In each of these questions, two equations I. and II are given. You have to solve both the equations and give answer
I. ` 8 x^(2) - 10 x + 3 = 0 `
II. ` 5 y ^(2) + 14 y - 3 = 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 step by step, we will first tackle each equation separately.

### Step 1: Solve the first equation \(8x^2 - 10x + 3 = 0\)

1. **Identify the coefficients**:
- \(a = 8\), \(b = -10\), \(c = 3\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-10)^2 - 4 \cdot 8 \cdot 3 = 100 - 96 = 4 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-(-10) \pm \sqrt{4}}{2 \cdot 8} = \frac{10 \pm 2}{16} \]

5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{10 + 2}{16} = \frac{12}{16} = \frac{3}{4} \]
- Second value:
\[ x_2 = \frac{10 - 2}{16} = \frac{8}{16} = \frac{1}{2} \]

### Step 2: Solve the second equation \(5y^2 + 14y - 3 = 0\)

1. **Identify the coefficients**:
- \(a = 5\), \(b = 14\), \(c = -3\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (14)^2 - 4 \cdot 5 \cdot (-3) = 196 + 60 = 256 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-14 \pm \sqrt{256}}{2 \cdot 5} = \frac{-14 \pm 16}{10} \]

5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{-14 + 16}{10} = \frac{2}{10} = \frac{1}{5} \]
- Second value:
\[ y_2 = \frac{-14 - 16}{10} = \frac{-30}{10} = -3 \]

### Summary of Solutions
- The values of \(x\) are \(\frac{3}{4}\) and \(\frac{1}{2}\).
- The values of \(y\) are \(\frac{1}{5}\) and \(-3\).

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

1. **Compare \(x_1 = \frac{1}{2}\) with both \(y\) values**:
- \(\frac{1}{2} > -3\) (True)
- \(\frac{1}{2} > \frac{1}{5}\) (True)

2. **Compare \(x_2 = \frac{3}{4}\) with both \(y\) values**:
- \(\frac{3}{4} > -3\) (True)
- \(\frac{3}{4} > \frac{1}{5}\) (True)

### Conclusion
From the comparisons, we can conclude that for both values of \(x\), \(x\) is greater than \(y\). Therefore, the final relation is:
\[ x > y \]