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) - 13 x + 40 = 0`
II ` 2y ^(2) - y - 15 = 0 `

Options

A. If ` x gt y`
B. If ` x ge y `
C. If ` x lt y `
D. If ` x le y`

Answer

To solve the given equations step by step, we will start with equation (I) and then move on to equation (II).

### Step 1: Solve Equation (I)
The first equation is:
\[ x^2 - 13x + 40 = 0 \]

To solve this quadratic equation, we will factor it. We need to find two numbers that multiply to \(40\) (the constant term) and add up to \(-13\) (the coefficient of \(x\)).

The two numbers that satisfy these conditions are \(-8\) and \(-5\).

Thus, we can factor the equation as:
\[ (x - 8)(x - 5) = 0 \]

### Step 2: Find the Values of \(x\)
Now, we set each factor to zero:
1. \( x - 8 = 0 \) → \( x = 8 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

So, the solutions for \(x\) are:
\[ x = 8 \quad \text{and} \quad x = 5 \]

### Step 3: Solve Equation (II)
The second equation is:
\[ 2y^2 - y - 15 = 0 \]

To solve this quadratic equation, we will use the factorization method. We need to find two numbers that multiply to \(2 \times (-15) = -30\) and add up to \(-1\) (the coefficient of \(y\)).

The two numbers that satisfy these conditions are \(5\) and \(-6\).

We can rewrite the equation as:
\[ 2y^2 + 5y - 6y - 15 = 0 \]

Now, we can group the terms:
\[ (2y^2 + 5y) + (-6y - 15) = 0 \]

Factoring by grouping:
\[ y(2y + 5) - 3(2y + 5) = 0 \]

This gives us:
\[ (2y + 5)(y - 3) = 0 \]

### Step 4: Find the Values of \(y\)
Now, we set each factor to zero:
1. \( 2y + 5 = 0 \) → \( 2y = -5 \) → \( y = -\frac{5}{2} \)
2. \( y - 3 = 0 \) → \( y = 3 \)

So, the solutions for \(y\) are:
\[ y = -\frac{5}{2} \quad \text{and} \quad y = 3 \]

### Step 5: Compare the Values of \(x\) and \(y\)
Now we have the values:
- For \(x\): \(8\) and \(5\)
- For \(y\): \(-\frac{5}{2}\) and \(3\)

Now we compare:
1. \(8 > 3\)
2. \(5 > 3\)
3. \(8 > -\frac{5}{2}\)
4. \(5 > -\frac{5}{2}\)

From these comparisons, we can conclude that:
\[ x > y \]

### Final Answer
The relation between \(x\) and \(y\) is:
\[ x > y \]

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