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.`3x^2-10x+8=0`
II. `2y^2-19y+35=0`

Options

A. `x gt y`
B. `x ge y`
C. `x le y`
D. `x lt y`

Answer

To solve the given equations step by step, we will start with each quadratic equation separately.

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

1. **Identify the coefficients**: In the equation \(3x^2 - 10x + 8 = 0\), the coefficients are:
- \(a = 3\)
- \(b = -10\)
- \(c = 8\)

2. **Use the quadratic formula**: The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 100 - 96 = 4 \]

4. **Substitute back into the formula**:
\[ x = \frac{10 \pm \sqrt{4}}{6} = \frac{10 \pm 2}{6} \]

5. **Find the two possible values for \(x\)**:
- \(x_1 = \frac{10 + 2}{6} = \frac{12}{6} = 2\)
- \(x_2 = \frac{10 - 2}{6} = \frac{8}{6} = \frac{4}{3}\)

So, the solutions for the first equation are \(x = 2\) and \(x = \frac{4}{3}\).

### Step 2: Solve the second equation \(2y^2 - 19y + 35 = 0\)

1. **Identify the coefficients**: In the equation \(2y^2 - 19y + 35 = 0\), the coefficients are:
- \(a = 2\)
- \(b = -19\)
- \(c = 35\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ y = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \cdot 2 \cdot 35}}{2 \cdot 2} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 361 - 280 = 81 \]

4. **Substitute back into the formula**:
\[ y = \frac{19 \pm \sqrt{81}}{4} = \frac{19 \pm 9}{4} \]

5. **Find the two possible values for \(y\)**:
- \(y_1 = \frac{19 + 9}{4} = \frac{28}{4} = 7\)
- \(y_2 = \frac{19 - 9}{4} = \frac{10}{4} = \frac{5}{2}\)

So, the solutions for the second equation are \(y = 7\) and \(y = \frac{5}{2}\).

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

- The values we found are:
- For \(x\): \(2\) and \(\frac{4}{3}\)
- For \(y\): \(7\) and \(\frac{5}{2}\)

Now we compare:
- \(7 > 2\)
- \(7 > \frac{4}{3}\)
- \(\frac{5}{2} = 2.5\) which is also greater than both values of \(x\).

### Conclusion

Since both values of \(y\) are greater than both values of \(x\), we conclude that \(y > x\).

### Final Answer

The correct relation is \(x < y\).

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