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.
 A: I think you've got a pretty good grasp of what might be meant by "solve by inspection."  As you point out, for example, the initial systems you post are certainly appropriate for "solving by inspection." 
The later systems require a little more thought, I agree, but lend themselves to making "educated guesses" at the very least: if not for immediate solutions, for how to proceed to readily find solutions. The task, I suspect, is aimed to get you to think about the system before just mechanically proceeding to solve, by rote means. 
For example: the system
$x + y = 6$
$2x + y = 8$
lends itself to a pretty easy solution $x = 2$ when we "mentally" subtract the first equation from the second, which then lends itself to a pretty immediate conclusion, "so $y = 4$." So certainly, "by inspection" doesn't mean "without thought."
So think of the task as a "meta-cognitive" task, of trying to "eye-up" - if not the immediate solutions - the way to proceed in solving it. 
A: 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.
A: "Solve by inspection" sometimes means "easiest way to eliminate". In the first and third sytems the coefficients of $\,y\,$ are equal $(=1),\,$ so simply subtracting the two equations eliminates $\,y\,$, easily yielding $\,x\,$. In the other systems the coef's are $\,\pm1\,$ and $\,2,\,$ so we need only add/subtract a doubled equation to eliminate $\,y.\,$ Generally one looks for "small" coefficients, and/or where one coef is a multiple of the other, so the arithmetic can easily be done mentally. 
Frequently, solving by inspection means employing intuition to see some optimization (or trick), before applying some rote or algorithmic technique - which would be much more tedious. For example, to solve the congruence $\ 6\color{#c00}x\equiv 1\pmod p\ $ for a prime $\ p > 3\ $ we could mechanicallly apply the extended Euclidean algorithm. Or we could apply intuition: every such prime is of the form $\ 6n\pm 1\ $ so, e.g. mod $\, p = 6n\!-\!1\!:\ 6n\!-\!1\equiv 0\,\Rightarrow\,6\color{#c00}n\equiv 1\,$ so $\, \color{#c00}{x\equiv n} = (p\!+\!1)/6.$
For another example, suppose you have a problem which requires knowing if the following matrix has nonzero determinant. Instead of rotely calculating the integer determinant, we employ parity, noticing that the odd entries are precisely  the diagonal entries. Thus,  mod $\,2,\,$ it is the identity matrix, so the determinat is $\equiv 1\pmod {2},\,$ i.e. the determinant is odd, hence nonzero.
$$\begin{bmatrix} 1234567 &1234568 &1234560 \\ 1234562 & 1234563 &1234564 \\ 1234566 &1234566 &1234568 \end{bmatrix} $$ 
