2. Find all real numbers $a$ and $b$ for which

$$
a+b=a b=\frac{a}{b}
$$
II/2. Let's first look at the equation $a \cdot b=\frac{a}{b}$. The number $b$ must not be equal to 0, so we can multiply the equation by $b$ and get $a b^{2}=a$ or $a b^{2}-a=0$, which can be written in the form $a\left(b^{2}-1\right)=0$ or $a(b-1)(b+1)=0$. The solutions to this equation are $a=0, b=1$ and $b=-1$.

If we consider the solution $a=0$ in $a+b=a b$, we get $b=0$, which is not possible. Similarly, the solution $b=1$ leads to $a+1=a$, which is a contradictory equation. The solution $b=-1$ results in $a-1=-a$ or $a=\frac{1}{2}$. The sought numbers are $\frac{1}{2}$ and -1.

Multiplying by $b$ and explicitly stating the condition $b \neq 0$

2 points

Equation $a(b-1)(b+1)=0$ or a similar one from which the solutions can be read ... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_fb5898f0472fc84ca82bg-08.jpg?height=49&width=1633&top_left_y=632&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_fb5898f0472fc84ca82bg-08.jpg?height=48&width=1631&top_left_y=684&top_left_x=218)

Analysis of the case $b=-1$: equation $a-1=-1$ and solution $a=\frac{1}{2}$ ... 2 points