Suppose a square formation of troops 50 meters deep is marching with a dog in the middle of its back rank. The dog runs to the front of the formation, turns around instantaneously and runs back to where it started, arriving exactly when the troops have marched 50 meters forward. How far did the dog run?
The only method of solution known to me requires setting several of the key unknowns to one (through an appropriate rescaling of units): Setting the distance and rate of the troops to one, the time required for their march is then one. Thus, if the rate of the dog is x, the distance the dog travels is also x and the solution becomes a simple quadratic equation.
The question is about the logical step of setting the unknowns relating to the distance, rate, and time of the troops to one. In struggling with this problem, that strategy did not occur to me, and even having seen how it works it feels handwavy.
So the question is in general, under what circumstances is this a valid and rigorous method of solution? Is it true that so long as the value of an answer can be rescaled back into meaningful (e.g., SI) units (through dimensional analysis and multiplication by known or given constants), then it matters not what unknowns of a system represented by variables in an equation are set to one?
You can assign any variable to any value so long as it doesn't contradict with your existing system of equations. Any variable for which such assignment is possible is called a free variable. So your question boils down to figuring out when a variable's value is inconsequential, and when it's value has already been determined.
In this particular problem, all is that is provided are distances. This means your velocity can have any reasonable magnitude so long as your time span is adjusted to match up with it. Likewise your time span can be any value so long as your velocity is adjusted to match up with it. You can treat either velocity or time as a free variable, but not both.
It is possible to solve this problem without substituting any values with $1$. I will do so and then make some observations to help illustrate my point.
First, let's define some variables.
Let the distance that the dog travels be $x$.
Let the distance that the troop travels be $y=50$.
Let $x_1$ and $y_1$ be the distances traveled by the dog and troop until the dog reaches the front, and let $x_2$ and $y_2$ be the distances traveled from that point until the dog reaches the back.
Let the size of the troop be $s=50$.
We have the following system of equations so far:$$ \begin{array}{} y &=& y_1+y_2 &=& 50\\ y &=& x_1-x_2 &=& 50\\ s &=& x_1-y_1 &=& 50 \\ x &=& x_1+x_2 \end{array} $$There are $4$ linear equations and $5$ unknown variables ($x,x_1,x_2,y_1,y_2$). We can choose $x_2$ to be our free variable and solve everything in terms of it:$$ \begin{array}{} x_1 &=& 50+x_2\\ y_1 &=& x_2\\ y_2 &=& 50-x_2 \end{array} $$Next, we need to make use of the fact that the speed of the dog and the troop remain the same before and after the dog reaches the front of the troop. Letting $t_1$ and $t_2$ represent the time it took the dog to get to the front and back repectively, we have:$$ \begin{array}{} \frac{y_1}{t_1} &=& \frac{y_2}{t_2} \\ \frac{x_1}{t_1} &=& \frac{x_2}{t_2} \end{array} $$We can convert $x_1$, $y_1$, and $y_2$ in terms of $x_2$:$$ \begin{array}{ccc} \frac{x_2}{t_1} &=& \frac{50-x_2}{t_2} &\implies& x_2 = \frac{t_1}{t_2}(50-x_2) \\ \frac{50+x_2}{t_1} &=& \frac{x_2}{t_2} &\implies& x_2 = \frac{t_2}{t_1}(50+x_2) \end{array} $$You can solve for $x_2$ by multiplying these equations with each other:$$ (x_2)^2 = 50^2-(x_2)^2 \implies x_2 = \sqrt{\frac{50^2}{2}} = 25\sqrt{2} $$Then we can solve for $x$:$$ x = x_1+x_2 = (50+x_2)+x_2 = 50+50\sqrt{2} $$
When I called $x_2$ free, it was only with respect to the system of equations which did not take into account equivalent speeds. When taking those equations into account, it causes $x_2$ to become a bound variable and to have a particular value.
The variables $t_1$ and $t_2$ were free variables, but we would only be able to assign an arbitrary value to one of them since their ratio $t_1/t_2$ has a particular value.
Typically you can avoid making arbitrary assignments to a free variable by finding its relationship to other variables and using substitution or elimination to get rid of it.