Find q in F = qVe + (Pe - Pa) * Ae with this simple formula

[.:Endeavour:.]

I have this formula: F = qV_e + (P_e - P_a) * A_e; I want to get q by its self. This what I did to get q by its self.

F = qV_e + (P_e - P_a) * A_e

$$\frac{F - (P_e - P_a)}{(A_e)} = \frac{(V_e * q)(A_e)}{(A_e)}$$

[( F - (P_e - P_a)) ÷ A_e] ÷ V_e = q

This is how I got q by itself in order to solve for q. I'm not sure if its correct, so please look over it to see if its correct. Thank you for your time.

 

[Office_Shredder]

It's wrong. When you move the (P_e-P_a)A_e term over to the side with F, you lost the A_e and multiplied V_eq by A_e instead.

 

[.:Endeavour:.]

...you lost the A_e and multiplied V_eq by A_e instead.

I don't get what you are saying by loosing A_e from (P_e - P_a). I get wat you are saying after the and part of the sentence. Can you change the place of A_e because of order of operations, you will have to multiply first before you can add: + A_e * (P_e - P_a)?

 

[Nabeshin]

I think what you did, correct me if I'm wrong, is subtract (Pe-Pa) from both sides and then divide everything by Ae.

The problem with this is that subtracting from the right hand side doesn't eliminate the (Pe-Pa) because it has an Ae attached to it. What Office_Shredder is saying is you have to move the entire term, Ae(Pe-P), over to the other side:

$$F=qV_{e}+(P_{e}-P_{a})A_{e}$$

Subtract $$(P_{e}-P_{a})A_{e}$$ from both sides:

$$F-(P_{e}-P_{a})A_{e}=qV_{e}+(P_{e}-P_{a})A_{e}-(P_{e}-P_{a})A_{e}$$

$$F-(P_{e}-P_{a})A_{e}=qV_{e}$$

$$\frac{F-(P_{e}-P_{a})A_{e}}{V_{e}}=q$$

To make explicit what you did:

$$F=qV_{e}+(P_{e}-P_{a})A_{e}$$

$$F-(P_{e}-P_{a})=qV_{e}+(P_{e}-P_{a})A_{e}-(P_{e}-P_{a})$$

And you see the terms don't drop out on the right hand side. Hope this clarifies.

 

[.:Endeavour:.]

Ok, yes it does clarify. A_e is meant to multiply with (P_e - P_a) not to divide F - (P_e - P_a), which is not dividing but multiplying. Thank you for your help.