In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option
I `5x^2-7x+2=0`
II `y^2-8y+15=0`

Options

A. `xgey`
B. `xlty`
C. `xgey`
D. no relationship can be established between x and y

Answer

To solve the given equations, we will go through each equation step by step.

### Step 1: Solve the first equation \(5x^2 - 7x + 2 = 0\)

1. **Identify the coefficients**: Here, \(a = 5\), \(b = -7\), and \(c = 2\).
2. **Use the quadratic formula**: The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 5 \cdot 2 = 49 - 40 = 9 \]
4. **Substitute values into the quadratic formula**:
\[ x = \frac{-(-7) \pm \sqrt{9}}{2 \cdot 5} = \frac{7 \pm 3}{10} \]
5. **Calculate the two possible values for \(x\)**:
- For \(x_1\):
\[ x_1 = \frac{7 + 3}{10} = \frac{10}{10} = 1 \]
- For \(x_2\):
\[ x_2 = \frac{7 - 3}{10} = \frac{4}{10} = \frac{2}{5} \]

### Step 2: Solve the second equation \(y^2 - 8y + 15 = 0\)

1. **Identify the coefficients**: Here, \(a = 1\), \(b = -8\), and \(c = 15\).
2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4 \]
4. **Substitute values into the quadratic formula**:
\[ y = \frac{-(-8) \pm \sqrt{4}}{2 \cdot 1} = \frac{8 \pm 2}{2} \]
5. **Calculate the two possible values for \(y\)**:
- For \(y_1\):
\[ y_1 = \frac{8 + 2}{2} = \frac{10}{2} = 5 \]
- For \(y_2\):
\[ y_2 = \frac{8 - 2}{2} = \frac{6}{2} = 3 \]

### Summary of Solutions
- From the first equation, we found \(x = 1\) and \(x = \frac{2}{5}\).
- From the second equation, we found \(y = 5\) and \(y = 3\).

### Step 3: Compare the values of \(x\) and \(y\)
- The values of \(x\) are \(1\) and \(\frac{2}{5}\).
- The values of \(y\) are \(5\) and \(3\).

### Conclusion
- Since both values of \(y\) (5 and 3) are greater than both values of \(x\) (1 and \(\frac{2}{5}\)), we can conclude that \(x < y\).

### Final Answer
The correct relation is \(x < y\).

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