Solve the given quadratic equations and mark the correct options based on your answer
I. ` x^(2) + 5x = 5 ( 2 x + 3 x)`
II. ` 3 y^(2) + 2y = 2 ( y + 6y)`

Options

A. `x gt y `
B. ` x ge y `
C. ` x lt y `
D. x = y or no relation can be established between x and y

Answer

Let's solve the given quadratic equations step by step.

### Step 1: Solve the first equation
The first equation is:
\[ x^2 + 5x = 5(2x + 3x) \]

**Hint:** Start by simplifying the right side of the equation.

#### Solution:
1. Simplify the right side:
\[ 5(2x + 3x) = 5(5x) = 25x \]

2. Now rewrite the equation:
\[ x^2 + 5x = 25x \]

3. Move all terms to one side:
\[ x^2 + 5x - 25x = 0 \]
\[ x^2 - 20x = 0 \]

4. Factor the quadratic:
\[ x(x - 20) = 0 \]

5. Set each factor to zero:
\[ x = 0 \quad \text{or} \quad x - 20 = 0 \]
\[ x = 20 \]

### Step 2: Solve the second equation
The second equation is:
\[ 3y^2 + 2y = 2(y + 6y) \]

**Hint:** Again, start by simplifying the right side of the equation.

#### Solution:
1. Simplify the right side:
\[ 2(y + 6y) = 2(7y) = 14y \]

2. Now rewrite the equation:
\[ 3y^2 + 2y = 14y \]

3. Move all terms to one side:
\[ 3y^2 + 2y - 14y = 0 \]
\[ 3y^2 - 12y = 0 \]

4. Factor the quadratic:
\[ 3y(y - 4) = 0 \]

5. Set each factor to zero:
\[ 3y = 0 \quad \text{or} \quad y - 4 = 0 \]
\[ y = 0 \quad \text{or} \quad y = 4 \]

### Step 3: Compare the values of x and y
From the solutions, we have:
- \( x = 20 \)
- \( y = 4 \)

**Hint:** Compare the values to determine the relationship.

#### Solution:
- Since \( 20 > 4 \), we conclude that \( x > y \).

### Final Answer
The correct option based on the comparison is:
- Option A: \( x > y \)