Finding values of $a$ with which a simple system has exactly 2 solutions 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.
 A: Note that the first pair of equations has gradient $1$ and the second pair has gradient $-1$. So the lines from the first pair will always meet lines from the second pair.
Now the lines from the first pair are clearly distinct (they are parallel with different intercepts), so will meet each of the lines in the second pair twice. So there needs to be just one line in what looks like the second pair.
In the second pair take the first equation away from the second term by them. you get $$2\sqrt {4a-1}=0$$
This is only true if $a=\frac 14$ and that is the only value for which you get a single line from the second pair of equations.

A second way of looking at the whole thing is as follows. It is clear from the equations that if $(x,y)$ is a solution so are $(-x,-y),(y,x),(-y-x)$ so in general there are four solutions. If there are two we must have $(x,y)$ equal to one of the others.
If a solution is equal to its negative, i.e. $(x,y)=(-x,-y)$ then it must be zero, and this would give $a=0$ from the first equation and a contradiction in the second.
Alternatively we can try $(x,y)=(y,x)$ or $x=y$ which gives $x^2=a$ from the first and tells us $-1=0$ when substituted into the second.
The final possibility gives $(x,y)=(-y,-x)$. This gives $x^2=a$ from the first and $4x^2=1$ in the second with $a=\frac 14$
