We are asked to simplify the complex fraction: $\frac{\frac{7}{x+y} + \frac{1}{6}}{\frac{x}{x+y} - 1}$

1. Problem Description

We are asked to simplify the complex fraction:

$\frac{\frac{7}{x+y} + \frac{1}{6}}{\frac{x}{x+y} - 1}$

2. Solution Steps

First, simplify the numerator:

$\frac{7}{x+y} + \frac{1}{6} = \frac{7(6) + 1(x+y)}{6(x+y)} = \frac{42 + x + y}{6(x+y)}$

Next, simplify the denominator:

$\frac{x}{x+y} - 1 = \frac{x - (x+y)}{x+y} = \frac{x - x - y}{x+y} = \frac{-y}{x+y}$

Now we have:

$\frac{\frac{42+x+y}{6(x+y)}}{\frac{-y}{x+y}}$

To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:

$\frac{42+x+y}{6(x+y)} \cdot \frac{x+y}{-y} = \frac{(42+x+y)(x+y)}{6(x+y)(-y)}$

We can cancel out the $(x+y)$ term in the numerator and denominator, assuming $x+y \neq 0$:

$\frac{42+x+y}{6(-y)} = \frac{42+x+y}{-6y} = -\frac{x+y+42}{6y}$

3. Final Answer

$-\frac{x+y+42}{6y}$

Or equivalently:

$\frac{-x-y-42}{6y}$