In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer
I. ` 7 x ^(2) - 17 x + 6 = 0 `
II. ` 5 y ^(2) - 24 y + 16 = 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 first tackle each equation separately.

### Step 1: Solve Equation I
**Equation I:** \( 7x^2 - 17x + 6 = 0 \)

1. **Identify coefficients:** Here, \( a = 7 \), \( b = -17 \), and \( c = 6 \).
2. **Calculate the discriminant:**
\[ D = b^2 - 4ac = (-17)^2 - 4 \cdot 7 \cdot 6 = 289 - 168 = 121 \]
3. **Find the roots using the quadratic formula:**
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{17 \pm \sqrt{121}}{2 \cdot 7} = \frac{17 \pm 11}{14} \]
4. **Calculate the two possible values for \( x \):**
- First root:
\[ x_1 = \frac{17 + 11}{14} = \frac{28}{14} = 2 \]
- Second root:
\[ x_2 = \frac{17 - 11}{14} = \frac{6}{14} = \frac{3}{7} \]

### Step 2: Solve Equation II
**Equation II:** \( 5y^2 - 24y + 16 = 0 \)

1. **Identify coefficients:** Here, \( a = 5 \), \( b = -24 \), and \( c = 16 \).
2. **Calculate the discriminant:**
\[ D = b^2 - 4ac = (-24)^2 - 4 \cdot 5 \cdot 16 = 576 - 320 = 256 \]
3. **Find the roots using the quadratic formula:**
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{24 \pm \sqrt{256}}{2 \cdot 5} = \frac{24 \pm 16}{10} \]
4. **Calculate the two possible values for \( y \):**
- First root:
\[ y_1 = \frac{24 + 16}{10} = \frac{40}{10} = 4 \]
- Second root:
\[ y_2 = \frac{24 - 16}{10} = \frac{8}{10} = \frac{4}{5} \]

### Step 3: Compare the Results
We have the values:
- From Equation I: \( x_1 = 2 \), \( x_2 = \frac{3}{7} \)
- From Equation II: \( y_1 = 4 \), \( y_2 = \frac{4}{5} \)

Now we compare:
1. **For \( x_1 = 2 \):**
- \( 2 < 4 \) (True)
- \( 2 > \frac{4}{5} \) (True)

2. **For \( x_2 = \frac{3}{7} \):**
- \( \frac{3}{7} < 4 \) (True)
- \( \frac{3}{7} < \frac{4}{5} \) (True)

### Conclusion
From the comparisons, we can conclude that:
- \( x_1 < y_1 \)
- \( x_1 > y_2 \)
- \( x_2 < y_1 \)
- \( x_2 < y_2 \)

Since there are cases where \( x \) is both less than and greater than \( y \), we cannot establish a consistent relationship between \( x \) and \( y \).

### Final Answer
No relation can be established between \( x \) and \( y \).

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