` 1/(x-3) + 1/(x+5) = 1/3`
`(y+2)(27-y)= 210`

Options

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

Explanation

To solve the equations \( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \) and \( (y+2)(27-y) = 210 \), we will proceed step by step.

### Step 1: Solve the first equation

1. Start with the equation:
\[ \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \]

2. Find a common denominator for the left side:
\[ \frac{(x+5) + (x-3)}{(x-3)(x+5)} = \frac{1}{3} \]

3. Simplify the numerator:
\[ \frac{2x + 2}{(x-3)(x+5)} = \frac{1}{3} \]

4. Cross-multiply:
\[ 3(2x + 2) = (x-3)(x+5) \]

5. Expand both sides:
\[ 6x + 6 = x^2 + 2x - 15 \]

6. Rearrange the equation to form a quadratic equation:
\[ x^2 - 4x - 21 = 0 \]

### Step 2: Factor the quadratic equation

1. Factor the quadratic:
\[ (x - 7)(x + 3) = 0 \]

2. Set each factor to zero:
\[ x - 7 = 0 \quad \text{or} \quad x + 3 = 0 \]

3. Solve for \( x \):
\[ x = 7 \quad \text{or} \quad x = -3 \]

### Step 3: Solve the second equation

1. Start with the equation:
\[ (y + 2)(27 - y) = 210 \]

2. Expand the left side:
\[ 27y - y^2 + 54 - 2y = 210 \]

3. Rearrange the equation:
\[ -y^2 + 25y + 54 - 210 = 0 \]
\[ -y^2 + 25y - 156 = 0 \]

4. Multiply through by -1:
\[ y^2 - 25y + 156 = 0 \]

### Step 4: Factor the quadratic equation

1. Factor the quadratic:
\[ (y - 12)(y - 13) = 0 \]

2. Set each factor to zero:
\[ y - 12 = 0 \quad \text{or} \quad y - 13 = 0 \]

3. Solve for \( y \):
\[ y = 12 \quad \text{or} \quad y = 13 \]

### Step 5: Determine the relationship between \( x \) and \( y \)

1. The values obtained are:
- \( x = 7 \) or \( x = -3 \)
- \( y = 12 \) or \( y = 13 \)

2. Compare the values:
- For \( x = 7 \), \( y = 12 \) or \( y = 13 \) (both are greater than 7).
- For \( x = -3 \), \( y = 12 \) or \( y = 13 \) (both are greater than -3).

Thus, in both cases, \( y > x \).

### Final Answer

The values of \( x \) are \( 7 \) and \( -3 \), and the values of \( y \) are \( 12 \) and \( 13 \). The relationship is \( y > x \).

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