In each of these questions, two equations I. and II. are given . you have to solve both the equations and answer the following questions
I. ` 1 = (1)/( x) ( 2 - (11)/( 36 x))`
II. `(( 14 y)/( 3) + (9)/( y)) = 13`

Options

A. ` x lt y `
B. ` x le y `
C. ` x gt y `
D. if x = y or relationship between x and y cannot be established

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:
\[ 1 = \frac{1}{x} \left( 2 - \frac{11}{36x} \right) \]

**Step 1.1:** Multiply both sides by \( x \) to eliminate the fraction:
\[ x = 2 - \frac{11}{36x} \]

**Step 1.2:** Multiply both sides by \( 36x \) to eliminate the fraction:
\[ 36x^2 = 72x - 11 \]

**Step 1.3:** Rearranging gives us a standard quadratic equation:
\[ 36x^2 - 72x + 11 = 0 \]

**Step 1.4:** Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 36, b = -72, c = 11 \):
- Calculate the discriminant:
\[ b^2 - 4ac = (-72)^2 - 4 \cdot 36 \cdot 11 = 5184 - 1584 = 3600 \]

- Now apply the quadratic formula:
\[ x = \frac{72 \pm \sqrt{3600}}{72} \]
\[ x = \frac{72 \pm 60}{72} \]

**Step 1.5:** Calculate the two possible values for \( x \):
1. \( x = \frac{132}{72} = \frac{11}{6} \approx 1.8333 \)
2. \( x = \frac{12}{72} = \frac{1}{6} \approx 0.1667 \)

### Step 2: Solve Equation II
The second equation is:
\[ \frac{14y}{3} + \frac{9}{y} = 13 \]

**Step 2.1:** Multiply through by \( 3y \) to eliminate the fractions:
\[ 14y^2 + 27 = 39y \]

**Step 2.2:** Rearranging gives us:
\[ 14y^2 - 39y + 27 = 0 \]

**Step 2.3:** Now we will use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 14, b = -39, c = 27 \):
- Calculate the discriminant:
\[ b^2 - 4ac = (-39)^2 - 4 \cdot 14 \cdot 27 = 1521 - 1512 = 9 \]

- Now apply the quadratic formula:
\[ y = \frac{39 \pm \sqrt{9}}{28} \]
\[ y = \frac{39 \pm 3}{28} \]

**Step 2.4:** Calculate the two possible values for \( y \):
1. \( y = \frac{42}{28} = \frac{3}{2} = 1.5 \)
2. \( y = \frac{36}{28} = \frac{9}{7} \approx 1.2857 \)

### Summary of Solutions
- From Equation I, we have:
- \( x_1 \approx 1.8333 \) and \( x_2 \approx 0.1667 \)
- From Equation II, we have:
- \( y_1 = 1.5 \) and \( y_2 \approx 1.2857 \)

### Step 3: Compare Values
Now we compare the values of \( x \) and \( y \):
1. For \( x_1 \approx 1.8333 \) and \( y_1 = 1.5 \):
- \( x_1 > y_1 \)

2. For \( x_1 \approx 1.8333 \) and \( y_2 \approx 1.2857 \):
- \( x_1 > y_2 \)

3. For \( x_2 \approx 0.1667 \) and \( y_1 = 1.5 \):
- \( x_2 < y_1 \)

4. For \( x_2 \approx 0.1667 \) and \( y_2 \approx 1.2857 \):
- \( x_2 < y_2 \)

### Conclusion
From the comparisons, we can conclude:
- \( x_1 > y_1 \)
- \( x_1 > y_2 \)
- \( x_2 < y_1 \)
- \( x_2 < y_2 \)

Thus, there is no consistent relationship between \( x \) and \( y \) across all values.

### Final Answer
The answer is that no relation can be established between \( x \) and \( y \).