In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer
I. ` 13 x^(2) + 9 x - 4 = 0 `
II. ` 2 y ^(2) + y - 3 = 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 follow the quadratic formula approach for both equations.

### Step 1: Solve the first equation \( 13x^2 + 9x - 4 = 0 \)

1. Identify the coefficients:
- \( a = 13 \)
- \( b = 9 \)
- \( c = -4 \)

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

3. Calculate the discriminant:
\[ b^2 - 4ac = 9^2 - 4 \cdot 13 \cdot (-4) = 81 + 208 = 289 \]

4. Substitute the values into the quadratic formula:
\[ x = \frac{-9 \pm \sqrt{289}}{2 \cdot 13} = \frac{-9 \pm 17}{26} \]

5. Calculate the two possible values for \( x \):
- First value:
\[ x_1 = \frac{-9 + 17}{26} = \frac{8}{26} = \frac{4}{13} \approx 0.3077 \]
- Second value:
\[ x_2 = \frac{-9 - 17}{26} = \frac{-26}{26} = -1 \]

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

1. Identify the coefficients:
- \( a = 2 \)
- \( b = 1 \)
- \( c = -3 \)

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

3. Calculate the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \]

4. Substitute the values into the quadratic formula:
\[ y = \frac{-1 \pm \sqrt{25}}{2 \cdot 2} = \frac{-1 \pm 5}{4} \]

5. Calculate the two possible values for \( y \):
- First value:
\[ y_1 = \frac{-1 + 5}{4} = \frac{4}{4} = 1 \]
- Second value:
\[ y_2 = \frac{-1 - 5}{4} = \frac{-6}{4} = -\frac{3}{2} = -1.5 \]

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

- Values obtained:
- \( x_1 = \frac{4}{13} \approx 0.3077 \)
- \( x_2 = -1 \)
- \( y_1 = 1 \)
- \( y_2 = -1.5 \)

- Now we compare:
- \( x_1 \approx 0.3077 \) and \( y_1 = 1 \): \( x_1 < y_1 \)
- \( x_1 \approx 0.3077 \) and \( y_2 = -1.5 \): \( x_1 > y_2 \)
- \( x_2 = -1 \) and \( y_1 = 1 \): \( x_2 < y_1 \)
- \( x_2 = -1 \) and \( y_2 = -1.5 \): \( x_2 > y_2 \)

### Conclusion

From the comparisons:
- \( x_1 \) is less than \( y_1 \) but greater than \( y_2 \).
- \( x_2 \) is less than \( y_1 \) but greater than \( y_2 \).

Thus, there is no consistent relationship established between \( x \) and \( y \) across both values.

### Final Answer
The answer is that there is no established relationship between \( x \) and \( y \).

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