In these questions two equations numbered I and II are given. You have to solve both the 'equations and mark the appropriate option. Give answer
`x^2-8x+15=0`
`y^2-3y+2=0`

Options

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

Answer

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

### Step 1: Solve the first equation \( x^2 - 8x + 15 = 0 \)

1. **Identify the coefficients**: The equation is in the standard quadratic form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -8 \), and \( c = 15 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( c = 15 \) and add up to \( b = -8 \). The numbers are \( -5 \) and \( -3 \).
3. **Write the factored form**: The equation can be factored as:
\[ (x - 5)(x - 3) = 0 \]
4. **Set each factor to zero**:
- \( x - 5 = 0 \) gives \( x = 5 \)
- \( x - 3 = 0 \) gives \( x = 3 \)

So, the solutions for \( x \) are \( x = 3 \) and \( x = 5 \).

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

1. **Identify the coefficients**: The equation is also in the standard quadratic form \( ay^2 + by + c = 0 \), where \( a = 1 \), \( b = -3 \), and \( c = 2 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( c = 2 \) and add up to \( b = -3 \). The numbers are \( -2 \) and \( -1 \).
3. **Write the factored form**: The equation can be factored as:
\[ (y - 2)(y - 1) = 0 \]
4. **Set each factor to zero**:
- \( y - 2 = 0 \) gives \( y = 2 \)
- \( y - 1 = 0 \) gives \( y = 1 \)

So, the solutions for \( y \) are \( y = 1 \) and \( y = 2 \).

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

Now we have the values:
- For \( x \): \( 3 \) and \( 5 \)
- For \( y \): \( 1 \) and \( 2 \)

We will compare these values:
1. If \( x = 3 \) and \( y = 1 \), then \( x > y \).
2. If \( x = 3 \) and \( y = 2 \), then \( x > y \).
3. If \( x = 5 \) and \( y = 1 \), then \( x > y \).
4. If \( x = 5 \) and \( y = 2 \), then \( x > y \).

In all cases, \( x \) is greater than \( y \).

### Conclusion

The final answer is that \( x \) is greater than \( y \) in every case.