Analysis linear ladder circuit
If I understand the diagram, I is the current through the 1 Ohm resistor at the far right of the schematic.

What they're saying is, assume I is 1 A. Now you know, for example, the voltage at the next node left is 3 V. And you have a 1 A current through the 3-Ohm resistor. Since you have 2 A through the last two rungs of the ladder, you must have 4 V across the 2-Ohm resistor 2nd-from-the-right (see how much easier this would be if you just put designators in your schematic?). And so on, working back to the left.

Now when you get back to the source, you will get some voltage other than 7 V. Say you get 21 V (I haven't calculated, I just made up a number).

Since the system is linear, you know that if you scale this voltage down to 7 V, the current I will also scale down by factor of 3, so the actual I is 333 mA, not the 1 A you originally assumed.