In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I `x^2=121`
II `y^2-23y+132=0`

Options

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

Answer

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

### Step 1: Solve Equation I
The first equation is:
\[ x^2 = 121 \]

To find the value of \( x \), we take the square root of both sides:
\[ x = \pm \sqrt{121} \]

Calculating the square root:
\[ x = \pm 11 \]

Thus, the possible values for \( x \) are:
\[ x = 11 \quad \text{or} \quad x = -11 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 - 23y + 132 = 0 \]

To solve this quadratic equation, we can factor it. We need to find two numbers that multiply to \( 132 \) and add up to \( -23 \). The numbers \( -11 \) and \( -12 \) fit this requirement:
\[ (y - 11)(y - 12) = 0 \]

Setting each factor to zero gives us:
1. \( y - 11 = 0 \) → \( y = 11 \)
2. \( y - 12 = 0 \) → \( y = 12 \)

Thus, the possible values for \( y \) are:
\[ y = 11 \quad \text{or} \quad y = 12 \]

### Step 3: Compare Values of \( x \) and \( y \)
Now we have the values:
- For \( x \): \( 11 \) or \( -11 \)
- For \( y \): \( 11 \) or \( 12 \)

We need to compare \( x \) and \( y \):
1. If \( x = 11 \), then \( y \) can be \( 11 \) or \( 12 \). In both cases, \( y \geq x \).
2. If \( x = -11 \), then both values of \( y \) (11 and 12) are greater than \( x \).

### Conclusion
In all cases, we find that \( y \) is greater than or equal to \( x \). Therefore, the answer is:
\[ x \leq y \]

### Final Answer
The correct option is: \( x \leq y \).

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