$x+y=3m$
$xy=2m^2$, $m$ is the parameter.
I came to this $2m^2-3mx+xy=0$. The solutions have to be:$(m,2m),(2m,m)$.
But I can't understand what is the role of this parameters, I don't know how to come to the solutions. Can someone help me I would appreciate that.
On
The easiest way is to think about this system of equations graphically. These are two intersecting curves (two points of intersection), one of which is linear, another is hyperbola. When you change your parameter you shift this curves (increasing parameter, shift linear curve up, hyperbola also up but faster).
In terms of solving, the easiest way would be just to get $y = 3m -x$ from the first equation and plug it to the second one: $x (3m -x) = 2m^2$, then you get quadratic equation for $x$: $x^2 - 3mx + 2m^2 = 0$ which gives exactly two solutions ($D = 9m^2 - 8m^2 = m^2$, then $x_1 = \dfrac{3m + m}{2} = 2m$ and $x_2 = \dfrac{3m-m}{2} = m$. Correspondingly you get two $y$s: $y_1 = 3m-x_1 = 3m-2m = m$ and $y_2 = 3m - m = 2m$.
Note that $x,y$ are the solutions of $$(t-x)(t-y)=0\Rightarrow t^2-(x+y)t+xy=0\Rightarrow t^2-3mt+2m^2=0.$$ So, we have $$t=\frac{3m\pm\sqrt{(-3m)^2-4\cdot 2m^2}}{2}=\frac{3m\pm m}{2}=m,2m.$$ Hence, we have $$(x,y)=(m,2m),(2m,m).$$