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

Domain of definition: $\frac{x+a}{y}>0$.

Let $\sqrt{\frac{x+a}{y}}=t$, where $t>0$. With respect to $t$, the equation takes the form $t+\frac{1}{t}=2, t^{2}-2 t+1=0, (t-1)^{2}=0$, from which $t=1$. Then $\sqrt{\frac{x+a}{y}}=1$, $\frac{x+a}{y}=1$, from which $y=x+a$. From the second equation, we get

$$
x+x+a=x(x+a)+a, x^{2}+(a-2) x=0, x(x+a-2)=0
$$

from which $x_{1}=0, x_{2}=2-a$. Then $y_{1}=a, y_{2}=2$.

Answer: if $a \neq 0$, then $x_{1}=0, y_{1}=a, x_{2}=2-a, y_{2}=2$; if $a=0$, then $x=y=2$.