In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option
I `3x^2+10x+7=0`
II `3y^2+4y+1=0`

Options

A. `xley`
B. x = y or no relation can be established between x and y
C. `xgty`
D. `xlty`

Answer

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

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 + 10x + 7 = 0 \]

To solve this quadratic equation, we can use the factorization method.

1. **Identify the coefficients**: Here, \( a = 3 \), \( b = 10 \), and \( c = 7 \).
2. **Multiply \( a \) and \( c \)**:
\[ ac = 3 \times 7 = 21 \]
3. **Find two numbers that multiply to \( ac \) (21) and add to \( b \) (10)**:
The numbers are \( 3 \) and \( 7 \).
4. **Rewrite the middle term**:
\[ 3x^2 + 3x + 7x + 7 = 0 \]
5. **Factor by grouping**:
\[ 3x(x + 1) + 7(x + 1) = 0 \]
\[ (3x + 7)(x + 1) = 0 \]
6. **Set each factor to zero**:
\[ 3x + 7 = 0 \quad \text{or} \quad x + 1 = 0 \]
- From \( 3x + 7 = 0 \):
\[ 3x = -7 \]
\[ x = -\frac{7}{3} \]
- From \( x + 1 = 0 \):
\[ x = -1 \]

Thus, the solutions for \( x \) are:
\[ x = -1 \quad \text{and} \quad x = -\frac{7}{3} \]

### Step 2: Solve Equation II
The second equation is:
\[ 3y^2 + 4y + 1 = 0 \]

We will also use the factorization method here.

1. **Identify the coefficients**: Here, \( a = 3 \), \( b = 4 \), and \( c = 1 \).
2. **Multiply \( a \) and \( c \)**:
\[ ac = 3 \times 1 = 3 \]
3. **Find two numbers that multiply to \( ac \) (3) and add to \( b \) (4)**:
The numbers are \( 3 \) and \( 1 \).
4. **Rewrite the middle term**:
\[ 3y^2 + 3y + 1y + 1 = 0 \]
5. **Factor by grouping**:
\[ 3y(y + 1) + 1(y + 1) = 0 \]
\[ (3y + 1)(y + 1) = 0 \]
6. **Set each factor to zero**:
\[ 3y + 1 = 0 \quad \text{or} \quad y + 1 = 0 \]
- From \( 3y + 1 = 0 \):
\[ 3y = -1 \]
\[ y = -\frac{1}{3} \]
- From \( y + 1 = 0 \):
\[ y = -1 \]

Thus, the solutions for \( y \) are:
\[ y = -1 \quad \text{and} \quad y = -\frac{1}{3} \]

### Step 3: Compare the Values of \( x \) and \( y \)
Now we have:
- For \( x \): \( -1 \) and \( -\frac{7}{3} \)
- For \( y \): \( -1 \) and \( -\frac{1}{3} \)

To compare:
- \( -1 \) is equal to \( -1 \)
- \( -\frac{7}{3} \) is less than both \( -1 \) and \( -\frac{1}{3} \)

### Conclusion
Since \( y \) can be equal to \( x \) or greater than \( x \), we conclude that:
\[ x \leq y \]