In each question, two equations numbered I and II have been given. you have to solve both the equations and mark the appropiate option.
I `6x^2-19x+15=0`
II `5y^2-22y+24=0`

Options

A. `xley`
B. x = yor no relation can be established between x and-y
C. `xgty`
D. `xlty`

Answer

To solve the given equations step by step, let's start with each equation separately.

### Step 1: Solve Equation I: \(6x^2 - 19x + 15 = 0\)

1. **Identify the coefficients**:
- \(a = 6\), \(b = -19\), \(c = 15\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-19)^2 - 4 \cdot 6 \cdot 15 = 361 - 360 = 1 \]

4. **Substitute values into the formula**:
\[ x = \frac{-(-19) \pm \sqrt{1}}{2 \cdot 6} = \frac{19 \pm 1}{12} \]

5. **Calculate the two possible values for \(x\)**:
- \(x_1 = \frac{20}{12} = \frac{5}{3}\)
- \(x_2 = \frac{18}{12} = \frac{3}{2}\)

### Step 2: Solve Equation II: \(5y^2 - 22y + 24 = 0\)

1. **Identify the coefficients**:
- \(a = 5\), \(b = -22\), \(c = 24\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-22)^2 - 4 \cdot 5 \cdot 24 = 484 - 480 = 4 \]

4. **Substitute values into the formula**:
\[ y = \frac{-(-22) \pm \sqrt{4}}{2 \cdot 5} = \frac{22 \pm 2}{10} \]

5. **Calculate the two possible values for \(y\)**:
- \(y_1 = \frac{24}{10} = \frac{12}{5}\)
- \(y_2 = \frac{20}{10} = 2\)

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

From the solutions:
- \(x_1 = \frac{5}{3} \approx 1.67\)
- \(x_2 = \frac{3}{2} = 1.5\)
- \(y_1 = \frac{12}{5} = 2.4\)
- \(y_2 = 2\)

### Step 4: Determine the relationship between \(x\) and \(y\)

- For \(x_1\) and \(y_1\):
- \(x_1 \approx 1.67 < y_1 = 2.4\)

- For \(x_2\) and \(y_2\):
- \(x_2 = 1.5 < y_2 = 2\)

In both cases, \(x < y\).

### Conclusion

The relationship is \(x < y\).