solving linear system "by inspection"?

Solution 1:

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.

Solution 2:

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.