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

Options

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

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:
\[ 2x^2 - 3x + 1 = 0 \]

We will use the factorization method to solve this quadratic equation.

1. **Multiply the coefficient of \(x^2\) (which is 2) with the constant term (which is 1)**:
\[ 2 \times 1 = 2 \]

2. **We need to find two numbers that multiply to 2 and add up to -3**. The numbers that satisfy these conditions are -2 and -1.

3. **Rewrite the middle term (-3x) using -2 and -1**:
\[ 2x^2 - 2x - x + 1 = 0 \]

4. **Group the terms**:
\[ (2x^2 - 2x) + (-x + 1) = 0 \]

5. **Factor by grouping**:
\[ 2x(x - 1) - 1(x - 1) = 0 \]

6. **Factor out the common term \((x - 1)\)**:
\[ (2x - 1)(x - 1) = 0 \]

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

8. **Solve for \(x\)**:
\[ 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \]
\[ x - 1 = 0 \Rightarrow x = 1 \]

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

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

We will also use the factorization method here.

1. **We need to find two numbers that multiply to 15 and add up to -8**. The numbers that satisfy these conditions are -3 and -5.

2. **Rewrite the equation using these numbers**:
\[ y^2 - 3y - 5y + 15 = 0 \]

3. **Group the terms**:
\[ (y^2 - 3y) + (-5y + 15) = 0 \]

4. **Factor by grouping**:
\[ y(y - 3) - 5(y - 3) = 0 \]

5. **Factor out the common term \((y - 3)\)**:
\[ (y - 3)(y - 5) = 0 \]

6. **Set each factor to zero**:
\[ y - 3 = 0 \quad \text{or} \quad y - 5 = 0 \]

7. **Solve for \(y\)**:
\[ y - 3 = 0 \Rightarrow y = 3 \]
\[ y - 5 = 0 \Rightarrow y = 5 \]

So, the solutions for \(y\) are:
\[ y = 3, \quad y = 5 \]

### Step 3: Compare the Values of \(x\) and \(y\)
Now we have:
- \(x\) can take values \( \frac{1}{2} \) and \( 1 \)
- \(y\) can take values \( 3 \) and \( 5 \)

We need to compare the values of \(x\) and \(y\):
- Both values of \(x\) (\(\frac{1}{2}\) and \(1\)) are less than both values of \(y\) (3 and 5).

Thus, we conclude that:
\[ x < y \]

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