We need to solve the following equation for $y$: $\frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} = \frac{y-3}{4-8y} + \frac{1}{6}$

1. Problem Description

We need to solve the following equation for $y$:

$\frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} = \frac{y-3}{4-8y} + \frac{1}{6}$

2. Solution Steps

First, let's rewrite the equation:

$\frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} = \frac{y-3}{4(1-2y)} + \frac{1}{6}$

We can factor out a 4 from the denominator of the third term on the left side. Note that $4-8y = -4(2y-1)$

$\frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} = \frac{y-3}{-4(2y-1)} + \frac{1}{6}$

$\frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} = -\frac{y-3}{4(2y-1)} + \frac{1}{6}$

To clear the fractions, we multiply both sides by $12(2y-1)$.

$12(2y-1) \left( \frac{y+1}{2y-1} - \frac{11y+5}{12(2y-1)} \right) = 12(2y-1) \left( -\frac{y-3}{4(2y-1)} + \frac{1}{6} \right)$

$12(y+1) - (11y+5) = -3(y-3) + 2(2y-1)$

$12y + 12 - 11y - 5 = -3y + 9 + 4y - 2$

$y + 7 = y + 7$

$0 = 0$

This implies that the equation is always true, meaning any value of $y$ is a solution, except for values that make the denominators zero. We have the following denominators:

$2y-1 \neq 0 \implies 2y \neq 1 \implies y \neq \frac{1}{2}$

$12(2y-1) \neq 0 \implies 2y-1 \neq 0 \implies y \neq \frac{1}{2}$

$4-8y \neq 0 \implies 4 \neq 8y \implies y \neq \frac{1}{2}$

Therefore, the solution is all real numbers except $y = \frac{1}{2}$.

3. Final Answer

All real numbers except $y = \frac{1}{2}$.