In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer
I.`3x^2 +10x+3=0` II.
`2y^2 +15y +27=0`

Options

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

Answer

To solve the given equations step by step, let's start with the first equation.

### Step 1: Solve the first equation \(3x^2 + 10x + 3 = 0\)

We will use the factorization method to solve this quadratic equation.

1. **Identify coefficients**: Here, \(a = 3\), \(b = 10\), and \(c = 3\).
2. **Multiply \(a\) and \(c\)**: \(3 \times 3 = 9\).
3. **Find two numbers that multiply to \(9\) and add to \(10\)**: The numbers are \(9\) and \(1\).
4. **Rewrite the equation**:
\[ 3x^2 + 9x + 1x + 3 = 0 \]
5. **Group the terms**:
\[ (3x^2 + 9x) + (1x + 3) = 0 \]
6. **Factor by grouping**:
\[ 3x(x + 3) + 1(x + 3) = 0 \]
7. **Factor out the common term**:
\[ (3x + 1)(x + 3) = 0 \]
8. **Set each factor to zero**:
\[ 3x + 1 = 0 \quad \text{or} \quad x + 3 = 0 \]
9. **Solve for \(x\)**:
- From \(3x + 1 = 0\):
\[ 3x = -1 \implies x = -\frac{1}{3} \]
- From \(x + 3 = 0\):
\[ x = -3 \]

Thus, the solutions for \(x\) are \(x = -\frac{1}{3}\) and \(x = -3\).

### Step 2: Solve the second equation \(2y^2 + 15y + 27 = 0\)

Now, we will solve the second quadratic equation using the factorization method.

1. **Identify coefficients**: Here, \(a = 2\), \(b = 15\), and \(c = 27\).
2. **Multiply \(a\) and \(c\)**: \(2 \times 27 = 54\).
3. **Find two numbers that multiply to \(54\) and add to \(15\)**: The numbers are \(6\) and \(9\).
4. **Rewrite the equation**:
\[ 2y^2 + 6y + 9y + 27 = 0 \]
5. **Group the terms**:
\[ (2y^2 + 6y) + (9y + 27) = 0 \]
6. **Factor by grouping**:
\[ 2y(y + 3) + 9(y + 3) = 0 \]
7. **Factor out the common term**:
\[ (2y + 9)(y + 3) = 0 \]
8. **Set each factor to zero**:
\[ 2y + 9 = 0 \quad \text{or} \quad y + 3 = 0 \]
9. **Solve for \(y\)**:
- From \(2y + 9 = 0\):
\[ 2y = -9 \implies y = -\frac{9}{2} \]
- From \(y + 3 = 0\):
\[ y = -3 \]

Thus, the solutions for \(y\) are \(y = -\frac{9}{2}\) and \(y = -3\).

### Summary of Solutions
- The solutions for \(x\) are \(x = -\frac{1}{3}\) and \(x = -3\).
- The solutions for \(y\) are \(y = -\frac{9}{2}\) and \(y = -3\).