In each question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answers
I `2x^2-15x+27=0`
II `y^2=9`

Options

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

Answer

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

### 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.

1. **Factor the quadratic**: We need to find two numbers that multiply to \(2 \times 27 = 54\) and add up to \(-15\). The numbers that satisfy this are \(-6\) and \(-9\).

Rewrite the equation:
\[ 2x^2 - 6x - 9x + 27 = 0 \]

2. **Group the terms**:
\[ (2x^2 - 6x) + (-9x + 27) = 0 \]

3. **Factor by grouping**:
\[ 2x(x - 3) - 9(x - 3) = 0 \]

4. **Factor out the common term**:
\[ (2x - 9)(x - 3) = 0 \]

5. **Set each factor to zero**:
\[ 2x - 9 = 0 \quad \text{or} \quad x - 3 = 0 \]

Solving these gives:
\[ x = \frac{9}{2} = 4.5 \quad \text{and} \quad x = 3 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 = 9 \]

1. **Take the square root of both sides**:
\[ y = \pm \sqrt{9} \]

2. **Find the values of y**:
\[ y = 3 \quad \text{or} \quad y = -3 \]

### Step 3: Compare the Values of x and y
From the solutions:
- For Equation I, we have \( x = 3 \) and \( x = 4.5 \).
- For Equation II, we have \( y = 3 \) and \( y = -3 \).

### Step 4: Determine the Relationship
Now we can compare the values:
1. When \( x = 3 \), \( y = 3 \) → \( x = y \)
2. When \( x = 4.5 \), \( y = 3 \) → \( x > y \)

Thus, we can conclude that the relationship between \( x \) and \( y \) is:
\[ x \geq y \]

### Final Answer
The correct option is:
\[ x \geq y \]

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