In the given questions, two equations (I) & (II) are given . You have to solve both the equations and mark the answer accordingly
I. ` x^(2) + 13 x + 40 = 0 `
II. ` y^(2) + 7y + 10 = 0`

Options

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

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:
\[ x^2 + 13x + 40 = 0 \]

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 factors of 40 that add up to 13 are 8 and 5. Therefore, we can rewrite the equation as:
\[ x^2 + 8x + 5x + 40 = 0 \]

Now, we can group the terms:
\[ (x^2 + 8x) + (5x + 40) = 0 \]

Factoring out the common terms:
\[ x(x + 8) + 5(x + 8) = 0 \]

Now, factor out \( (x + 8) \):
\[ (x + 8)(x + 5) = 0 \]

Setting each factor to zero gives us:
1. \( x + 8 = 0 \) → \( x = -8 \)
2. \( x + 5 = 0 \) → \( x = -5 \)

So, the solutions for Equation I are:
\[ x = -8 \quad \text{and} \quad x = -5 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 + 7y + 10 = 0 \]

Similarly, we will factor this quadratic equation. We need to find two numbers that multiply to 10 and add up to 7.

The factors of 10 that add up to 7 are 5 and 2. Therefore, we can rewrite the equation as:
\[ y^2 + 5y + 2y + 10 = 0 \]

Now, we can group the terms:
\[ (y^2 + 5y) + (2y + 10) = 0 \]

Factoring out the common terms:
\[ y(y + 5) + 2(y + 5) = 0 \]

Now, factor out \( (y + 5) \):
\[ (y + 5)(y + 2) = 0 \]

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

So, the solutions for Equation II are:
\[ y = -5 \quad \text{and} \quad y = -2 \]

### Step 3: Compare the Solutions
Now we have the solutions:
- For \( x \): \( -8 \) and \( -5 \)
- For \( y \): \( -5 \) and \( -2 \)

Now we can compare the values:
1. \( -8 < -5 \)
2. \( -5 = -5 \)
3. \( -5 > -2 \)

From the comparisons, we can conclude:
- \( x \) can be less than or equal to \( y \).

### Final Conclusion
Thus, the relationship between \( x \) and \( y \) is:
\[ x \leq y \]