In each of these questions , two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 5 x^(2) + 17 x + 6 = 0`
II ` 2 y^(2) + 11 y + 12 = 0`

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation \( 5x^2 + 17x + 6 = 0 \)

1. **Identify coefficients**: The coefficients are \( a = 5 \), \( b = 17 \), and \( c = 6 \).
2. **Multiply \( a \) and \( c \)**: Calculate \( ac = 5 \times 6 = 30 \).
3. **Find two numbers that multiply to \( ac \) and add to \( b \)**: We need two numbers that multiply to 30 and add to 17. The numbers are 15 and 2.
4. **Rewrite the equation**: Rewrite \( 5x^2 + 15x + 2x + 6 = 0 \).
5. **Factor by grouping**:
- Group the first two terms: \( 5x^2 + 15x \)
- Group the last two terms: \( 2x + 6 \)
- Factor out common terms:
- From \( 5x^2 + 15x \), factor out \( 5x \): \( 5x(x + 3) \)
- From \( 2x + 6 \), factor out \( 2 \): \( 2(x + 3) \)
6. **Combine the factors**: The equation becomes \( (5x + 2)(x + 3) = 0 \).
7. **Set each factor to zero**:
- \( 5x + 2 = 0 \) gives \( x = -\frac{2}{5} = -0.4 \)
- \( x + 3 = 0 \) gives \( x = -3 \)

### Step 2: Solve the second equation \( 2y^2 + 11y + 12 = 0 \)

1. **Identify coefficients**: The coefficients are \( a = 2 \), \( b = 11 \), and \( c = 12 \).
2. **Multiply \( a \) and \( c \)**: Calculate \( ac = 2 \times 12 = 24 \).
3. **Find two numbers that multiply to \( ac \) and add to \( b \)**: We need two numbers that multiply to 24 and add to 11. The numbers are 8 and 3.
4. **Rewrite the equation**: Rewrite \( 2y^2 + 8y + 3y + 12 = 0 \).
5. **Factor by grouping**:
- Group the first two terms: \( 2y^2 + 8y \)
- Group the last two terms: \( 3y + 12 \)
- Factor out common terms:
- From \( 2y^2 + 8y \), factor out \( 2y \): \( 2y(y + 4) \)
- From \( 3y + 12 \), factor out \( 3 \): \( 3(y + 4) \)
6. **Combine the factors**: The equation becomes \( (2y + 3)(y + 4) = 0 \).
7. **Set each factor to zero**:
- \( 2y + 3 = 0 \) gives \( y = -\frac{3}{2} = -1.5 \)
- \( y + 4 = 0 \) gives \( y = -4 \)

### Summary of Solutions
- From the first equation, we have \( x = -0.4 \) and \( x = -3 \).
- From the second equation, we have \( y = -1.5 \) and \( y = -4 \).

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

1. Compare \( x = -0.4 \) with \( y = -1.5 \):
- \( -0.4 > -1.5 \) (so \( x > y \))

2. Compare \( x = -0.4 \) with \( y = -4 \):
- \( -0.4 > -4 \) (so \( x > y \))

3. Compare \( x = -3 \) with \( y = -1.5 \):
- \( -3 < -1.5 \) (so \( x < y \))

4. Compare \( x = -3 \) with \( y = -4 \):
- \( -3 > -4 \) (so \( x > y \))

### Conclusion
From the comparisons, we see that \( x \) is greater than \( y \) in two cases and less than \( y \) in one case. Therefore, we conclude that there is no consistent relationship established between \( x \) and \( y \).