In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I`3x^2-4x+1=0`
II `15y^2-8y+1=0`

Options

A. `xgty`
B. `xley`
C. x = y or relation can’t be established
D. `xgey`

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 - 4x + 1 = 0 \]

We can use the quadratic formula to solve this equation, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3 \), \( b = -4 \), and \( c = 1 \).

#### Step 1.1: Calculate the Discriminant
First, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot 1 = 16 - 12 = 4 \]

#### Step 1.2: Apply the Quadratic Formula
Now we can substitute the values into the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{4}}{2 \cdot 3} \]
\[ x = \frac{4 \pm 2}{6} \]

This gives us two solutions:
1. \( x_1 = \frac{4 + 2}{6} = \frac{6}{6} = 1 \)
2. \( x_2 = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3} \)

### Step 2: Solve Equation II
The second equation is:
\[ 15y^2 - 8y + 1 = 0 \]

Again, we will use the quadratic formula where \( a = 15 \), \( b = -8 \), and \( c = 1 \).

#### Step 2.1: Calculate the Discriminant
Calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-8)^2 - 4 \cdot 15 \cdot 1 = 64 - 60 = 4 \]

#### Step 2.2: Apply the Quadratic Formula
Now substitute the values into the quadratic formula:
\[ y = \frac{-(-8) \pm \sqrt{4}}{2 \cdot 15} \]
\[ y = \frac{8 \pm 2}{30} \]

This gives us two solutions:
1. \( y_1 = \frac{8 + 2}{30} = \frac{10}{30} = \frac{1}{3} \)
2. \( y_2 = \frac{8 - 2}{30} = \frac{6}{30} = \frac{1}{5} \)

### Step 3: Compare the Values of x and y
From the solutions we have:
- \( x_1 = 1 \), \( x_2 = \frac{1}{3} \)
- \( y_1 = \frac{1}{3} \), \( y_2 = \frac{1}{5} \)

We need to compare the values of \( x \) and \( y \):
- \( x_1 = 1 \) is greater than both \( y_1 = \frac{1}{3} \) and \( y_2 = \frac{1}{5} \).
- \( x_2 = \frac{1}{3} \) is equal to \( y_1 = \frac{1}{3} \) and greater than \( y_2 = \frac{1}{5} \).

### Conclusion
Thus, we can conclude that:
\[ x \geq y \]

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

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