Simple Algebra problem, but done with reference frames

[QuietMind]

Homework Statement

Sonia walks up an escalator which is going up. When she walks at one step per second, it takes her 20 steps to get to the top. If she walks at two steps per second, it takes her 32 steps to get to the top. She never skips over any steps. How many steps does the escalator have?

Homework Equations

No equations, but the relevant example from the text (Art and Craft of Problem Solving, Zeitz) is an example of a person swimming upstream a river for an hour. (Here I paraphrase the problem) After realizing she dropped her hat as she started swimming, she has to turn around and retrieve the hat. How long will the swim back towards the hat take? The answer is 1 hour, regardless of the speed of the current, because from the point of view from the hat, the girl swims away for an hour, so must swim back for an hour

The Attempt at a Solution

I did the algebra and got 80 steps, but I'm looking for guidance on a solution that uses reference frames to do the problem simply without (or minimal) algebra. Let's examine the reference frame of the escalator's steps. To the escalator, the girl is moving up at one step per second in the first case and takes 20 seconds. In the second case, she moves two steps per second and takes 16 seconds.

I'm struggling to see this with reference frames. From the point of view of the steps, she is moving twice as fast in the second case but for a different amount of time. Is there an intuitive way to think about this?

(I am under the assumption there is a way to do this with changing the point of view, as that was the pedagogical emphasis of the example)

 

[FactChecker]

I assume that you have an algebraic solution for the stationary reference frame. You need to show that. The solution in the escalator reference frame should be closely related.

 

[QuietMind]

My algebraic solution:

Let D be the number of steps total and let s be the speed of the escalator.
It took 20 seconds in the first case and 16 in the second case.

$D = (s + 1) 20$
$D = (s+2) 16$

$(s+1)20 = (s+2) 16$
$4s = 12$
$s = 3$

Then $D = 4*20 = 80$

To do this in a different reference frame, would I define s' = 0, and define D' = D - s*time?