` x^(2) + 30x + 221 = 0`
` y^(2) - 53y+196 = 0`

Options

A. `x gt y`
B. ` x lt y`
C. ` x ge y`
D. ` x le y`

Explanation

To solve the equations \( x^2 + 30x + 221 = 0 \) and \( y^2 - 53y + 196 = 0 \), we will follow these steps:

### Step 1: Solve for \( x \)
We start with the equation:
\[ x^2 + 30x + 221 = 0 \]
To factor this quadratic equation, we need to express \( 30x \) as the sum of two numbers that multiply to \( 221 \). We can rewrite \( 30x \) as:
\[ x^2 + 13x + 17x + 221 = 0 \]
Now, we group the terms:
\[ x(x + 13) + 17(x + 13) = 0 \]
Factoring out the common term \( (x + 13) \):
\[ (x + 13)(x + 17) = 0 \]
Setting each factor to zero gives us:
\[ x + 13 = 0 \quad \Rightarrow \quad x = -13 \]
\[ x + 17 = 0 \quad \Rightarrow \quad x = -17 \]
Thus, the solutions for \( x \) are:
\[ x = -13 \quad \text{and} \quad x = -17 \]

### Step 2: Solve for \( y \)
Next, we solve the equation:
\[ y^2 - 53y + 196 = 0 \]
To factor this quadratic equation, we need to express \( -53y \) as the sum of two numbers that multiply to \( 196 \). We can rewrite \( -53y \) as:
\[ y^2 - 49y - 4y + 196 = 0 \]
Now, we group the terms:
\[ y(y - 49) - 4(y - 49) = 0 \]
Factoring out the common term \( (y - 49) \):
\[ (y - 49)(y - 4) = 0 \]
Setting each factor to zero gives us:
\[ y - 49 = 0 \quad \Rightarrow \quad y = 49 \]
\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]
Thus, the solutions for \( y \) are:
\[ y = 49 \quad \text{and} \quad y = 4 \]

### Step 3: Determine the relation between \( x \) and \( y \)
Now we have the values:
- \( x = -13 \) and \( x = -17 \)
- \( y = 49 \) and \( y = 4 \)

We can compare the values:
- The largest \( x \) value is \( -13 \) and the smallest \( y \) value is \( 4 \).
- Since \( -13 < 4 \) and \( -17 < 4 \), we conclude that:
\[ x < y \]

### Final Conclusion
The relation between \( x \) and \( y \) is:
\[ x < y \]