Solve $$\begin{array}{|l}\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2} \\ 2x+5y=1\end{array}$$
As always: $$\begin{array}{|l} x-y \ne0 \\ x+y \ne 0 \\ x^2-y^2\ne0\end{array} \Leftrightarrow \begin{array}{|l} x \ne y \\ x\ne -y \\ (x-y)(x+y) \ne 0\end{array} \Leftrightarrow \begin{array}{|l} x \ne y \\ x\ne -y \end{array}$$ We can solve the second equation for $y: y=-\dfrac{2}{5}x+\dfrac{1}{5}$ and substitute into the first equation and solve for $x$. I don't have experience with systems and I would like to ask you if there is a simpler solution for this system.
$$\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2}$$
$$\dfrac{xy+x+y+y^2+x^2+2x-xy-2y}{x^2-y^2} = \dfrac{x^2+y^2+10}{x^2-y^2}$$
Since $(x^2-y^2) \ne 0$ , we have :
$$x+y+2x-2y = 10\implies 3x-y =10 \quad \color{#2d0}{\text{(1.)}}$$
Also $$2x+5y=1 \quad \color{#d05}{\text{(2.)}}$$
Multiplying $(1.)$ by $5$ and adding $(2.)$ ,
$$17x = 51 \implies x = 3$$
And we get $y = -1.$
So the solution is $\boxed{(x,y) = (3,-1)}$