In the following questions, two equations numbered I and II are given. You have to solve both the equations and
Give answer If
`9x^2-18x+8=0`
`3y^2-10y+8=0`

Options

A. `xlty`
B. `xgty`
C. `xley`
D. `xgey`

Answer

To solve the given equations step by step, we will follow these steps:

### Step 1: Solve the first equation \(9x^2 - 18x + 8 = 0\)

1. **Identify the coefficients**:
- \(a = 9\)
- \(b = -18\)
- \(c = 8\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-18)^2 - 4 \cdot 9 \cdot 8 = 324 - 288 = 36 \]

3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{18 \pm \sqrt{36}}{2 \cdot 9} = \frac{18 \pm 6}{18} \]
- First root:
\[ x_1 = \frac{18 + 6}{18} = \frac{24}{18} = \frac{4}{3} \approx 1.33 \]
- Second root:
\[ x_2 = \frac{18 - 6}{18} = \frac{12}{18} = \frac{2}{3} \approx 0.67 \]

### Step 2: Solve the second equation \(3y^2 - 10y + 8 = 0\)

1. **Identify the coefficients**:
- \(a = 3\)
- \(b = -10\)
- \(c = 8\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 8 = 100 - 96 = 4 \]

3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{10 \pm \sqrt{4}}{2 \cdot 3} = \frac{10 \pm 2}{6} \]
- First root:
\[ y_1 = \frac{10 + 2}{6} = \frac{12}{6} = 2 \]
- Second root:
\[ y_2 = \frac{10 - 2}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.33 \]

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

- The values obtained are:
- \(x_1 = \frac{4}{3} \approx 1.33\)
- \(x_2 = \frac{2}{3} \approx 0.67\)
- \(y_1 = 2\)
- \(y_2 = \frac{4}{3} \approx 1.33\)

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

1. **For \(x_1\) and \(y_1\)**:
- \(x_1 = 1.33\) and \(y_1 = 2\) → \(x_1 < y_1\)

2. **For \(x_1\) and \(y_2\)**:
- \(x_1 = 1.33\) and \(y_2 = 1.33\) → \(x_1 = y_2\)

3. **For \(x_2\) and \(y_1\)**:
- \(x_2 = 0.67\) and \(y_1 = 2\) → \(x_2 < y_1\)

4. **For \(x_2\) and \(y_2\)**:
- \(x_2 = 0.67\) and \(y_2 = 1.33\) → \(x_2 < y_2\)

### Conclusion

From the comparisons, we can conclude that:
- In all cases, \(x\) is either less than or equal to \(y\).

Thus, the final answer is:
**Option 3: \(x \leq y\)**.