9. Find all values of $a$ for which the system of equations $2 y-2=a(x-1), \quad \frac{2 x}{|y|+y}=\sqrt{x}$ has at least one solution, and solve it for each $a$.

#
# Solution.

Domain of definition: $y>0, x \geq 0$.

In the domain of definition, the second equation of the system takes the form: $x=y \sqrt{x}$.
I. $x=0, y=1-\frac{a}{2}>0$, hence $a<2$, and $a \leq 0$.
II. $x>0, y=\sqrt{x} ; 2(\sqrt{x}-1)=a(\sqrt{x}-1)(\sqrt{x}+1)$.

1) $\sqrt{x}=1, x=1, y=1, a \in \mathbb{R}$.
2) $2=a(\sqrt{x}+1), \sqrt{x}=\frac{2}{a}-1=\frac{2-a}{a}>0,0<a<2$. The solution found $x=\left(\frac{2-a}{a}\right)^{2}, y=\frac{2-a}{a}$ coincides with the previous one if $1=\frac{2}{a}-1, a=1$. Thus, for $a \in(0 ; 1) \cup(1 ; 2) \quad x=\left(\frac{2-a}{a}\right)^{2}, y=\frac{2-a}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_50f0bbe00f908a027b9ag-15.jpg?height=220&width=1268&top_left_y=244&top_left_x=357)

## Answer:

$$
\begin{aligned}
& a \in(-\infty ; 0] \cup\{1\}, x_{1}=0, y_{1}=1-\frac{a}{2} ; x_{2}=1, y_{2}=1 ; \\
& a \in(0 ; 1) \cup(1 ; 2), x_{1}=0, y_{1}=1-\frac{a}{2} ; x_{2}=1, y_{2}=1 ; x_{3}=\left(\frac{2-a}{a}\right)^{2}, y_{3}=\frac{2-a}{a} ; \\
& a \in[2 ;+\infty), x=1, y=1
\end{aligned}
$$