In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer
I. ` 15 x^(2) + 10 x - 5 = 0 `
II. ` 6 y^(2) + 2y - 4 = 0`

Options

A. If ` x gt y `
B. If ` x ge y`
C. If ` x lt y `
D. if x = y or relationship between x and y cannot be established

Answer

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

### Step 1: Solve the first equation \( 15x^2 + 10x - 5 = 0 \)

1. **Identify coefficients**:
- \( a = 15 \)
- \( b = 10 \)
- \( c = -5 \)

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 10^2 - 4 \cdot 15 \cdot (-5) = 100 + 300 = 400 \]

4. **Calculate the roots**:
\[ x = \frac{-10 \pm \sqrt{400}}{2 \cdot 15} = \frac{-10 \pm 20}{30} \]
- First root:
\[ x_1 = \frac{-10 + 20}{30} = \frac{10}{30} = \frac{1}{3} \approx 0.33 \]
- Second root:
\[ x_2 = \frac{-10 - 20}{30} = \frac{-30}{30} = -1 \]

### Step 2: Solve the second equation \( 6y^2 + 2y - 4 = 0 \)

1. **Identify coefficients**:
- \( a = 6 \)
- \( b = 2 \)
- \( c = -4 \)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 2^2 - 4 \cdot 6 \cdot (-4) = 4 + 96 = 100 \]

4. **Calculate the roots**:
\[ y = \frac{-2 \pm \sqrt{100}}{2 \cdot 6} = \frac{-2 \pm 10}{12} \]
- First root:
\[ y_1 = \frac{-2 + 10}{12} = \frac{8}{12} = \frac{2}{3} \approx 0.67 \]
- Second root:
\[ y_2 = \frac{-2 - 10}{12} = \frac{-12}{12} = -1 \]

### Summary of Solutions
- The solutions for \( x \) are \( x_1 = \frac{1}{3} \) (approximately 0.33) and \( x_2 = -1 \).
- The solutions for \( y \) are \( y_1 = \frac{2}{3} \) (approximately 0.67) and \( y_2 = -1 \).

### Step 3: Compare the values of \( x \) and \( y \)
- Comparing \( x_1 \) and \( y_1 \):
- \( 0.33 < 0.67 \) (thus \( x_1 < y_1 \))

- Comparing \( x_1 \) and \( y_2 \):
- \( 0.33 > -1 \) (thus \( x_1 > y_2 \))

- Comparing \( x_2 \) and \( y_1 \):
- \( -1 < 0.67 \) (thus \( x_2 < y_1 \))

- Comparing \( x_2 \) and \( y_2 \):
- \( -1 = -1 \) (thus \( x_2 = y_2 \))

### Conclusion
- The relationships established are:
- \( x_1 < y_1 \)
- \( x_1 > y_2 \)
- \( x_2 = y_2 \)

### Final Answer
Since we have both equalities and inequalities, we conclude that there is no consistent relationship between \( x \) and \( y \) that can be established.

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