In each of these questions, two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` x = (1331) ^((1)/(3))`
II.` 2 y^(2) - 17 y + 36 = 0 `

Options

A. If ` x gt y `
B. If ` x lt y `
C. If ` x ge Y`
D. If ` x le y `

Answer

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

### Step 1: Solve Equation I
The first equation is:
\[ x = (1331)^{\frac{1}{3}} \]

To find the value of \( x \), we need to calculate the cube root of 1331.

**Calculation:**
\[ 1331 = 11^3 \]
Thus,
\[ x = (11^3)^{\frac{1}{3}} = 11 \]

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 - 17y + 36 = 0 \]

This is a quadratic equation in the standard form \( ay^2 + by + c = 0 \). We can solve it using the factorization method.

**Finding Factors:**
We need to find two numbers that multiply to \( 2 \times 36 = 72 \) and add up to \( -17 \).

The factors of 72 that satisfy this condition are \( -8 \) and \( -9 \).

**Rewriting the Equation:**
We can rewrite the equation as:
\[ 2y^2 - 8y - 9y + 36 = 0 \]

Now, we can group the terms:
\[ (2y^2 - 8y) + (-9y + 36) = 0 \]

Factoring by grouping:
\[ 2y(y - 4) - 9(y - 4) = 0 \]

Now, we can factor out \( (y - 4) \):
\[ (2y - 9)(y - 4) = 0 \]

**Finding Values of \( y \):**
Setting each factor to zero gives us:
1. \( 2y - 9 = 0 \) → \( 2y = 9 \) → \( y = \frac{9}{2} = 4.5 \)
2. \( y - 4 = 0 \) → \( y = 4 \)

### Step 3: Summary of Solutions
Now we have:
- From Equation I: \( x = 11 \)
- From Equation II: \( y = 4 \) and \( y = 4.5 \)

### Step 4: Compare \( x \) and \( y \)
Now we compare the values of \( x \) and \( y \):
- For \( y = 4 \): \( x = 11 > 4 \)
- For \( y = 4.5 \): \( x = 11 > 4.5 \)

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

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

---