The problem is to solve the following equation for $y$: $\frac{4y-1}{y-4} - \frac{5y}{3y-12} - \frac{6y-4}{5y-20} = 1 + \frac{y+1}{2y-8}$

1. Problem Description

The problem is to solve the following equation for $y$:

$\frac{4y-1}{y-4} - \frac{5y}{3y-12} - \frac{6y-4}{5y-20} = 1 + \frac{y+1}{2y-8}$

2. Solution Steps

First, we factor the denominators:

$3y - 12 = 3(y-4)$

$5y - 20 = 5(y-4)$

$2y - 8 = 2(y-4)$

So the equation becomes:

$\frac{4y-1}{y-4} - \frac{5y}{3(y-4)} - \frac{6y-4}{5(y-4)} = 1 + \frac{y+1}{2(y-4)}$

Multiply both sides of the equation by $30(y-4)$ to eliminate the fractions, assuming $y \neq 4$:

$30(4y-1) - 10(5y) - 6(6y-4) = 30(y-4) + 15(y+1)$

Expand:

$120y - 30 - 50y - 36y + 24 = 30y - 120 + 15y + 15$

Combine like terms on each side:

$34y - 6 = 45y - 105$

Subtract $34y$ from both sides:

$-6 = 11y - 105$

Add $105$ to both sides:

$99 = 11y$

Divide both sides by $11$:

$y = 9$

3. Final Answer

The solution to the equation is $y=9$.

Final Answer: The final answer is $\boxed{9}$