I. ` x^(2) - 6x = 7" "`
II. ` 2y^(2) + 13y + 15 = 0`

Options

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

Explanation

To solve the given equations step by step, we will first handle the equation for \( x \) and then the equation for \( y \).

### Step 1: Solve for \( x \)
The equation given is:
\[ x^2 - 6x = 7 \]
We will rearrange this equation to set it to zero:
\[ x^2 - 6x - 7 = 0 \]

### Step 2: Factor the quadratic equation
Next, we will factor the quadratic equation \( x^2 - 6x - 7 \). We need two numbers that multiply to \(-7\) (the constant term) and add to \(-6\) (the coefficient of \( x \)). The numbers \(-7\) and \(1\) satisfy this:
\[ (x - 7)(x + 1) = 0 \]

### Step 3: Set each factor to zero
Now we will set each factor equal to zero:
1. \( x - 7 = 0 \) → \( x = 7 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

Thus, the solutions for \( x \) are:
\[ x = 7 \quad \text{and} \quad x = -1 \]

### Step 4: Solve for \( y \)
Now we will solve the equation for \( y \):
\[ 2y^2 + 13y + 15 = 0 \]

### Step 5: Factor the quadratic equation for \( y \)
We need to find two numbers that multiply to \( 2 \times 15 = 30 \) and add to \( 13 \). The numbers \( 10 \) and \( 3 \) work:
\[ 2y^2 + 10y + 3y + 15 = 0 \]
Grouping the terms gives:
\[ (2y^2 + 10y) + (3y + 15) = 0 \]
Factoring out the common terms:
\[ 2y(y + 5) + 3(y + 5) = 0 \]
Factoring further:
\[ (2y + 3)(y + 5) = 0 \]

### Step 6: Set each factor to zero
Now we will set each factor equal to zero:
1. \( 2y + 3 = 0 \) → \( y = -\frac{3}{2} \)
2. \( y + 5 = 0 \) → \( y = -5 \)

Thus, the solutions for \( y \) are:
\[ y = -\frac{3}{2} \quad \text{and} \quad y = -5 \]

### Step 7: Compare the values of \( x \) and \( y \)
Now we have:
- For \( x \): \( 7 \) and \( -1 \)
- For \( y \): \( -\frac{3}{2} \) (which is approximately \(-1.5\)) and \( -5 \)

Comparing these values:
- \( 7 > -1 \)
- \( 7 > -\frac{3}{2} \)
- \( 7 > -5 \)
- \( -1 > -\frac{3}{2} \)
- \( -1 > -5 \)

Thus, we can conclude that \( x \) is greater than both values of \( y \).

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

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