To solve the given complex fraction step by step, we will first analyze the numerator and denominator separately and then divide them.

Problem:

\frac{\frac{\frac{1}{2y-2} + \frac{3}{5}}{\frac{2}{y+1} + \frac{1}{y-1}}}

Step 1: Simplify the numerator

The numerator of the large fraction is: $\frac{1}{2y-2} + \frac{3}{5}.$

To add these fractions, find a common denominator. Factor $2y - 2$ as $2(y - 1)$. The common denominator will be $2 \cdot 5 \cdot (y - 1)$.

Rewrite the fractions with this common denominator: $\frac{1}{2(y-1)} = \frac{5}{10(y-1)},$ $\frac{3}{5} = \frac{6(y-1)}{10(y-1)}.$

Now combine: $\frac{1}{2(y-1)} + \frac{3}{5} = \frac{5 + 6(y-1)}{10(y-1)} = \frac{6y - 1}{10(y-1)}.$

Step 2: Simplify the denominator

The denominator of the large fraction is: $\frac{2}{y+1} + \frac{1}{y-1}.$

To add these fractions, find the common denominator $(y+1)(y-1)$: $\frac{2}{y+1} = \frac{2(y-1)}{(y+1)(y-1)},$ $\frac{1}{y-1} = \frac{y+1}{(y+1)(y-1)}.$

Combine: $\frac{2}{y+1} + \frac{1}{y-1} = \frac{2(y-1) + (y+1)}{(y+1)(y-1)} = \frac{2y - 2 + y + 1}{(y+1)(y-1)} = \frac{3y - 1}{(y+1)(y-1)}.$

Step 3: Divide the simplified numerator by the simplified denominator

We now have: $\frac{\frac{6y - 1}{10(y-1)}}{\frac{3y - 1}{(y+1)(y-1)}}.$

To divide, multiply the first fraction by the reciprocal of the second: $\frac{6y - 1}{10(y-1)} \cdot \frac{(y+1)(y-1)}{3y - 1}.$

Cancel out $(y-1)$ from numerator and denominator: $\frac{(6y - 1)(y+1)}{10(3y - 1)}.$

Final Answer:

$\frac{(6y - 1)(y+1)}{10(3y - 1)}.$

Let me know if you'd like a deeper explanation or have questions!

Related Questions:

1. How do you simplify a complex fraction involving multiple terms?
2. What is the importance of finding a common denominator in fraction operations?
3. How do factoring techniques help simplify algebraic fractions?
4. How do you identify and cancel out common factors in a rational expression?
5. Can this expression have restrictions on the variable $y$?

Tip:

Always factor expressions completely before performing operations; this helps in canceling out common terms efficiently.