I. ` 2x^(2) + 5x+1 = x^(2) + 2x - 1`
II. ` 2y^(2) - 8y+1 =- 1`

Options

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

Explanation

Let's solve the given equations step by step.

### Step 1: Solve the first equation for x
The first equation is:
\[ 2x^2 + 5x + 1 = x^2 + 2x - 1 \]

**Rearranging the equation:**
Subtract \(x^2\), \(2x\), and \(-1\) from both sides:
\[ 2x^2 + 5x + 1 - x^2 - 2x + 1 = 0 \]
This simplifies to:
\[ (2x^2 - x^2) + (5x - 2x) + (1 + 1) = 0 \]
\[ x^2 + 3x + 2 = 0 \]

**Factoring the quadratic:**
Now, we can factor the quadratic:
\[ x^2 + 3x + 2 = (x + 2)(x + 1) = 0 \]

**Finding the roots:**
Setting each factor to zero gives:
1. \( x + 2 = 0 \) → \( x = -2 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

### Step 2: Solve the second equation for y
The second equation is:
\[ 2y^2 - 8y + 1 = -1 \]

**Rearranging the equation:**
Add 1 to both sides:
\[ 2y^2 - 8y + 1 + 1 = 0 \]
This simplifies to:
\[ 2y^2 - 8y + 2 = 0 \]

**Dividing by 2:**
To simplify, divide the entire equation by 2:
\[ y^2 - 4y + 1 = 0 \]

**Using the quadratic formula:**
The quadratic formula is:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -4\), and \(c = 1\).

Substituting the values:
\[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ y = \frac{4 \pm \sqrt{16 - 4}}{2} \]
\[ y = \frac{4 \pm \sqrt{12}}{2} \]
\[ y = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ y = 2 \pm \sqrt{3} \]

### Summary of Solutions
- The values of \(x\) are: \(x = -2\) and \(x = -1\)
- The values of \(y\) are: \(y = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\)