6.201. $\left\{\begin{array}{l}\frac{4}{x+y-1}-\frac{5}{2 x-y+3}+\frac{5}{2}=0, \\ \frac{3}{x+y-1}+\frac{1}{2 x-y+3}+\frac{7}{5}=0 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+y \neq 1 \\ 2 x-y \neq-3\end{array}\right.$

Let $\left\{\begin{array}{l}\frac{1}{x+y-1}=u \\ \frac{1}{2 x-y+3}=v\end{array}\right.$

With respect to $u$ and $v$, the system takes the form

$\left\{\begin{array}{l}4 u-5 v=-\frac{5}{2}, \\ 3 u+v=-\frac{7}{5},\end{array} \Rightarrow\left\{\begin{array}{l}u=-\frac{1}{2} \\ v=\frac{1}{10}\end{array}\right.\right.$

Then

$$
\left\{\begin{array} { l } 
{ \frac { 1 } { x + y - 1 } = - \frac { 1 } { 2 } , } \\
{ \frac { 1 } { 2 x - y + 3 } = \frac { 1 } { 1 0 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=-1 \\
2 x-y=7, \text { from which } x=2, y=-3
\end{array}\right.\right.
$$

Answer: $(2 ;-3)$.