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. `99 x^(2) + 149 x + 56 = 0`
II. ` 156 y^(2) + 287 y + 132 = 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 problem, we need to analyze the two quadratic equations given and find the relationship between the variables \( x \) and \( y \).

### Step 1: Solve the first equation for \( x \)

The first equation is:
\[ 99x^2 + 149x + 56 = 0 \]

To factor this equation, we can look for two numbers that multiply to \( 99 \times 56 = 5544 \) and add up to \( 149 \). After factoring, we can rewrite the equation as:
\[ 99x^2 + 77x + 72x + 56 = 0 \]

Now, we can group the terms:
\[ (99x^2 + 77x) + (72x + 56) = 0 \]

Factoring out the common terms:
\[ 11x(9x + 7) + 8(9x + 7) = 0 \]

This gives us:
\[ (9x + 7)(11x + 8) = 0 \]

Setting each factor to zero, we find:
1. \( 9x + 7 = 0 \) → \( x = -\frac{7}{9} \)
2. \( 11x + 8 = 0 \) → \( x = -\frac{8}{11} \)

### Step 2: Solve the second equation for \( y \)

The second equation is:
\[ 156y^2 + 287y + 132 = 0 \]

Similarly, we look for two numbers that multiply to \( 156 \times 132 = 20592 \) and add up to \( 287 \). We can rewrite the equation as:
\[ 156y^2 + 144y + 143y + 132 = 0 \]

Grouping the terms:
\[ (156y^2 + 144y) + (143y + 132) = 0 \]

Factoring out the common terms:
\[ 12y(13y + 12) + 11(13y + 12) = 0 \]

This gives us:
\[ (13y + 12)(12y + 11) = 0 \]

Setting each factor to zero, we find:
1. \( 13y + 12 = 0 \) → \( y = -\frac{12}{13} \)
2. \( 12y + 11 = 0 \) → \( y = -\frac{11}{12} \)

### Step 3: Compare the values of \( x \) and \( y \)

Now we have the values:
- For \( x \): \( -\frac{7}{9} \) and \( -\frac{8}{11} \)
- For \( y \): \( -\frac{12}{13} \) and \( -\frac{11}{12} \)

We need to compare these values to determine the relationship between \( x \) and \( y \).

1. Comparing \( -\frac{7}{9} \) with \( -\frac{12}{13} \):
- Convert to a common denominator (117):
- \( -\frac{7}{9} = -\frac{91}{117} \)
- \( -\frac{12}{13} = -\frac{108}{117} \)
- Since \( -\frac{91}{117} > -\frac{108}{117} \), we have \( x > y \).

2. Comparing \( -\frac{7}{9} \) with \( -\frac{11}{12} \):
- Convert to a common denominator (108):
- \( -\frac{7}{9} = -\frac{84}{108} \)
- \( -\frac{11}{12} = -\frac{99}{108} \)
- Since \( -\frac{84}{108} > -\frac{99}{108} \), we have \( x > y \).

3. Now comparing \( -\frac{8}{11} \) with \( -\frac{12}{13} \):
- Convert to a common denominator (143):
- \( -\frac{8}{11} = -\frac{104}{143} \)
- \( -\frac{12}{13} = -\frac{132}{143} \)
- Since \( -\frac{104}{143} > -\frac{132}{143} \), we have \( x > y \).

4. Finally, comparing \( -\frac{8}{11} \) with \( -\frac{11}{12} \):
- Convert to a common denominator (132):
- \( -\frac{8}{11} = -\frac{96}{132} \)
- \( -\frac{11}{12} = -\frac{121}{132} \)
- Since \( -\frac{96}{132} > -\frac{121}{132} \), we have \( x > y \).

### Conclusion

From all comparisons, we conclude that \( x > y \).

### Final Answer

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