# Equation; PA - P atm A = K

1. Jul 18, 2009

### Auto Engineer

The equation;

PA - P atm A = K, change in X

If I wish to factor this equation to get "p" on its own, then;

PA - P atm A = K, change in X
= A ( P - P atm)

so the conclusion I draw is;

(P - P atm)A = K, change in X

My confusion is that I cannot work out how or where the second "A" went?

Could somebody put some light on the topic for me

David

2. Jul 18, 2009

### HallsofIvy

Re: Factorisation

If it is "P" you want to get on its own, why don' you take P out rather than A?

so the conclusion I draw is;

(P - P atm)A = K, change in X

My confusion is that I cannot work out how or where the second "A" went?

Could somebody put some light on the topic for me

David[/QUOTE]
PA- P atm A= K
P(A- atm A)= K

P= k/(A- atm A)= k/(A(1- atm)).

12 - 6= 3(4)- 3(2)= 3(4- 2). Where did the second "3" go? That is the distributive law: ax+ ay= a(x+ y).

3. Jul 18, 2009

### Auto Engineer

Re: Factorisation

PA- P atm A= K
P(A- atm A)= K

P= k/(A- atm A)= k/(A(1- atm)).

12 - 6= 3(4)- 3(2)= 3(4- 2). Where did the second "3" go? That is the distributive law: ax+ ay= a(x+ y).[/QUOTE]

The above distributive law is ok, but my example I feel you have misunderstood it

If "P" referred to "pressure" and "A" referred to "area", and "atm" was atmosphere, then one cannot remove "P" in that way and leave "atm" because "P" is the pressure of the atmosphere?
If using your method we said; PA - P atm A = K, and then you said; P(A - atm A) = K, the atmosphere cannot exist without "P" pressure, so A - A = 0, so P = K, the rest does not make sense to me, sorry

When the equation ends up with; (P - P atm)A = K, change in X, then I could work out from that the change in length of X?, this is because P - P atm means that say something (600 000 - 100 0000 = 500 000 x A = K, change in X, if you follow me with what I mean?

The equation is being factored to show the result obtained, I just need to understand how A seems to go missing?

David

4. Jul 18, 2009

### HallsofIvy

Re: Factorisation

Apparently I did misunderstand. Perhaps if you told me what all this (particularly "the atmosphere cannot exist without "P" pressure") has to do with "Linear and Abstract Algebra", it would help.

5. Jul 18, 2009

### Auto Engineer

Re: Factorisation

Sorry that is my fault

P = pressure, but the equation is using more than one pressure which is why there are two "P"'s present, so (P - P atm) is saying that one pressure could be say liquid and then the other P atm is the atmospheric pressure. Also "A" is considered in this equation to be the area of a property, so a cylinder would be an example. K is understood to be a constant, and X is the change in length of a property like a spring, but the right hand side of the equation can be ignored when re aranging for "P". I am to understand the method is to factorise it, but I can only do so much of it.

So to recap;

PA - P atm A = K, change in X

= (P - P atm)A

This is all I know, what confuses me is where the second "A" went?

David