In each of these questions , two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` x^(2) - 8 x + 15 = 0`
II ` y^(2) - 3y + 2 = 0`

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:
\[ x^2 - 8x + 15 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to 15 (the constant term) and add up to -8 (the coefficient of x).

The two numbers that satisfy this condition are -5 and -3, since:
- (-5) * (-3) = 15
- (-5) + (-3) = -8

Now we can rewrite the equation as:
\[ (x - 5)(x - 3) = 0 \]

Setting each factor to zero gives us:
1. \( x - 5 = 0 \) → \( x = 5 \)
2. \( x - 3 = 0 \) → \( x = 3 \)

Thus, the solutions for Equation I are:
\[ x = 5 \quad \text{and} \quad x = 3 \]

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

Similarly, we will factor this quadratic equation. We need to find two numbers that multiply to 2 (the constant term) and add up to -3 (the coefficient of y).

The two numbers that satisfy this condition are -2 and -1, since:
- (-2) * (-1) = 2
- (-2) + (-1) = -3

Now we can rewrite the equation as:
\[ (y - 2)(y - 1) = 0 \]

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

Thus, the solutions for Equation II are:
\[ y = 2 \quad \text{and} \quad y = 1 \]

### Step 3: Compare the Solutions
Now we have the solutions:
- From Equation I: \( x = 5 \) and \( x = 3 \)
- From Equation II: \( y = 2 \) and \( y = 1 \)

We can compare the values:
- \( 5 > 2 \)
- \( 5 > 1 \)
- \( 3 > 2 \)
- \( 3 > 1 \)

### Conclusion
In both cases, the values of x are greater than the values of y. Therefore, we can conclude:
\[ x > y \]