` x^(2) - 1 = 0, y^(2) + 4y + 3 = 0 `

Options

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

Explanation

To solve the equations \( x^2 - 1 = 0 \) and \( y^2 + 4y + 3 = 0 \), we will follow these steps:

### Step 1: Solve the equation \( x^2 - 1 = 0 \)

1. Start with the equation:
\[ x^2 - 1 = 0 \]
2. Rearranging gives:
\[ x^2 = 1 \]
3. Taking the square root of both sides results in:
\[ x = \pm 1 \]
So, the solutions for \( x \) are:
\[ x = 1 \quad \text{and} \quad x = -1 \]

### Step 2: Solve the equation \( y^2 + 4y + 3 = 0 \)

1. Start with the equation:
\[ y^2 + 4y + 3 = 0 \]
2. We will factor the quadratic. We need two numbers that multiply to \( 3 \) (the constant term) and add to \( 4 \) (the coefficient of \( y \)). The numbers \( 3 \) and \( 1 \) satisfy this:
\[ (y + 3)(y + 1) = 0 \]
3. Setting each factor to zero gives:
\[ y + 3 = 0 \quad \Rightarrow \quad y = -3 \]
\[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]
So, the solutions for \( y \) are:
\[ y = -3 \quad \text{and} \quad y = -1 \]

### Step 3: Determine the relationship between \( x \) and \( y \)

1. We have the values:
- \( x = 1 \) or \( x = -1 \)
- \( y = -3 \) or \( y = -1 \)

2. We need to check the relationship \( x \geq y \):
- For \( x = 1 \) and \( y = -3 \):
\[ 1 \geq -3 \quad \text{(True)} \]
- For \( x = 1 \) and \( y = -1 \):
\[ 1 \geq -1 \quad \text{(True)} \]
- For \( x = -1 \) and \( y = -3 \):
\[ -1 \geq -3 \quad \text{(True)} \]
- For \( x = -1 \) and \( y = -1 \):
\[ -1 \geq -1 \quad \text{(True)} \]

Since all conditions hold true, we can conclude that \( x \geq y \).

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

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