9. Find all values of $a$ for which the system of equations $y-1=a(x-1), \quad \frac{2 x}{|y|+y}=\sqrt{x}$ 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 becomes: $x=y \sqrt{x}$.
I. $x=0, y=1-a>0$, hence $a<1$.
II. $x>0, y=\sqrt{x} ; \sqrt{x}-1=a(\sqrt{x}-1)(\sqrt{x}+1)$.

1) $\sqrt{x}=1, x=1, y=1, a \in \mathbb{R}$.

$$
\text { 2) } 1=a(\sqrt{x}+1), \sqrt{x}=\frac{1}{a}-1=\frac{1-a}{a}>0,0<a<1 \text {. The found solution } x=\left(\frac{1-a}{a}\right)^{2}, y=\frac{1-a}{a}
$$

coincides with the previous one if $1=\frac{1}{a}-1, a=\frac{1}{2}$. Thus, for $a \in\left(0 ; \frac{1}{2}\right) \cup\left(\frac{1}{2} ; 1\right) \quad x=\left(\frac{1-a}{a}\right)^{2}, y=\frac{1-a}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_50f0bbe00f908a027b9ag-07.jpg?height=246&width=1396&top_left_y=662&top_left_x=433)

Answer:

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