6.090. $\left\{\begin{array}{l}a x+\frac{b}{y}=2, \\ \frac{b}{x}+a y=2 a b .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0 \\ y \neq 0\end{array}\right.$

Rewrite the system as $\left\{\begin{array}{l}a x y+b=2 y, \\ b+a x y=2 a b x\end{array} \Rightarrow 2 y=2 a b x, y=a b x\right.$.

From the first equation of the system, we get $a^{2} x^{2}-2 a x+1=0$ (for $b \neq 0$) or $(a x-1)^{2}=0$, from which $x_{1}=x_{2}=\frac{1}{a}$. Then $y=a \cdot b \cdot \frac{1}{a}=b$, where $a \neq 0$.

Answer: if $a b=0$, then there are no solutions; if $a b \neq 0$, then $x=\frac{1}{a}, y=b$.