Two equations (I) and (II) are given in each questions . On the basis of these equations you have to decide the relation between 'x' and 'y' and give answer
I. ` 77 x^(2) + 58 x + 8 = 0 `
II. ` 42 y^(2) + 59 y + 20 = 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 and determine the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 77x^2 + 58x + 8 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \( 77 \times 8 = 616 \) and add to \( 58 \).

After breaking down \( 58x \) into \( 44x + 14x \), we rewrite the equation:
\[ 77x^2 + 44x + 14x + 8 = 0 \]

Now, we can group the terms:
\[ (77x^2 + 44x) + (14x + 8) = 0 \]

Factoring out the common terms:
\[ 11x(7x + 4) + 2(7x + 4) = 0 \]

Now, we can factor out \( (7x + 4) \):
\[ (7x + 4)(11x + 2) = 0 \]

Setting each factor to zero gives us:
1. \( 7x + 4 = 0 \) → \( x = -\frac{4}{7} \)
2. \( 11x + 2 = 0 \) → \( x = -\frac{2}{11} \)

### Step 2: Solve Equation II
The second equation is:
\[ 42y^2 + 59y + 20 = 0 \]

We will factor this quadratic equation similarly. We need to find two numbers that multiply to \( 42 \times 20 = 840 \) and add to \( 59 \).

Breaking down \( 59y \) into \( 35y + 24y \), we rewrite the equation:
\[ 42y^2 + 35y + 24y + 20 = 0 \]

Now, we group the terms:
\[ (42y^2 + 35y) + (24y + 20) = 0 \]

Factoring out the common terms:
\[ 7y(6y + 5) + 4(6y + 5) = 0 \]

Now, we can factor out \( (6y + 5) \):
\[ (6y + 5)(7y + 4) = 0 \]

Setting each factor to zero gives us:
1. \( 6y + 5 = 0 \) → \( y = -\frac{5}{6} \)
2. \( 7y + 4 = 0 \) → \( y = -\frac{4}{7} \)

### Step 3: Compare Values of \( x \) and \( y \)
Now we have the values for \( x \) and \( y \):
- \( x = -\frac{4}{7} \) or \( x = -\frac{2}{11} \)
- \( y = -\frac{5}{6} \) or \( y = -\frac{4}{7} \)

We can compare these values:
1. Comparing \( x = -\frac{4}{7} \) with \( y = -\frac{4}{7} \):
- Here, \( x = y \).

2. Comparing \( x = -\frac{2}{11} \) with \( y = -\frac{5}{6} \):
- To compare, we can convert both fractions to a common denominator:
- \( -\frac{2}{11} \) is approximately \( -0.1818 \)
- \( -\frac{5}{6} \) is approximately \( -0.8333 \)
- Thus, \( -\frac{2}{11} > -\frac{5}{6} \) or \( x > y \).

### Final Relation
Combining the results:
- From the first comparison, we have \( x = y \) when \( x = -\frac{4}{7} \).
- From the second comparison, we have \( x > y \) when \( x = -\frac{2}{11} \).

Thus, the final relation is:
\[ x \geq y \]

### Conclusion
The answer is:
**Option B: \( x \geq y \)**