In each of the following questions two equations are given. Solve these equations and give answer
I. ` 5 x^(2) + 11 x + 2 = 0 `
II. ` 4y ^(2) + 13 y + 3 = 0 `

Options

A. ` x lt y `
B. ` x gt y `
C. ` x le y`
D. ` x = y ` or no relation

Answer

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

### Step 1: Solve the first equation
The first equation is:
\[ 5x^2 + 11x + 2 = 0 \]

We can use the factorization method. We need to break down the middle term (11x) into two terms that multiply to \(5 \cdot 2 = 10\) and add up to \(11\). We can break it down as \(10x + 1x\).

Rewriting the equation:
\[ 5x^2 + 10x + 1x + 2 = 0 \]

Now, we can group the terms:
\[ (5x^2 + 10x) + (1x + 2) = 0 \]

Factoring out the common terms:
\[ 5x(x + 2) + 1(x + 2) = 0 \]

Now we can factor by grouping:
\[ (5x + 1)(x + 2) = 0 \]

### Step 2: Set each factor to zero
Now we set each factor to zero:
1. \(5x + 1 = 0\)
2. \(x + 2 = 0\)

Solving these gives:
1. \(5x + 1 = 0 \Rightarrow 5x = -1 \Rightarrow x = -\frac{1}{5}\)
2. \(x + 2 = 0 \Rightarrow x = -2\)

So, the solutions for \(x\) are:
\[ x = -\frac{1}{5} \quad \text{and} \quad x = -2 \]

### Step 3: Solve the second equation
The second equation is:
\[ 4y^2 + 13y + 3 = 0 \]

Again, we will factor this equation. We need to break down the middle term (13y) into two terms that multiply to \(4 \cdot 3 = 12\) and add up to \(13\). We can break it down as \(12y + 1y\).

Rewriting the equation:
\[ 4y^2 + 12y + 1y + 3 = 0 \]

Now, we can group the terms:
\[ (4y^2 + 12y) + (1y + 3) = 0 \]

Factoring out the common terms:
\[ 4y(y + 3) + 1(y + 3) = 0 \]

Now we can factor by grouping:
\[ (4y + 1)(y + 3) = 0 \]

### Step 4: Set each factor to zero
Now we set each factor to zero:
1. \(4y + 1 = 0\)
2. \(y + 3 = 0\)

Solving these gives:
1. \(4y + 1 = 0 \Rightarrow 4y = -1 \Rightarrow y = -\frac{1}{4}\)
2. \(y + 3 = 0 \Rightarrow y = -3\)

So, the solutions for \(y\) are:
\[ y = -\frac{1}{4} \quad \text{and} \quad y = -3 \]

### Step 5: Compare the values of \(x\) and \(y\)
Now we have the values:
- For \(x\): \(x = -\frac{1}{5}, -2\)
- For \(y\): \(y = -\frac{1}{4}, -3\)

Now we compare:
1. \(x = -\frac{1}{5}\) with \(y = -\frac{1}{4}\):
\(-\frac{1}{5} < -\frac{1}{4}\) (since \(-0.2 < -0.25\))

2. \(x = -\frac{1}{5}\) with \(y = -3\):
\(-\frac{1}{5} > -3\) (since \(-0.2 > -3\))

3. \(x = -2\) with \(y = -\frac{1}{4}\):
\(-2 < -\frac{1}{4}\) (since \(-2 < -0.25\))

4. \(x = -2\) with \(y = -3\):
\(-2 > -3\) (since \(-2 > -3\))

### Conclusion
From the comparisons:
- \(x = -\frac{1}{5}\) is greater than \(y = -3\) but less than \(y = -\frac{1}{4}\).
- \(x = -2\) is greater than \(y = -3\) but less than \(y = -\frac{1}{4}\).

Thus, we do not have a consistent relationship between \(x\) and \(y\) across all comparisons.

### Final Answer
The relation between \(x\) and \(y\) is that there is no consistent relationship.

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