option is that \(x\) is greater than or equal to \(y\).

---