In the following questions two equations numbered I and IIare given. You have to solve both the equations and give answer
I `2x^2+15x+27=0`
II `2y^2+7y+6=0`

Options

A. if`xgty`
B. if`xgey`
C. if`xlty`
D. if`xley`

Answer

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

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 + 15x + 27 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(2 \times 27 = 54\) and add up to \(15\).

### Step 2: Factorization
The two numbers that satisfy this condition are \(9\) and \(6\). We can rewrite the equation as:
\[ 2x^2 + 9x + 6x + 27 = 0 \]

Now, we can group the terms:
\[ (2x^2 + 9x) + (6x + 27) = 0 \]

Factoring out common terms:
\[ x(2x + 9) + 3(2x + 9) = 0 \]

Now, factor out \((2x + 9)\):
\[ (2x + 9)(x + 3) = 0 \]

### Step 3: Find the Values of x
Setting each factor to zero gives us:
1. \(2x + 9 = 0 \Rightarrow x = -\frac{9}{2}\)
2. \(x + 3 = 0 \Rightarrow x = -3\)

So, the solutions for \(x\) are:
\[ x = -\frac{9}{2} \quad \text{and} \quad x = -3 \]

### Step 4: Solve Equation II
Now, let's solve the second equation:
\[ 2y^2 + 7y + 6 = 0 \]

Again, we will factor this equation. We need two numbers that multiply to \(2 \times 6 = 12\) and add up to \(7\).

### Step 5: Factorization
The two numbers that satisfy this condition are \(4\) and \(3\). We can rewrite the equation as:
\[ 2y^2 + 4y + 3y + 6 = 0 \]

Now, we can group the terms:
\[ (2y^2 + 4y) + (3y + 6) = 0 \]

Factoring out common terms:
\[ 2y(y + 2) + 3(y + 2) = 0 \]

Now, factor out \((y + 2)\):
\[ (y + 2)(2y + 3) = 0 \]

### Step 6: Find the Values of y
Setting each factor to zero gives us:
1. \(y + 2 = 0 \Rightarrow y = -2\)
2. \(2y + 3 = 0 \Rightarrow y = -\frac{3}{2}\)

So, the solutions for \(y\) are:
\[ y = -2 \quad \text{and} \quad y = -\frac{3}{2} \]

### Step 7: Compare x and y
Now, we have the values:
- For \(x\): \(-\frac{9}{2} \approx -4.5\) and \(-3\)
- For \(y\): \(-2\) and \(-\frac{3}{2} \approx -1.5\)

Now we can compare the values:
1. Comparing \(x = -\frac{9}{2}\) with \(y = -2\):
\(-\frac{9}{2} < -2\)

2. Comparing \(x = -3\) with \(y = -2\):
\(-3 < -2\)

3. Comparing \(x = -\frac{9}{2}\) with \(y = -\frac{3}{2}\):
\(-\frac{9}{2} < -\frac{3}{2}\)

### Conclusion
In all comparisons, \(x\) is less than \(y\). Therefore, the answer is:
\[ x < y \]