In each of the following questions a pair of equations is given. You have to find out the value of x and y and give answer -----
(A)If x `lt` y (B) If x `le`y
( C ) If x = y ( D ) If x `gt` y
(E ) If x `ge` y
(1) `9x^2` - 18x + 5 = 0
(2) `2y^2` - 9y + 10 = 0

Answer

To solve the given equations and determine the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation for \( x \)

The first equation is:
\[ 9x^2 - 18x + 5 = 0 \]

We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 9 \), \( b = -18 \), and \( c = 5 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-18)^2 - 4 \cdot 9 \cdot 5 = 324 - 180 = 144 \]

Now substituting into the quadratic formula:
\[ x = \frac{18 \pm \sqrt{144}}{2 \cdot 9} \]
\[ x = \frac{18 \pm 12}{18} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{30}{18} = \frac{5}{3} \)
2. \( x = \frac{6}{18} = \frac{1}{3} \)

Thus, the values of \( x \) are:
\[ x_1 = \frac{5}{3}, \quad x_2 = \frac{1}{3} \]

### Step 2: Solve the second equation for \( y \)

The second equation is:
\[ 2y^2 - 9y + 10 = 0 \]

Using the quadratic formula again:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = -9 \), and \( c = 10 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-9)^2 - 4 \cdot 2 \cdot 10 = 81 - 80 = 1 \]

Now substituting into the quadratic formula:
\[ y = \frac{9 \pm \sqrt{1}}{2 \cdot 2} \]
\[ y = \frac{9 \pm 1}{4} \]

Calculating the two possible values for \( y \):
1. \( y = \frac{10}{4} = \frac{5}{2} \)
2. \( y = \frac{8}{4} = 2 \)

Thus, the values of \( y \) are:
\[ y_1 = \frac{5}{2}, \quad y_2 = 2 \]

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

Now we have:
- For \( x \): \( \frac{5}{3} \) and \( \frac{1}{3} \)
- For \( y \): \( \frac{5}{2} \) and \( 2 \)

Now we will compare each \( x \) value with each \( y \) value.

1. **Comparing \( x_1 = \frac{5}{3} \) with \( y_1 = \frac{5}{2} \)**:
\[ \frac{5}{3} \approx 1.67 \quad \text{and} \quad \frac{5}{2} = 2 \]
Thus, \( \frac{5}{3} < \frac{5}{2} \).

2. **Comparing \( x_1 = \frac{5}{3} \) with \( y_2 = 2 \)**:
\[ \frac{5}{3} < 2 \]

3. **Comparing \( x_2 = \frac{1}{3} \) with \( y_1 = \frac{5}{2} \)**:
\[ \frac{1}{3} < \frac{5}{2} \]

4. **Comparing \( x_2 = \frac{1}{3} \) with \( y_2 = 2 \)**:
\[ \frac{1}{3} < 2 \]

### Conclusion

In all comparisons, \( x \) is less than \( y \). Therefore, the answer is:
**(A) If \( x < y \)**.