In each of these questions, two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 2 x ^(2) + 3 x + 1 = 0 `
II. ` 12 y ^(2) + 7y + 1 = 0 `

Options

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

Answer

To solve the given equations step by step, let's start with each equation one by one.

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 + 3x + 1 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(2 \times 1 = 2\) and add to \(3\).

We can rewrite the middle term:
\[ 2x^2 + 2x + x + 1 = 0 \]

Now, we can group the terms:
\[ (2x^2 + 2x) + (x + 1) = 0 \]

Factoring out the common terms:
\[ 2x(x + 1) + 1(x + 1) = 0 \]

Now, we can factor by grouping:
\[ (2x + 1)(x + 1) = 0 \]

Setting each factor to zero gives us:
1. \(2x + 1 = 0\) → \(2x = -1\) → \(x = -\frac{1}{2}\)
2. \(x + 1 = 0\) → \(x = -1\)

Thus, the solutions for \(x\) are:
\[ x_1 = -\frac{1}{2}, \quad x_2 = -1 \]

### Step 2: Solve Equation II
The second equation is:
\[ 12y^2 + 7y + 1 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \(12 \times 1 = 12\) and add to \(7\).

We can rewrite the middle term:
\[ 12y^2 + 4y + 3y + 1 = 0 \]

Now, we can group the terms:
\[ (12y^2 + 4y) + (3y + 1) = 0 \]

Factoring out the common terms:
\[ 4y(3y + 1) + 1(3y + 1) = 0 \]

Now, we can factor by grouping:
\[ (4y + 1)(3y + 1) = 0 \]

Setting each factor to zero gives us:
1. \(4y + 1 = 0\) → \(4y = -1\) → \(y = -\frac{1}{4}\)
2. \(3y + 1 = 0\) → \(3y = -1\) → \(y = -\frac{1}{3}\)

Thus, the solutions for \(y\) are:
\[ y_1 = -\frac{1}{4}, \quad y_2 = -\frac{1}{3} \]

### Step 3: Compare the Values of x and y
Now we have the solutions:
- For \(x\): \(x_1 = -\frac{1}{2}, \quad x_2 = -1\)
- For \(y\): \(y_1 = -\frac{1}{4}, \quad y_2 = -\frac{1}{3}\)

Now we will compare the values of \(x\) and \(y\):
1. Comparing \(x_1 = -\frac{1}{2}\) with \(y_1 = -\frac{1}{4}\):
\(-\frac{1}{2} < -\frac{1}{4}\) (So, \(x_1 < y_1\))

2. Comparing \(x_1 = -\frac{1}{2}\) with \(y_2 = -\frac{1}{3}\):
\(-\frac{1}{2} < -\frac{1}{3}\) (So, \(x_1 < y_2\))

3. Comparing \(x_2 = -1\) with \(y_1 = -\frac{1}{4}\):
\(-1 < -\frac{1}{4}\) (So, \(x_2 < y_1\))

4. Comparing \(x_2 = -1\) with \(y_2 = -\frac{1}{3}\):
\(-1 < -\frac{1}{3}\) (So, \(x_2 < y_2\))

### Conclusion
In all comparisons, we find that \(x\) is less than \(y\). Therefore, the relation between \(x\) and \(y\) is:
\[ x < y \]