9. Find all values of $a$ for which the system of equations $2 y-2=a(x-2), \frac{4 y}{|x|+x}=\sqrt{y}$ has at least one solution, and solve it for each $a$.

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# Solution:

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

In the domain of definition, the second equation of the system takes the form: $2 y=x \sqrt{y}$.
I. $y=0, x=2-\frac{2}{a}=\frac{2(a-1)}{a}>0$, hence $\left[\begin{array}{l}a1 \text {. }\end{array}\right.$

II. $y>0, y=x^{2} / 4, x>0 ;(x-2)(x+2)=2 a(x-2)$.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-29.jpg?height=225&width=1468&top_left_y=1778&top_left_x=337)

## Answer:

$a \in(-\infty ; 0) \bigcup\{2\}, x_{1}=2-2 / a, y_{1}=0 ; x_{2}=2, y_{2}=1 ;$

$a \in[0 ; 1], x=2, y=1$

$a \in(1 ; 2) \cup(2 ;+\infty), x_{1}=2-2 / a, y_{1}=0 ; x_{2}=2, y_{2}=1 ; x_{3}=2 a-2, y_{3}=(a-1)^{2}$.