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 < br> (E ) If x `ge` y
(1) `4x^2` = 49
(2) `9y^2` - 66y + 121 = 0

Answer

To solve the given problem, we need to find the values of \( x \) and \( y \) from the equations provided and then determine the relationship between \( x \) and \( y \).

### Step 1: Solve the first equation for \( x \)
The first equation is:
\[ 4x^2 = 49 \]
To isolate \( x^2 \), divide both sides by 4:
\[ x^2 = \frac{49}{4} \]
Now, take the square root of both sides:
\[ x = \pm \sqrt{\frac{49}{4}} = \pm \frac{7}{2} \]
Thus, the possible values for \( x \) are:
\[ x = \frac{7}{2} \quad \text{or} \quad x = -\frac{7}{2} \]

### Step 2: Solve the second equation for \( y \)
The second equation is:
\[ 9y^2 - 66y + 121 = 0 \]
To solve this quadratic equation, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 9 \), \( b = -66 \), and \( c = 121 \). First, calculate the discriminant:
\[ b^2 - 4ac = (-66)^2 - 4 \cdot 9 \cdot 121 = 4356 - 4356 = 0 \]
Since the discriminant is 0, there is one real solution:
\[ y = \frac{66}{2 \cdot 9} = \frac{66}{18} = \frac{11}{3} \]

### Step 3: Compare the values of \( x \) and \( y \)
Now we have the values:
- \( x = \frac{7}{2} \) or \( x = -\frac{7}{2} \)
- \( y = \frac{11}{3} \)

Next, we need to compare these values:
1. Calculate \( \frac{7}{2} \) in decimal form:
\[ \frac{7}{2} = 3.5 \]
2. Calculate \( \frac{11}{3} \) in decimal form:
\[ \frac{11}{3} \approx 3.67 \]

Now we can compare:
- For \( x = \frac{7}{2} \) (3.5) and \( y = \frac{11}{3} \) (approximately 3.67):
\[ x < y \]
- For \( x = -\frac{7}{2} \) (approximately -3.5) and \( y = \frac{11}{3} \):
\[ x < y \]

### Conclusion
In both cases, \( x < y \). Therefore, the answer is:
**(A) If \( x < y \)**.