In the following questions, there are two equations in x and y . You have to solve both the equations and give answer
I. ` 6 x^(2) + 5 x + 1 = 0`
II. ` 15 y ^(2) + 11 y + 2 = 0 `

Options

A. If ` x gt y `
B. If ` x lt y `
C. If ` x ge y `
D. If x = y or there is no relation between x and y

Answer

To solve the equations given in the question, we will tackle each equation step by step.

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

We will use the quadratic formula to find the values of \( x \). The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \( a = 6 \), \( b = 5 \), and \( c = 1 \).

#### Step 1.1: Calculate the discriminant

First, we calculate the discriminant \( D = b^2 - 4ac \):

\[ D = 5^2 - 4 \cdot 6 \cdot 1 = 25 - 24 = 1 \]

#### Step 1.2: Apply the quadratic formula

Now, we can substitute the values into the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{1}}{2 \cdot 6} = \frac{-5 \pm 1}{12} \]

This gives us two possible solutions for \( x \):

1. \( x_1 = \frac{-5 + 1}{12} = \frac{-4}{12} = -\frac{1}{3} \)
2. \( x_2 = \frac{-5 - 1}{12} = \frac{-6}{12} = -\frac{1}{2} \)

Thus, the solutions for \( x \) are:

\[ x = -\frac{1}{3} \quad \text{and} \quad x = -\frac{1}{2} \]

### Step 2: Solve the second equation \( 15y^2 + 11y + 2 = 0 \)

Again, we will use the quadratic formula where \( a = 15 \), \( b = 11 \), and \( c = 2 \).

#### Step 2.1: Calculate the discriminant

Calculate the discriminant \( D = b^2 - 4ac \):

\[ D = 11^2 - 4 \cdot 15 \cdot 2 = 121 - 120 = 1 \]

#### Step 2.2: Apply the quadratic formula

Now we substitute the values into the quadratic formula:

\[ y = \frac{-11 \pm \sqrt{1}}{2 \cdot 15} = \frac{-11 \pm 1}{30} \]

This gives us two possible solutions for \( y \):

1. \( y_1 = \frac{-11 + 1}{30} = \frac{-10}{30} = -\frac{1}{3} \)
2. \( y_2 = \frac{-11 - 1}{30} = \frac{-12}{30} = -\frac{2}{5} \)

Thus, the solutions for \( y \) are:

\[ y = -\frac{1}{3} \quad \text{and} \quad y = -\frac{2}{5} \]

### Final Solutions

The final solutions for the equations are:

- For \( x \): \( x = -\frac{1}{3} \) and \( x = -\frac{1}{2} \)
- For \( y \): \( y = -\frac{1}{3} \) and \( y = -\frac{2}{5} \)