In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer
I. `3x^2+13x+12=0`
II. `4y^2+5y+1=0`

Options

A. `xgty`
B. `xgey`
C. the relationship can’t be established
D. `xlty`

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve Equation I: \(3x^2 + 13x + 12 = 0\)

1. **Rewrite the equation**: We have the quadratic equation in standard form.
\[ 3x^2 + 13x + 12 = 0 \]

2. **Factor the quadratic**: We need to express \(13x\) as the sum of two numbers that multiply to \(3 \times 12 = 36\) and add to \(13\). The numbers are \(9\) and \(4\).
\[ 3x^2 + 9x + 4x + 12 = 0 \]

3. **Group the terms**: Group the first two and the last two terms.
\[ (3x^2 + 9x) + (4x + 12) = 0 \]

4. **Factor by grouping**: Factor out the common terms from each group.
\[ 3x(x + 3) + 4(x + 3) = 0 \]

5. **Factor out the common binomial**:
\[ (x + 3)(3x + 4) = 0 \]

6. **Set each factor to zero**:
\[ x + 3 = 0 \quad \text{or} \quad 3x + 4 = 0 \]

7. **Solve for \(x\)**:
- From \(x + 3 = 0\):
\[ x = -3 \]
- From \(3x + 4 = 0\):
\[ 3x = -4 \implies x = -\frac{4}{3} \]

### Step 2: Solve Equation II: \(4y^2 + 5y + 1 = 0\)

1. **Rewrite the equation**: We have the quadratic equation in standard form.
\[ 4y^2 + 5y + 1 = 0 \]

2. **Factor the quadratic**: We need to express \(5y\) as the sum of two numbers that multiply to \(4 \times 1 = 4\) and add to \(5\). The numbers are \(4\) and \(1\).
\[ 4y^2 + 4y + 1y + 1 = 0 \]

3. **Group the terms**: Group the first two and the last two terms.
\[ (4y^2 + 4y) + (1y + 1) = 0 \]

4. **Factor by grouping**: Factor out the common terms from each group.
\[ 4y(y + 1) + 1(y + 1) = 0 \]

5. **Factor out the common binomial**:
\[ (y + 1)(4y + 1) = 0 \]

6. **Set each factor to zero**:
\[ y + 1 = 0 \quad \text{or} \quad 4y + 1 = 0 \]

7. **Solve for \(y\)**:
- From \(y + 1 = 0\):
\[ y = -1 \]
- From \(4y + 1 = 0\):
\[ 4y = -1 \implies y = -\frac{1}{4} \]

### Final Values:
- The solutions for \(x\) are \(x = -3\) and \(x = -\frac{4}{3}\).
- The solutions for \(y\) are \(y = -1\) and \(y = -\frac{1}{4}\).

### Comparison:
Now we compare the values of \(x\) and \(y\):
- \(x\) values: \(-3\), \(-\frac{4}{3}\)
- \(y\) values: \(-1\), \(-\frac{1}{4}\)

From the values:
- \(-1 > -3\)
- \(-\frac{1}{4} > -\frac{4}{3}\)

Thus, \(y\) is greater than \(x\).

### Conclusion:
The correct answer is that \(x < y\).

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