In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. Give answer.
I `6x^2-7x+2=0`
II `2y^2-5y+3=0`

Options

A. `xgty`
B. `xlty`
C. `xley`
D. `xgey`

Answer

To solve the given quadratic equations, we will follow these steps:

### Step 1: Solve the first equation \(6x^2 - 7x + 2 = 0\)

1. **Factor the equation**: We can rewrite the equation as:
\[ 6x^2 - 3x - 4x + 2 = 0 \]
Grouping the terms:
\[ (6x^2 - 3x) + (-4x + 2) = 0 \]
Factoring out common terms:
\[ 3x(2x - 1) - 2(2x - 1) = 0 \]
This gives us:
\[ (3x - 2)(2x - 1) = 0 \]

2. **Set each factor to zero**:
\[ 3x - 2 = 0 \quad \text{or} \quad 2x - 1 = 0 \]

3. **Solve for \(x\)**:
- From \(3x - 2 = 0\):
\[ 3x = 2 \implies x = \frac{2}{3} \]
- From \(2x - 1 = 0\):
\[ 2x = 1 \implies x = \frac{1}{2} \]

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

### Step 2: Solve the second equation \(2y^2 - 5y + 3 = 0\)

1. **Factor the equation**: We can rewrite the equation as:
\[ 2y^2 - 2y - 3y + 3 = 0 \]
Grouping the terms:
\[ (2y^2 - 2y) + (-3y + 3) = 0 \]
Factoring out common terms:
\[ 2y(y - 1) - 3(y - 1) = 0 \]
This gives us:
\[ (2y - 3)(y - 1) = 0 \]

2. **Set each factor to zero**:
\[ 2y - 3 = 0 \quad \text{or} \quad y - 1 = 0 \]

3. **Solve for \(y\)**:
- From \(2y - 3 = 0\):
\[ 2y = 3 \implies y = \frac{3}{2} \]
- From \(y - 1 = 0\):
\[ y = 1 \]

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

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

The solutions we found are:
- For \(x\): \(\frac{1}{2}\) and \(\frac{2}{3}\)
- For \(y\): \(1\) and \(\frac{3}{2}\)

Now, we compare the maximum values:
- The maximum \(x\) value is \(\frac{2}{3}\).
- The minimum \(y\) value is \(1\).

Since \(\frac{2}{3} < 1\), we conclude that:
\[ x < y \]

### Final Answer:
The correct relation is \(2x < y\).