I'm trying to write this expression as one fraction:
\begin{align} \frac{5}{y+1}\cdot\frac{1}{5}+\frac{y}{\frac{y+1}{3}} &= \frac{5y}{5y+5}+\frac{3y}{y+1} \\ &= \frac{5y^2+5+15y^2+15y}{5y^2+10y+5} \\ &= \frac{20y^2+15y+5}{5y^2+10y+5} \\ &= \frac{4y^2+3y+1}{2y}. \end{align}
Am I doing this in the right way? Any hints about what to do next, please?
Key idea to remember is that $$\frac{1}{\frac{a}{b}} = \frac{b}{a}.$$ So you get $$ \frac{5}{y+1} \cdot \frac{1}{5} + \frac{y}{\frac{y+1}{3}} = \frac{1}{y+1} + \frac{3y}{y+1} = \frac{1+3y}{y+1}. $$
Your mistake was in simplifying from step 3 to 4. You cannot simplify each term like that. You can only simplify factors.
For example $\frac{5}{13}$ could be rewritten as $\frac{2+3}{4+9}$ Using your method this would simplify to $\frac{1+1}{2+3}$ or $\frac{2}{5}$ But we know that $\frac{5}{13}$ isn't equal to $\frac{2}{5}$
You can only simplify by factoring and canceling out factors. YOu should factor out the extra (y+1) that you have in the numerator and denominator. Alternatively you could simplify the first fraction before you add it making your life so much simpler.
You made a lot of errors. If you do calculations keep your numbers as small as possible:
If you calculte $\frac{5}{y+1}\cdot\frac{1}{5}$ observe tha the 5 can be canceled before multiplication, leaving $\frac{1}{y+1}$.
If you add two fraction as in $(2)$ try to split nominator and denominator in factors an to cancel.
The smallest common denominator is the least common multiple and not the product.
You should try to find your errors by doing checks like the following
To check if the calculation
$$ \begin{align} \frac{5}{y+1}\cdot\frac{1}{5}+\frac{y}{\frac{y+1}{3}} = \tag{1}\\ \frac{5y}{5y+5}+\frac{3y}{y+1} =\tag{2}\\ \frac{5y^2+5+15y^2+15y}{5y^2+10y+5} =\tag{3}\\ \frac{20y^2+15y+5}{5y^2+10y+5} =\tag{4}\\ \frac{4y^2+3y+1}{2y}\tag{5} \end{align} $$ you can substitute $y$ by arbitrary numbers and check if corresponding terms are equal. If they are not equal you have an error in your calculation. If they are equal your calculaion is right ot the terms are equal by accident. You can try different numbers for $y$ to decrease the chance that terms are equal by accident.
I substitute $y$ by $0$ because calculations with $0$ are easy. If we substitute $y$ by $0$ in $(1)$ get $0$ . If we substitute $y$ by $0$ in $(5)$ we get $\frac{1}{0}$. So you made an error in your calculation.
Where do we make the error? Let's substitute $y$ by $0$ in $(3)$ and we get $1$. So there is at least an error if you caculate $(5)$ from $(7)$. Because the value of $(4)$ is $1$ after substitution is $1$, you made an error when transforming $(4)$ to $(5)$. I don't know how you get from $(4)$ to $(5)$. Maybe you assume that the steps 1-2-3-4 are correct because the value of (4) is the same as the value of $(1)$ if you substitute $y$ by $0$, but I f you substitute $y$ by $2$ the value of $(4)$ is $\frac{23}{9}$ or 2.556 f you use a calculator and the value of $(1)$ is $\frac{7}{3}$ or 2.333.
I hope with this method you can fix your errors.
