In questions two equations numbered I and II are given you have to solve both equations and give answer
`I.12x^2+7x+1=0 II. 6y^2+5y+1=0`

Options

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

Answer

To solve the given equations, we need to find the values of \( x \) and \( y \) from the equations:

1. \( 12x^2 + 7x + 1 = 0 \) (Equation I)
2. \( 6y^2 + 5y + 1 = 0 \) (Equation II)

### Step 1: Solve Equation I

We start with the first equation:

\[ 12x^2 + 7x + 1 = 0 \]

To factor this quadratic equation, we look for two numbers that multiply to \( 12 \times 1 = 12 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( 7 \) (the coefficient of \( x \)). The numbers \( 3 \) and \( 4 \) satisfy these conditions.

Now we can rewrite the equation:

\[ 12x^2 + 4x + 3x + 1 = 0 \]

Next, we group the terms:

\[ (12x^2 + 4x) + (3x + 1) = 0 \]

Factoring out the common terms gives us:

\[ 4x(3x + 1) + 1(3x + 1) = 0 \]

Now, we can factor by grouping:

\[ (4x + 1)(3x + 1) = 0 \]

Setting each factor to zero gives us the possible values for \( x \):

1. \( 4x + 1 = 0 \) → \( x = -\frac{1}{4} \)
2. \( 3x + 1 = 0 \) → \( x = -\frac{1}{3} \)

### Step 2: Solve Equation II

Next, we solve the second equation:

\[ 6y^2 + 5y + 1 = 0 \]

We look for two numbers that multiply to \( 6 \times 1 = 6 \) and add up to \( 5 \). The numbers \( 3 \) and \( 2 \) work here.

Rewriting the equation:

\[ 6y^2 + 3y + 2y + 1 = 0 \]

Now, we group the terms:

\[ (6y^2 + 3y) + (2y + 1) = 0 \]

Factoring out the common terms gives us:

\[ 3y(2y + 1) + 1(2y + 1) = 0 \]

Factoring by grouping results in:

\[ (3y + 1)(2y + 1) = 0 \]

Setting each factor to zero gives us the possible values for \( y \):

1. \( 3y + 1 = 0 \) → \( y = -\frac{1}{3} \)
2. \( 2y + 1 = 0 \) → \( y = -\frac{1}{2} \)

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

Now we have the values:

- For \( x \): \( -\frac{1}{4} \) and \( -\frac{1}{3} \)
- For \( y \): \( -\frac{1}{3} \) and \( -\frac{1}{2} \)

We can convert these fractions to decimals for easier comparison:

- \( -\frac{1}{4} = -0.25 \)
- \( -\frac{1}{3} \approx -0.33 \)
- \( -\frac{1}{2} = -0.5 \)

Now we compare the values:

1. Comparing \( x = -0.25 \) with \( y = -0.33 \):
- \( -0.25 > -0.33 \) → \( x > y \)

2. Comparing \( x = -0.25 \) with \( y = -0.5 \):
- \( -0.25 > -0.5 \) → \( x > y \)

3. Comparing \( x = -0.33 \) with \( y = -0.33 \):
- \( -0.33 = -0.33 \) → \( x = y \)

4. Comparing \( x = -0.33 \) with \( y = -0.5 \):
- \( -0.33 > -0.5 \) → \( x > y \)

### Conclusion

From the comparisons, we see that \( x \) is either greater than or equal to \( y \). Therefore, the final answer is:

**Answer:** \( x \geq y \)