2. [4 points] Find all pairs of real parameters $a$ and $b$, for each of which the system of equations

$$
\left\{\begin{array}{l}
2(a-b) x+6 y=a \\
3 b x+(a-b) b y=1
\end{array}\right.
$$

has infinitely many solutions.
Answer: $(-1 ; 2),(-2 ; 1),\left(\frac{3-\sqrt{17}}{2} ;-\frac{3+\sqrt{17}}{2}\right),\left(\frac{3+\sqrt{17}}{2} ; \frac{\sqrt{17}-3}{2}\right)$.

Solution. If $b=0$, then the second equation of the system takes the form $1=0$, and therefore the system has no solutions. For all other values of the parameters, in each of the equations at least one of the coefficients of the variables is different from zero. Consequently, both equations of the system define certain lines, and for the existence of an infinite number of solutions, these lines must coincide. This is possible with the proportionality of the coefficients of the equations, i.e.,

$$
\frac{2(a-b)}{3 b}=\frac{6}{(a-b) b}=\frac{a}{1}
$$

It is also worth noting that the case where the coefficients of one of the variables or the free terms turn to zero in both equations is impossible ${ }^{2}$. Considering the first equality from (??), we get $(a-b)^{2}=9, a=b \pm 3$. Substituting this into the second equality (??): - if $a=b+3$, then $\frac{2}{b}=b+3, b^{2}+3 b-2=0, b=\frac{-3 \pm \sqrt{17}}{2}$; from here we obtain two solutions of the system: $\left(\frac{3-\sqrt{17}}{2} ;-\frac{3+\sqrt{17}}{2}\right),\left(\frac{3+\sqrt{17}}{2} ; \frac{\sqrt{17}-3}{2}\right)$[^1]- if $a=b-3$, then $-\frac{2}{b}=b-3, b^{2}-3 b+2=0, b=1$ or $b=2$; from here we obtain two more solutions of the system: $(-2 ; 1),(-1 ; 2)$.