Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between 'x' and 'y' and given answer
I. `( x)/( x + 7) + ( x + 7)/( x) = 12 `
II ` ( y)/( y + 8) + ( y + 8)/( y) = 16 `

Options

A. If ` x gt y `
B. If ` x le y `
C. If ` x lt y `
D. If x = y or Relation cannot be established

Answer

To solve the given equations and determine the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve Equation I
The first equation is given as:
\[ \frac{x}{x + 7} + \frac{x + 7}{x} = 12 \]
To simplify, we will find a common denominator:
\[ \frac{x^2 + (x + 7)^2}{x(x + 7)} = 12 \]
Expanding \( (x + 7)^2 \):
\[ \frac{x^2 + (x^2 + 14x + 49)}{x(x + 7)} = 12 \]
This simplifies to:
\[ \frac{2x^2 + 14x + 49}{x(x + 7)} = 12 \]
Cross-multiplying gives:
\[ 2x^2 + 14x + 49 = 12x^2 + 84x \]
Rearranging the equation:
\[ 0 = 10x^2 + 70x - 49 \]
Thus, we have:
\[ 10x^2 + 70x - 49 = 0 \]

### Step 2: Apply the Quadratic Formula for Equation I
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 10 \), \( b = 70 \), and \( c = -49 \).
Calculating the discriminant:
\[ b^2 - 4ac = 70^2 - 4 \cdot 10 \cdot (-49) = 4900 + 1960 = 6860 \]
Now substituting into the formula:
\[ x = \frac{-70 \pm \sqrt{6860}}{20} \]
Calculating the square root:
\[ \sqrt{6860} \approx 82.8 \]
Thus:
\[ x = \frac{-70 \pm 82.8}{20} \]
Calculating the two possible values for \( x \):
1. \( x_1 = \frac{12.8}{20} = 0.64 \)
2. \( x_2 = \frac{-152.8}{20} = -7.64 \)

### Step 3: Solve Equation II
The second equation is:
\[ \frac{y}{y + 8} + \frac{y + 8}{y} = 16 \]
Finding a common denominator:
\[ \frac{y^2 + (y + 8)^2}{y(y + 8)} = 16 \]
Expanding \( (y + 8)^2 \):
\[ \frac{y^2 + (y^2 + 16y + 64)}{y(y + 8)} = 16 \]
This simplifies to:
\[ \frac{2y^2 + 16y + 64}{y(y + 8)} = 16 \]
Cross-multiplying gives:
\[ 2y^2 + 16y + 64 = 16y^2 + 128y \]
Rearranging:
\[ 0 = 14y^2 + 112y - 64 \]
Thus, we have:
\[ 14y^2 + 112y - 64 = 0 \]

### Step 4: Apply the Quadratic Formula for Equation II
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 14 \), \( b = 112 \), and \( c = -64 \).
Calculating the discriminant:
\[ b^2 - 4ac = 112^2 - 4 \cdot 14 \cdot (-64) = 12544 + 3584 = 16128 \]
Now substituting into the formula:
\[ y = \frac{-112 \pm \sqrt{16128}}{28} \]
Calculating the square root:
\[ \sqrt{16128} \approx 126.9 \]
Thus:
\[ y = \frac{-112 \pm 126.9}{28} \]
Calculating the two possible values for \( y \):
1. \( y_1 = \frac{14.9}{28} \approx 0.53 \)
2. \( y_2 = \frac{-238.9}{28} \approx -8.53 \)

### Step 5: Compare Values of \( x \) and \( y \)
From the results:
- \( x \) can be \( 0.64 \) or \( -7.64 \)
- \( y \) can be \( 0.53 \) or \( -8.53 \)

### Conclusion
1. If we take the positive values: \( x = 0.64 \) and \( y = 0.53 \), then \( x > y \).
2. If we take the negative values: \( x = -7.64 \) and \( y = -8.53 \), then \( x < y \).

Since we have both \( x > y \) and \( x < y \), we cannot establish a definitive relationship between \( x \) and \( y \).