<s>[INST] solving linear system "by inspection"? A text question is asking to solve some linear systems by inspection. My interpretation of "by inspection" is "by looking". For a linear system like
$$
\begin{cases}
3x + 4y = 28\\
3x + 4y = 83
\end{cases}
$$
you could say that by inspection there is no solution because "two (of the same) numbers can't have different sums."
or given the system
$$
\begin{cases}
y = 3x + 5\\
y = 2x + 5
\end{cases}
$$
you could say that by inspection, the solution is (0, 5) because both lines have the same y-intercept but different slopes.
But the exercises given in my text don't seem (to me) to lend themselves to solution by inspection (hence my question here).
The four systems are:
$$
\begin{cases}
x + y = 6\\
2x + y = 8
\end{cases}
$$
$$
\begin{cases}
x - y = 1\\
5x + 2y = 5
\end{cases}
$$
$$
\begin{cases}
x + y = 8\\
2x + y = -11
\end{cases}
$$
$$
\begin{cases}
2x + y = 13\\
x + 2y = 7
\end{cases}
$$
For each one, it seems that at the very least some mental calculation is required — which goes beyond the concept of "by inspection" in my opinion. Or does it? Or is there something else I'm missing?
Thank you.
 [/INST] There is something you are missing for this particular situation. "By inspection", as noted by others, means "by happening to think of a solution that works", which is allowed to include mental calculation. 
The thing is, in different situations, there should be different things you are looking for to help you think of such a solution, and to focus your attention on the right sorts of mental calculations.
In the case of linear equations, there is something very useful to focus your attention: the fact that the equations can describe linear combinations of vectors. Then trying to think of a combination of x and y that works is actually trying to think of a combination of vectors that works.
So to your examples:

$$
x -y = 1\\
5x+2y = 5\\
$$
becomes
$$
x\begin{pmatrix} 1\\ 5 \end{pmatrix} + y \begin{pmatrix}-1\\ 2\end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}
$$
And it's now clear that $x=1$ and $y=0$ will give the solution.

$$
x+y = 6\\
2x+y = 8\\
$$
becomes
$$
x\begin{pmatrix} 1\\ 2 \end{pmatrix} + y \begin{pmatrix}1\\ 1\end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}
$$
This final vector is not a multiple of one of the others like in the one before. So I'll try subtracting one of the vectors off it. Taking off $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ a couple of times gives $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$ which is 4 of my second vector, so it must be $x=2$, $y=4$.

The same one where the vector is $\begin{pmatrix} -8 \\ -11 \end{pmatrix}$ is a bit trickier. If I add $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ it will change how far apart the two coordinates are, so I'll try to make the coordinates the same so that it's a multiple of $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Adding three times will give $\begin{pmatrix} -5 \\ -5 \end{pmatrix}$. So it looks like we need $x=-3$ and $y=-5$. 

The final one is:
$$
x\begin{pmatrix} 2 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 13 \\ 7 \end{pmatrix}
$$
The answer-vector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ away from $\begin{pmatrix} 12 \\ 6 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is a third of the sum of the x-vector and the y-vector. So if we do $6\frac{1}{3}$ of the x-vector and $\frac{1}{3}$ of the y-vector we'll get the right answer. That is, $x = 6\frac{1}{3} = \frac{19}{3}$ and $y = \frac{1}{3}$. 

As you can see, there's no "method" per se, but thinking of it as a vector equation can help.
</s>