In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
I `6x^2+17x+12=0`
II `3y^2+11y+10=0`

Options

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

Answer

To solve the given equations, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 6x^2 + 17x + 12 = 0 \]

To solve this quadratic equation, we will factor it. We need to find two numbers that multiply to \(6 \times 12 = 72\) and add up to \(17\). The factors of \(72\) that add up to \(17\) are \(9\) and \(8\).

Now, we can rewrite the equation:
\[ 6x^2 + 9x + 8x + 12 = 0 \]

Next, we group the terms:
\[ (6x^2 + 9x) + (8x + 12) = 0 \]

Now, we factor by grouping:
\[ 3x(2x + 3) + 4(2x + 3) = 0 \]

This gives us:
\[ (3x + 4)(2x + 3) = 0 \]

Setting each factor to zero gives:
1. \(3x + 4 = 0 \Rightarrow x = -\frac{4}{3}\)
2. \(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)

So, the solutions for \(x\) are:
\[ x_1 = -\frac{4}{3} \quad \text{and} \quad x_2 = -\frac{3}{2} \]

### Step 2: Solve Equation II
The second equation is:
\[ 3y^2 + 11y + 10 = 0 \]

Similarly, we will factor this equation. We need two numbers that multiply to \(3 \times 10 = 30\) and add up to \(11\). The factors of \(30\) that add up to \(11\) are \(6\) and \(5\).

Now, we can rewrite the equation:
\[ 3y^2 + 6y + 5y + 10 = 0 \]

Next, we group the terms:
\[ (3y^2 + 6y) + (5y + 10) = 0 \]

Now, we factor by grouping:
\[ 3y(y + 2) + 5(y + 2) = 0 \]

This gives us:
\[ (3y + 5)(y + 2) = 0 \]

Setting each factor to zero gives:
1. \(3y + 5 = 0 \Rightarrow y = -\frac{5}{3}\)
2. \(y + 2 = 0 \Rightarrow y = -2\)

So, the solutions for \(y\) are:
\[ y_1 = -\frac{5}{3} \quad \text{and} \quad y_2 = -2 \]

### Step 3: Compare Values of x and y
Now we have the values:
- For \(x\): \(-\frac{4}{3} \approx -1.33\) and \(-\frac{3}{2} \approx -1.5\)
- For \(y\): \(-\frac{5}{3} \approx -1.66\) and \(-2\)

Now we will compare:
1. Comparing \(x_1 = -\frac{4}{3}\) and \(y_1 = -\frac{5}{3}\):
\(-\frac{4}{3} > -\frac{5}{3}\) (since -1.33 is greater than -1.66)

2. Comparing \(x_1 = -\frac{4}{3}\) and \(y_2 = -2\):
\(-\frac{4}{3} > -2\) (since -1.33 is greater than -2)

3. Comparing \(x_2 = -\frac{3}{2}\) and \(y_1 = -\frac{5}{3}\):
\(-\frac{3}{2} < -\frac{5}{3}\) (since -1.5 is less than -1.66)

4. Comparing \(x_2 = -\frac{3}{2}\) and \(y_2 = -2\):
\(-\frac{3}{2} > -2\) (since -1.5 is greater than -2)

### Conclusion
From the comparisons, we can conclude that in all cases where \(x\) is compared with \(y\), \(x\) is greater than \(y\).

### Final Answer
Thus, the answer is:
\[ x > y \]