In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
I `2x^2-11x+15=0`
II `y^2-3y+2=0`

Options

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

Answer

To solve the given equations step by step, we will first solve Equation I and then Equation II.

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 - 11x + 15 = 0 \]

To factor this quadratic equation, we need to find two numbers that multiply to \(2 \times 15 = 30\) and add up to \(-11\). The numbers that satisfy this condition are \(-6\) and \(-5\).

Now, we can rewrite the equation:
\[ 2x^2 - 6x - 5x + 15 = 0 \]

Next, we group the terms:
\[ (2x^2 - 6x) + (-5x + 15) = 0 \]

Now, factor by grouping:
\[ 2x(x - 3) - 5(x - 3) = 0 \]

This can be factored further:
\[ (2x - 5)(x - 3) = 0 \]

Now, we set each factor to zero:
1. \(2x - 5 = 0\) → \(2x = 5\) → \(x = \frac{5}{2}\)
2. \(x - 3 = 0\) → \(x = 3\)

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

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

To factor this quadratic equation, we need to find two numbers that multiply to \(2\) and add up to \(-3\). The numbers that satisfy this condition are \(-2\) and \(-1\).

Now, we can rewrite the equation:
\[ y^2 - 2y - y + 2 = 0 \]

Next, we group the terms:
\[ (y^2 - 2y) + (-y + 2) = 0 \]

Now, factor by grouping:
\[ y(y - 2) - 1(y - 2) = 0 \]

This can be factored further:
\[ (y - 1)(y - 2) = 0 \]

Now, we set each factor to zero:
1. \(y - 1 = 0\) → \(y = 1\)
2. \(y - 2 = 0\) → \(y = 2\)

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

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

We will compare \(x\) and \(y\):
1. Compare \(x = \frac{5}{2} = 2.5\) with \(y = 1\):
- \(2.5 > 1\) → \(x > y\)

2. Compare \(x = \frac{5}{2} = 2.5\) with \(y = 2\):
- \(2.5 > 2\) → \(x > y\)

3. Compare \(x = 3\) with \(y = 1\):
- \(3 > 1\) → \(x > y\)

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

In all cases, \(x\) is greater than \(y\).

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