In each of these questions, two equations I. and II are given. You have to solve both the equations and give answer
I. ` 3 x ^(2) + 13 x + 12 = 0 `
II. ` y^(2) + 9 y + 20 = 0 `

Options

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

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 + 13x + 12 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(3 \times 12 = 36\) and add up to \(13\).

The numbers that satisfy these conditions are \(9\) and \(4\).

Now, we can rewrite the middle term using these numbers:
\[ 3x^2 + 9x + 4x + 12 = 0 \]

Next, we group the terms:
\[ (3x^2 + 9x) + (4x + 12) = 0 \]

Factoring out the common factors:
\[ 3x(x + 3) + 4(x + 3) = 0 \]

Now, we can factor out \((x + 3)\):
\[ (3x + 4)(x + 3) = 0 \]

Setting each factor to zero gives us the solutions:
1. \(3x + 4 = 0 \Rightarrow 3x = -4 \Rightarrow x = -\frac{4}{3}\)
2. \(x + 3 = 0 \Rightarrow x = -3\)

So, the solutions for \(x\) are:
\[ x = -\frac{4}{3} \quad \text{and} \quad x = -3 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 + 9y + 20 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \(20\) and add up to \(9\).

The numbers that satisfy these conditions are \(4\) and \(5\).

Now, we can rewrite the equation:
\[ y^2 + 4y + 5y + 20 = 0 \]

Next, we group the terms:
\[ (y^2 + 4y) + (5y + 20) = 0 \]

Factoring out the common factors:
\[ y(y + 4) + 5(y + 4) = 0 \]

Now, we can factor out \((y + 4)\):
\[ (y + 4)(y + 5) = 0 \]

Setting each factor to zero gives us the solutions:
1. \(y + 4 = 0 \Rightarrow y = -4\)
2. \(y + 5 = 0 \Rightarrow y = -5\)

So, the solutions for \(y\) are:
\[ y = -4 \quad \text{and} \quad y = -5 \]

### Summary of Solutions
- For Equation I, \(x = -\frac{4}{3}\) and \(x = -3\).
- For Equation II, \(y = -4\) and \(y = -5\).

### Step 3: Compare the Values
Now we can compare the values of \(x\) and \(y\):
1. Compare \(x = -3\) with \(y = -4\):
\(-3 > -4\) (so \(x > y\))

2. Compare \(x = -\frac{4}{3}\) (which is approximately -1.33) with \(y = -5\):
\(-\frac{4}{3} > -5\) (so \(x > y\))

### Conclusion
In both cases, we find that \(x\) is greater than \(y\).

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