In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option
I. ` x^(2) + 41 x + 420 = 0 `
II. ` 6 y^(2) - 11 y - 10 = 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 equations given in the question, we will follow these steps:

### Step 1: Solve the first equation
The first equation is:
\[ x^2 + 41x + 420 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \( 420 \) and add up to \( 41 \).

The numbers \( 21 \) and \( 20 \) satisfy these conditions because:
\[ 21 \times 20 = 420 \]
\[ 21 + 20 = 41 \]

Now we can rewrite the equation as:
\[ x^2 + 21x + 20x + 420 = 0 \]

Grouping the terms:
\[ (x^2 + 21x) + (20x + 420) = 0 \]

Factoring by grouping:
\[ x(x + 21) + 20(x + 21) = 0 \]

Now we can factor out \( (x + 21) \):
\[ (x + 21)(x + 20) = 0 \]

Setting each factor to zero gives us:
1. \( x + 21 = 0 \) → \( x = -21 \)
2. \( x + 20 = 0 \) → \( x = -20 \)

### Step 2: Solve the second equation
The second equation is:
\[ 6y^2 - 11y - 10 = 0 \]

We will factor this quadratic equation as well. We need to find two numbers that multiply to \( 6 \times -10 = -60 \) and add up to \( -11 \).

The numbers \( -15 \) and \( 4 \) satisfy these conditions because:
\[ -15 \times 4 = -60 \]
\[ -15 + 4 = -11 \]

Now we can rewrite the equation as:
\[ 6y^2 - 15y + 4y - 10 = 0 \]

Grouping the terms:
\[ (6y^2 - 15y) + (4y - 10) = 0 \]

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

Now we can factor out \( (2y - 5) \):
\[ (2y - 5)(3y + 2) = 0 \]

Setting each factor to zero gives us:
1. \( 2y - 5 = 0 \) → \( y = \frac{5}{2} \)
2. \( 3y + 2 = 0 \) → \( y = -\frac{2}{3} \)

### Step 3: Compare the values of \( x \) and \( y \)
We have the following values:
- For \( x \): \( -21 \) and \( -20 \)
- For \( y \): \( \frac{5}{2} \) and \( -\frac{2}{3} \)

Now we can compare these values:
- Compare \( -21 \) with \( \frac{5}{2} \) and \( -\frac{2}{3} \):
- \( -21 < \frac{5}{2} \) (True)
- \( -21 < -\frac{2}{3} \) (True)

- Compare \( -20 \) with \( \frac{5}{2} \) and \( -\frac{2}{3} \):
- \( -20 < \frac{5}{2} \) (True)
- \( -20 < -\frac{2}{3} \) (True)

### Conclusion
In both cases, \( x \) is less than \( y \). Therefore, the correct option is:
**C: \( x < y \)**