6.198. $\left\{\begin{array}{l}x+y+\frac{1}{x-y}=\frac{a b+1}{b}, \\ x-y+\frac{1}{x+y}=\frac{a b+1}{a} .\end{array}\right.$
## Solution.

Domain of definition: $x \neq \pm y, a \neq 0, b \neq 0$.

Multiplying the first equation of the system by $x-y \neq 0$, and the second by $x+y \neq 0$, we get

$\left\{\begin{array}{l}x^{2}-y^{2}+1=\frac{a b+1}{b}(x-y), \\ x^{2}-y^{2}+1=\frac{a b+1}{a}(x+y) .\end{array} \Rightarrow 1=\frac{a}{b} \cdot \frac{(x-y)}{(x+y)}\right.$,

$y=\frac{a-b}{a+b} x$, where $a \neq-b \Rightarrow x^{2}-\frac{(a-b)^{2}}{(a+b)^{2}} x^{2}+1=\frac{a b+1}{a}\left(x+\frac{a-b}{a+b} x\right) \Rightarrow$

$\Rightarrow 4 a b x^{2}-2(a b+1)(a+b) x+(a+b)^{2}=0$, from which

$x_{1,2}=\frac{(a b+1)(a+b) \pm \sqrt{(a b+1)^{2}(a+b)^{2}-4 a b(a+b)^{2}}}{4 a b}=$

$=\frac{(a b+1)(a+b) \pm \sqrt{(a+b)^{2}\left((a b+1)^{2}-4 a b\right)}}{4 a b}=$

$=\frac{(a b+1)(a+b) \pm(a+b) \sqrt{(a b-1)^{2}}}{4 a b}=\frac{(a b+1)(a+b) \pm(a+b)(a b-1)}{4 a b}=$

$=\frac{(a+b)(a b+1 \pm(a b-1))}{4 a b}$.

Then

$$
\begin{aligned}
& x_{1}=\frac{(a+b)(a b+1-a b+1)}{4 a b}=\frac{a+b}{2 a b}, x_{2}=\frac{(a+b)(a b+1+a b-1)}{4 a b}=\frac{a+b}{2} \\
& y_{1}=\frac{a-b}{a+b} \cdot \frac{a+b}{2 a b}=\frac{a-b}{2 a b}, y_{2}=\frac{a-b}{a+b} \cdot \frac{a+b}{2}=\frac{a-b}{2}
\end{aligned}
$$

Answer: if $a b+1=0$, then $y= \pm \sqrt{x^{2}+1}, x$ - any; if $a b+1 \neq 0$, then $x_{1}=\frac{a+b}{2 a b}, y_{1}=\frac{a-b}{2 a b} ; x_{2}=\frac{a+b}{2}, y_{2}=\frac{a-b}{2}$.