We need to solve the following equation for $y$: $\frac{1}{15y-10} - \frac{5-y}{27y^3 - 54y^2 + 36y - 8} = \frac{1.2y - 1}{18y^2 - 24y + 8}$.

1. Problem Description

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

$\frac{1}{15y-10} - \frac{5-y}{27y^3 - 54y^2 + 36y - 8} = \frac{1.2y - 1}{18y^2 - 24y + 8}$.

2. Solution Steps

First, we simplify the denominators:

$15y - 10 = 5(3y - 2)$

$27y^3 - 54y^2 + 36y - 8 = (3y)^3 - 3(3y)^2(2) + 3(3y)(2^2) - 2^3 = (3y-2)^3$

$18y^2 - 24y + 8 = 2(9y^2 - 12y + 4) = 2(3y-2)^2$

Now the equation becomes:

$\frac{1}{5(3y-2)} - \frac{5-y}{(3y-2)^3} = \frac{1.2y - 1}{2(3y-2)^2}$

Multiply both sides by $10(3y-2)^3$:

$2(3y-2)^2 - 10(5-y) = 5(1.2y - 1)(3y-2)$

$2(9y^2 - 12y + 4) - 50 + 10y = 5(3.6y^2 - 2.4y - 3y + 2)$

$18y^2 - 24y + 8 - 50 + 10y = 5(3.6y^2 - 5.4y + 2)$

$18y^2 - 14y - 42 = 18y^2 - 27y + 10$

$-14y - 42 = -27y + 10$

$27y - 14y = 10 + 42$

$13y = 52$

$y = \frac{52}{13}$

$y = 4$

We need to check if $y=4$ is a valid solution by checking if the denominators are zero:

$15y-10 = 15(4)-10 = 60-10 = 50 \ne 0$

$27y^3 - 54y^2 + 36y - 8 = (3(4)-2)^3 = (12-2)^3 = 10^3 = 1000 \ne 0$

$18y^2 - 24y + 8 = 2(3y-2)^2 = 2(3(4)-2)^2 = 2(10)^2 = 200 \ne 0$

3. Final Answer

$y = 4$