Calculation of linear equation system

Question

I know that $$\frac{1}{x+2y+1 } +\frac{1}{2x+y-6 }=1$$

How can I calculate $x+y$?

Answer 1

$$\frac{1}{a}+\frac{1}{b}=1 ,a,b\in \mathbb{N} \to \frac12+\frac12=1 \to a=b=2$$so one of the solution come from $$\frac{1}{\underbrace{x+2y+1}_{2} } +\frac{1}{\underbrace{2x+y-6}_{2} }=1\\ \to \begin{cases}2x+y-6=2\\x+2y+1=2\end{cases}\to x=5,y=-2$$but you said $x,y,z >0$

Answer 2

Observe that

$$\frac1a+\frac1b=1$$ can be written $$(a-1)(b-1)=1$$

and the only integer solutions are $a=b=0$ and $a=b=2$.

But as $x,y$ are positive, $x+2y+1>2$. There are no solutions.