In each of the these question two equation (I) and (II) are given. You have to solve both the equtions and give answer accordingly
I`12x^2+11x+2=0`
II `12y^2+7y+1=0`

Options

A. `xgty`
B. `xgey`
C. `xley`
D. no relationship established b/w x and y

Answer

To solve the given quadratic equations step by step, we will follow the standard method of factoring or using the quadratic formula.

### Step 1: Solve the first equation \(12x^2 + 11x + 2 = 0\)

1. **Identify coefficients**:
- \(a = 12\), \(b = 11\), \(c = 2\)

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 11^2 - 4 \cdot 12 \cdot 2 = 121 - 96 = 25 \]

4. **Calculate the roots**:
\[ x = \frac{-11 \pm \sqrt{25}}{2 \cdot 12} = \frac{-11 \pm 5}{24} \]
- For \(x_1\):
\[ x_1 = \frac{-11 + 5}{24} = \frac{-6}{24} = -\frac{1}{4} \]
- For \(x_2\):
\[ x_2 = \frac{-11 - 5}{24} = \frac{-16}{24} = -\frac{2}{3} \]

### Step 2: Solve the second equation \(12y^2 + 7y + 1 = 0\)

1. **Identify coefficients**:
- \(a = 12\), \(b = 7\), \(c = 1\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 12 \cdot 1 = 49 - 48 = 1 \]

4. **Calculate the roots**:
\[ y = \frac{-7 \pm \sqrt{1}}{2 \cdot 12} = \frac{-7 \pm 1}{24} \]
- For \(y_1\):
\[ y_1 = \frac{-7 + 1}{24} = \frac{-6}{24} = -\frac{1}{4} \]
- For \(y_2\):
\[ y_2 = \frac{-7 - 1}{24} = \frac{-8}{24} = -\frac{1}{3} \]

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

- The values obtained are:
- From the first equation: \(x_1 = -\frac{1}{4}\), \(x_2 = -\frac{2}{3}\)
- From the second equation: \(y_1 = -\frac{1}{4}\), \(y_2 = -\frac{1}{3}\)

### Step 4: Determine the relationship

- Comparing the values:
- \(y_1 = -\frac{1}{4}\) is equal to \(x_1 = -\frac{1}{4}\)
- \(y_2 = -\frac{1}{3}\) is greater than \(x_2 = -\frac{2}{3}\)

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

### Final Answer:
The correct option is \(x \leq y\).