In each question, two equations numbered I and II have been given. you have to solve both the equations and mark the appropiate option.
I `4x^2-14x+6=0`
II `4y^2-8y+3=0`

Options

A. `xgey`
B. `xlty`
C. `xley`
D. x = yor no relationship can be established between x and-y

Answer

To solve the given equations step by step, we will first tackle each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ 4x^2 - 14x + 6 = 0 \]

To solve this quadratic equation, we can factor it. We will look for two numbers that multiply to \(4 \times 6 = 24\) and add up to \(-14\).

We can rewrite \(-14x\) as \(-12x - 2x\):
\[ 4x^2 - 12x - 2x + 6 = 0 \]

Now, we can group the terms:
\[ (4x^2 - 12x) + (-2x + 6) = 0 \]

Factoring out the common factors:
\[ 4x(x - 3) - 2(x - 3) = 0 \]

Now, we can factor out \((x - 3)\):
\[ (x - 3)(4x - 2) = 0 \]

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

So, the solutions for \(x\) are:
\[ x = 3 \quad \text{and} \quad x = \frac{1}{2} \]

### Step 2: Solve Equation II
The second equation is:
\[ 4y^2 - 8y + 3 = 0 \]

Again, we will factor this quadratic equation. We need two numbers that multiply to \(4 \times 3 = 12\) and add up to \(-8\).

We can rewrite \(-8y\) as \(-6y - 2y\):
\[ 4y^2 - 6y - 2y + 3 = 0 \]

Now, we can group the terms:
\[ (4y^2 - 6y) + (-2y + 3) = 0 \]

Factoring out the common factors:
\[ 2y(2y - 3) - 1(2y - 3) = 0 \]

Now, we can factor out \((2y - 3)\):
\[ (2y - 3)(2y - 1) = 0 \]

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

So, the solutions for \(y\) are:
\[ y = \frac{3}{2} \quad \text{and} \quad y = \frac{1}{2} \]

### Step 3: Compare the Values of x and y
Now we have the values:
- For \(x\): \(3\) and \(\frac{1}{2}\)
- For \(y\): \(\frac{3}{2}\) and \(\frac{1}{2}\)

We need to compare these values:
- \(3 > \frac{3}{2}\)
- \(3 > \frac{1}{2}\)
- \(\frac{1}{2} < \frac{3}{2}\)

From this comparison, we can conclude:
\[ x \text{ can be greater than or equal to } y \]

### Final Answer
The correct relation is:
\[ x \geq y \]