1 ) I have this equation and don't know how $(1 + 2e^y)$ and $(1 + 2e^x)$ from each side of equation cancelled each other out and get the final answer $x = y$.
$$\frac{e^x}{1+2e^x}=\frac{e^y}{1+2e^y}$$
$$e^x(1 + 2e^y) = e^y(1 + 2e^x)$$
$$e^x = e^y$$
$$x = y$$
Question 2 - I don't know how the equation turned into fraction in the last step. Appreciate much for fast reply :)
$$y =\frac{e^x}{1+2e^x}$$
$$e^x = y + 2ye^x$$
$$e^x =\frac{y}{1-2y}$$ (How does $y + 2ye^x$ turned into this?)
If you need I can add in the missing steps. For question 1, you have
$e^x(1+2e^y)=e^y(1+2e^x)$
Clear the parentheses to obtain
$e^x+2e^{x+y}=e^y+2e^{x+y}$
Then just subtract $2e^{x+y}$. Now for problem 2, you have
$e^x=y+2ye^x$
$e^x-2ye^x=y$
$e^x(1-2y)=y$
Now just divide and you have your last line.