Contradictory equations Question: Find whether the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is a singleton.

My attempt: Rewriting the first equation will give us $x = \frac{1}{2}$.

The other two equations can be written as $x - y = 5$ and $x - y = 3$.

Now, the solution of these equations is $\emptyset$. 
So we can say that the solution set of $$\begin{cases}2x = 1\\ y + 5 = x\\ x = y + 3\end{cases}$$ is not singleton.

Kindly verify and throw some insights.
The equations you gave represent three lines in $\mathbb R^2$. When we have three (or more) lines in a plane, we can have  one of the following situations:


They have infinitely many common points.
They have one point in common.
They have no points in common. 


The lines $2x=1$ and $y+5=x$ are clearly not parallel (can you see this?). So they intersect at one point, which means we are not dealing with case one. The lines will intersect at the point $(x,y)=\left(\frac12;-4\frac12\right)$.  

Now the three lines will have one point in common if $\left(\frac12;-4\frac12\right)$ lies on the third line and no points in common when it doesn't. I will leave it to you to figure out if it is on the third line or not.



After edit from OP: We can solve this even quicker if we just look at the bottom two equations: $$\begin{cases} x=y+5\\ x=y+3\end{cases}$$ These lines are obviously parallel (and don't coincide). You could also argue that this implies $y+5=y+3$, which is a contradiction.
A vertical line ( first equation ) cuts two parallel lines ( last two ) transversely, creating two points of intersection or two solutions. 

If there were three (non-concurrent) lines then there would be three points of intersection enclosing a triangle.