The problem is:
Find such values of $a$ with which the system
will have exactly two solutions
I understand the solution provided at the Resuhege.ru website (problem no. 484630):
First we are substracting and summing up the equations respectively, yielding
Next we are noting that $a$ should not be below $\frac{1}{4}$, and we derive four equations describing linear graphs:
Next we plot the graphs and observe that there are 4 solutions in most cases, but that if $a=\frac{1}{4}$ then the system (2) graphs will merge into one, giving two solutions. Hence, the answer is $a=\frac{1}{4}$.
But how to come to the same conclusion non-graphically? I would first equalize the right parts of each of the equations from (1) with each of the equations from (2):
$$x-1=-x-\sqrt{4a-1}$$
From this I would get four pairs of coordinates ($x, y$) for transections. For example, the first pair of coordinates:
$$ \begin{cases} x=\frac{-\sqrt{4a-1}+1}{2} \\ y=\frac{-\sqrt{4a-1}+1}{2}-1 \end{cases} $$
But how to discover algebraically, without graphing, the presence of situations in which a couple of transections will precisely equal another couple of transections, so that there will only be 2 solutions for the system?
I need just a general idea. It may take some time for it to sink in, though, my mind boggles yet at systems of equations.