Alraxite

You are just not giving me a direct explanation! I just gave you a reason why they can't be action-reaction pairs and now you have brought maths into this. Your explanation didn't include any maths to begin with.

If B decides not to push, then A will experience a force by the reaction force caused by him. You are saying that if B decides to push, then for some reason that deliberate push will become the reaction force. You can see this is wrong because if B wants he can apply 150 N instead which isn't equal to the force applied by A and hence these two deliberate forces can't be third law pairs. I know you are trying to defend that the force should be 100 N, but I just don't find your argument correct.

@zephyr I'll wholeheartedly accept the answer which explains to me properly why my reasoning is incorrect. And the explanation given by this answer doesn't quite do that. In fact, according to the answer, the only forces considered are the ones that are applied deliberately and not the reaction forces. So not only I find this answer not an explanation as to why my reasoning is incorrect but also wrong.

Why are you saying that B becomes part of A's mass? When two objects collide, do you say that they become one mass and then apply equations? No. If this is problematic, then imagine that both of them are pushing with their hands on each others chest. So B experiences a force by A's chest due to the third law and another force by A's hands.

People have upvoted your answer unthinkingly. If the upvoters are reading this, would they mind explaining why this answer makes any sense at all?

What do you mean if you look at the system as a whole? See, when A pushes on B, B's body by virtue of having mass, will exert an opposite force on A. This is not him (or rather his hands) providing the force. This happens by virtue of B's mass or inertia. The catch is that at the same time another force is being applied on A, which as it so happens is by B again (by B's hands if you will). The solution is clear. I don't know why you're having so much trouble grasping this.

Okay, forget the equations. Your equations come from the fact that there are only two forces which is not the case. According to you it's because they do form a Newton's third law pair. I've given my reason why they don't. Could you now give yours?

And how do they form a pair? Didn't I mention that it's entirely B's choice that he is pushing? There are four forces in this situation. Two of which comes from each of them which they did deliberately, and the other two are the reaction forces to each of them as a consequence of the third law. And now you've just brought in two redundant equations into this discussion. What are x and y exactly?

Since these two do not form an action-reaction pair of each other, there must be two other reaction forces for these two. So now I ask, what are the they? Think about it and then reply.

I absolutely don't follow. Your wording is too vague. Look at this way: You agree that by the third law, forces always come in action-reaction pairs, right? Now, consider the same situation where the two people are pushing on each other. Let's consider these two forces: the force exerted by 'A' on 'B' and the force exerted by 'B' on 'A'. Note here, that I'm talking about the deliberate forces that they exert on each other. These do not form the third law's action-reaction pair. It's entirely B's choice that he is pushing. continued

What you are saying doesn't make any sense. You are saying that something can't experience a force as long as something else doesn't apply an external force on it. That is correct. But this doesn't mean that the object which wants to get pushed can't make other objects push it. That is what Newton's third law is all about. You can push a rock in space and the rock will push you too because you pushed it. Here the rock is a person who is pushing as well. And note, that this 'pushing' isn't a reaction to the original force. This is a different force.

I understand better what the problem with the question as stated is

@arctictern okay thank you for your time

makes no sense

it's like asking whether the function on the set of real numbers described by x^2 + x = 5 is linear or not (here linear having a different but analogous definition for functions on the set of real numbers)

if the the equation is meant to be understood as describing a system and also giving a particular value for the output then giving the second piece of information wouldn't make sense if you only want to find whether the system is linear

it asks us to simply tell whether the system is linear or non-linear

I think the problem as our teacher has given us is incorrect

yeah

(like 0 in the case of H(x,y) = 0)

because then the equation is not only describing a system it also is giving us a particular value in the range of the system

for operator(input) to equal otherstuff?

and usually problems of this sort would proceed to ask us what input can be?

okay

or are you just trying to guess what our teacher could possibly have meant?

Is this a standard way to describe systems?

So it defines a system and also gives us a particular value of the system for some x and y?

I think this problem itself as our teacher has given us is wrong

describe*

doesn't that equation describes a relationship between x and y

why would anyone write that to describe $H(x, y) = y' + 3ty - t^2$?

which would give $y' + 3ty - t^2 x = 0$

I meant this^

$H(y, x) = y' + 3ty - t^2 x$

oops

from that equation

I mean, how do you get $H(y, x) = y' + 3ty = t^2 x$

Then why would one write $y' + 3ty = t^2 x$

If we do interpret that DE as describing a two function system

we haven't talked or studied anything about that in class. so far we have only been dealing with single function systems

I mean two functions

and as for H being a system of two variables

our teacher has always been using $x$ as the input function and $y$ as the output function

Because

okay

x and y are variables of t

Are you thinking of $t$ as a number or some function?

How is $H(y) = x + t^2$ different from what I wrote?

The system here is clearly defined

And we can easily talk about whether this is linear or not